IA ao vivo / IME 2027 · 2ª fase · Física

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Questão 5 (Física)

Questão discursiva: a banca não divulga gabarito oficial. ← 4 (Física) 6 (Física) →

A questão

Resposta de referência

a) T=450\ \text{K}. b) \dot Q=90\ \text{kW}. c) A proposta não é realística: as residências exigem 116{,}7\ \text{kW} e o aquecimento da água, 33{,}3\ \text{kW}. Somando a rejeição de 45\ \text{kW} à fonte de 450\ \text{K}, seriam necessários 195\ \text{kW}>150\ \text{kW}, contrariando a primeira lei. Mantido o aquecimento, termodinamicamente viável, restam apenas 71{,}7\ \text{kW}, suficientes para cerca de 215 residências.

Segundo a resolução dos cursinhos Farias Brito e Poliedro. A banca não publica gabarito das questões discursivas.

Respostas das IAs

GPT-6 Astra

Respondeu · bate com a resposta de referência dos cursinhos

a) 450\ \text{K}; b) 90\ \text{kW}; c) proposta inviável: com o aquecimento, a potência útil é de, no máximo, 71{,}7\ \text{kW}, suficiente para 215 residências.

levou 38 s

Como pensou

a) Temperatura da fonte fria da empresa A

A potência útil da máquina é

P_A=80\cdot\frac34=60\ \text{kW}.

Logo, seu rendimento vale

\eta_A=\frac{P_A}{\dot Q_{\text{quente}}} =\frac{60}{150}=0{,}40.

O rendimento máximo admissível é o de Carnot. Como a máquina apresenta 80\% desse rendimento,

0{,}40=0{,}80\left(1-\frac{T}{900}\right).

Portanto,

\boxed{T=450\ \text{K}}.

b) Taxa de calor rejeitada pela máquina A

Pela primeira lei da Termodinâmica, aplicada ao ciclo,

\dot Q_{\text{quente}}=P_A+\dot Q_A.

Assim,

\dot Q_A=150-60 \quad\Rightarrow\quad \boxed{\dot Q_A=90\ \text{kW}}.

c) Viabilidade da proposta da empresa B

A máquina B rejeita para a fonte a 450\ \text{K}:

\dot Q_{450}=\frac{\dot Q_A}{2}=45\ \text{kW}.

A potência elétrica média necessária para abastecer as 350 residências é

P_{\text{necessária}} =\frac{350\cdot240\ \text{kWh}}{720\ \text{h}} =\frac{350}{3}\ \text{kW} \approx116{,}7\ \text{kW}.

Para o aquecimento da água, a vazão mássica é

\dot m=\frac{50\cdot1}{60}=\frac56\ \text{kg/s}.

A potência térmica necessária vale

\dot Q_{\text{água}} =\dot m c\Delta T =\frac56\cdot4000\cdot10 =\frac{100}{3}\ \text{kW}.

No cenário mais favorável, todo o calor rejeitado à fonte de 300\ \text{K} é aproveitado no aquecimento. Pela primeira lei, a potência útil disponível seria, no máximo,

P_B\leq150-45-\frac{100}{3} =\frac{215}{3}\ \text{kW} \approx71{,}7\ \text{kW}.

Isso permitiria atender apenas

N_{\max}=\frac{(215/3)\cdot720}{240} =\boxed{215\ \text{residências}}.

A segunda lei também impõe um limite à potência. Para o funcionamento cíclico,

\dot S_{\text{gerada}} =\frac{45}{450}+\frac{\dot Q_{300}}{300} -\frac{150}{900}\geq0,

com as taxas de calor em kW. Daí,

\dot Q_{300}\geq20\ \text{kW} \quad\Rightarrow\quad P_B\leq150-45-20=85\ \text{kW}.

Portanto, mesmo sem a exigência de aquecer a água, a máquina não poderia abastecer as 350 residências. Com o aquecimento proposto, o limite energético cai para 215 residências, tornando a promessa inviável.

Resposta final: a) 450\ \text{K}; b) 90\ \text{kW}; c) proposta inviável: com o aquecimento, a potência útil é de, no máximo, 71{,}7\ \text{kW}, suficiente para 215 residências.

GPT-6.1 Sol

Respondeu · bate com a resposta de referência dos cursinhos

a) 450\ \text{K}; b) 90\ \text{kW}; c) proposta inviável: exigiria 195\ \text{kW} de entrada; mantido o aquecimento da água, atenderia no máximo 215 residências.

levou 32 s

Como pensou

**Formulating equations for efficiency** I need to address the issue with artifacts and formulate equations using \dot Q notation. I've got specific values for a, b, and power requirements. If all heat is utilized, Q is exactly 33333, but excess heat may lower power to 71.67. The Clausius relation could demonstrate thermodynamic limits, but there's more depth to explore. I want to ensure the final presentation is clear, concise, and covers essential equations. We have a token budget, so I’ll keep it straightforward while addressing nuances.**Planning energy sections** I need to include energy discussions and the second law, with a maximum target of 85. Two paragraphs should do the trick, and I’m thinking about aiming for around 600 words total, perhaps collating the relevant details to reach a target of about 450 for the main content. It’s essential to balance the flow of information while ensuring clarity. I’ll focus on getting the main points across effectively.

a) Temperatura da fonte fria da empresa A

A potência útil da máquina A é

P_A=80\cdot\frac34=60\ \text{kW}.

Logo, seu rendimento vale

\eta_A=\frac{P_A}{\dot Q_h}=\frac{60}{150}=0{,}40.

Como esse rendimento corresponde a 80\% do máximo admissível, dado pelo ciclo de Carnot,

0{,}40=0{,}80\left(1-\frac{T}{900}\right).

Assim,

1-\frac{T}{900}=\frac12 \quad\Rightarrow\quad \boxed{T=450\ \text{K}}.

b) Taxa de calor rejeitada pela máquina A

Pela primeira lei da Termodinâmica, aplicada ao ciclo,

\dot Q_h=P_A+\dot Q_A.

Portanto,

\dot Q_A=150-60 \quad\Rightarrow\quad \boxed{\dot Q_A=90\ \text{kW}}.

c) Viabilidade da proposta da empresa B

A potência elétrica média necessária para abastecer as 350 residências é

P_{\mathrm{res}}=\frac{350\cdot240}{720} =\frac{350}{3}\ \text{kW} \approx116{,}7\ \text{kW}.

Para aquecer a água, a vazão mássica é

\dot m=\frac{50\cdot1}{60}=\frac56\ \text{kg/s}.

A taxa de calor necessária, admitindo aproveitamento integral, vale

\dot Q_{\mathrm{água}}=\dot m\,c\,\Delta T =\frac56\cdot4000\cdot10 =\frac{100}{3}\ \text{kW} \approx33{,}3\ \text{kW}.

Além disso, a máquina B rejeita para a fonte a 450\ \text{K}:

\dot Q_{450}=\frac{\dot Q_A}{2}=45\ \text{kW}.

Para cumprir simultaneamente as duas promessas, a potência térmica de entrada teria de ser, no mínimo,

\dot Q_{h,\mathrm{necessária}} =P_{\mathrm{res}}+\dot Q_{450}+\dot Q_{\mathrm{água}} =\frac{350}{3}+45+\frac{100}{3} =195\ \text{kW}.

Entretanto, a entrada disponível é de apenas 150\ \text{kW}. Portanto, a proposta viola a primeira lei da Termodinâmica.

Mantido o aquecimento da água, a potência útil seria, no máximo,

P_B=150-45-\frac{100}{3} =\frac{215}{3}\ \text{kW} \approx71{,}7\ \text{kW},

suficiente para apenas

N_{\max}=\frac{P_B\cdot720}{240} =\boxed{215\ \text{residências}}.

A segunda lei também impõe uma restrição: para a máquina cíclica,

\frac{45}{450}+\frac{\dot Q_{300}}{300} -\frac{150}{900}\geq0 \quad\Rightarrow\quad \dot Q_{300}\geq20\ \text{kW}.

Consequentemente, mesmo sem exigir o aquecimento da água especificado,

P_B\leq150-45-20=85\ \text{kW}<116{,}7\ \text{kW}.

Logo, a máquina B não conseguiria abastecer as 350 residências nem nessa condição.

Resposta final: a) 450\ \text{K}; b) 90\ \text{kW}; c) proposta inviável: exigiria 195\ \text{kW} de entrada; mantido o aquecimento da água, atenderia no máximo 215 residências.

GPT-6 Luna

Respondeu · bate com a resposta de referência dos cursinhos

a) 450\ \text{K}; b) 90\ \text{kW}; c) não é realística: a máquina B forneceria cerca de 71{,}7\ \text{kW}, insuficientes para os 116{,}7\ \text{kW} demandados por 350 residências.

levou 30 s

Como pensou

**Calculating energy balance** I'm analyzing an energy output of 71.67 kW for a system versus 116.67 kW for a household, which seems to assert an impossible energy balance. There might be efficiency or entropy constraints at play. For a certain cycle, when applying the Clausius inequality, it indicates a positive entropy balance, making it a feasible engine. However, if heating water is deemed a byproduct, the total rejection becomes significant. I need to calculate demand and compare it realistically with the output.**Assessing heating water claims** I'm considering whether the claimed ability to heat water aligns with the available energy rejection. Using a thermal rate of 33.3 kW, I need to compute work via the first law, but it seems insufficient for electrical demand. The second law suggests that simultaneous claims might be impossible due to energy conservation. For 350 houses, I calculate a power requirement of 116.7 kW. However, with the given inputs, the generated electricity supports around 215 homes, not the expected 350. So, I need to demonstrate this discrepancy further.**Assessing thermodynamic feasibility** I need to consider that it's not just an issue of insufficient power but also to check the feasibility of heat transfer into water and calculate the entropy criterion. It's vital to analyze B's claimed figures to see if they are thermodynamically possible. The argument states the electric load may be false, but I should also examine whether heating water at 50L/min is feasible, especially given the heat rejection involved. I want to parse through B's multi-source cycle recommendations to understand the heat transfer process better.**Analyzing product claims** I'm evaluating claims that a product can simultaneously supply residential needs for 350 units and heat water at 50L/min by rejecting heat at 300K. I infer certain cold rates and calculate work produced, which leads to a home demand that's greater than what seems plausible. Using the second law of thermodynamics, the multi-source configuration appears acceptable, but the energy requirements don't make sense. If the minimum input needed exceeds expectations, it clearly violates the first law of thermodynamics, especially when factoring in heat rejection.**Evaluating energy assertions** It looks like the simultaneous claims violate the principle of energy conservation. I should mention that while the process itself could be possible at a 71.7 kW electrical output, the positive entropy generation indicates issues. This output means it could service about 215 homes, but answering the question about the realism of these aspects must be a firm "no." While heating might be feasible, the electric claim seems impossible. The second law could be essential for a thorough thermodynamic analysis.**Calculating energy balance** I’m analyzing a system with a rejection temperature of 300K, focusing on heat transfer and water temperature. They handle a reservoir at 300K while maintaining a temperature difference of 10K. I need to be careful with unit conversions, such as 1HP equaling 0.75kW. For household demand, that's about 1/3 kW per home, and my calculations show potential to support 215 homes based on energy balance. I see a discrepancy between inputs and outputs that I should note.**Analyzing thermal energy requirements** I see that 195 kW is clearly greater than 150 kW, and it looks like 195 kW is indeed necessary. However, heat rejection at 45 kW to T is an issue. The multireservoir engine’s entropy generating positively complicates things without violating the second law. The claims about heating for 350 homes seem unrealistic, especially if each home is using an average of 240 kWh per month. This calculation could lead us to a more precise number, like 215 homes.**Evaluating entropy and second law** I need to check the Q residual and Sgen. The incoming reservoir shows a decrease in entropy of 150/900, which equals about 0.1667. Meanwhile, the outgoing sinks gain an entropy of 45/450, which equals 0.1, and then 33.333/300 is about 0.11111. This gives us a net increase of approximately 0.04444. So yes, our approach seems robust, but I wonder if there's a latent question about whether it’s realistic under the second law?

a) Temperatura da fonte fria da empresa A

A potência útil da máquina é

\dot W_A=80\ \text{HP}\times \frac{3}{4}\frac{\text{kW}}{\text{HP}} =60\ \text{kW}.

Logo, sua eficiência é

\eta_A=\frac{\dot W_A}{\dot Q_H} =\frac{60}{150}=0{,}40.

Como essa eficiência corresponde a 80\% da eficiência máxima de Carnot,

\eta_A=0{,}8\left(1-\frac{T_c}{900}\right).

Assim,

0{,}40=0{,}8\left(1-\frac{T_c}{900}\right) \quad\Rightarrow\quad T_c=450\ \text{K}.

b) Taxa de calor rejeitada pela empresa A

Pela conservação de energia no ciclo,

\dot Q_H=\dot W_A+\dot Q_c,

portanto,

\dot Q_c=150-60=\boxed{90\ \text{kW}}.

c) Análise da proposta da empresa B

Na proposta B, a rejeição para a fonte a 450\ \text{K} é metade da rejeição da empresa A:

\dot Q_{c,450}=\frac{90}{2}=45\ \text{kW}.

A vazão de água é 50\ \text{kg/min}=5/6\ \text{kg/s}. A potência térmica necessária para aquecê-la em 10\ \text{K} é

\dot Q_{c,300}=\dot m c\,\Delta T =\frac{5}{6}\times 4000\times 10 =33\,333\ \text{W} \approx 33{,}3\ \text{kW}.

Então, pela primeira lei, a potência útil da máquina B seria

\dot W_B=150-45-33{,}3 \approx 71{,}7\ \text{kW}.

Cada residência consome, em média,

\frac{240\ \text{kW·h}}{720\ \text{h}}=\frac{1}{3}\ \text{kW}.

Logo, 350 residências demandam

350\times\frac{1}{3}\approx116{,}7\ \text{kW},

valor superior aos 71{,}7\ \text{kW} disponíveis. Essa potência sustentaria, em média, cerca de 215 residências, não 350.

A rejeição de calor indicada não viola, por si só, a segunda lei, pois

\frac{45}{450}+\frac{33{,}3}{300}-\frac{150}{900} \approx 0{,}0444\ \text{kW/K}>0.

Assim, a recuperação de calor para aquecer água é compatível com o balanço térmico, mas não pode ocorrer simultaneamente com o fornecimento de energia elétrica prometido para 350 residências. A alegação conjunta do representante não é realística, por violar a conservação de energia.

Resposta final: a) 450\ \text{K}; b) 90\ \text{kW}; c) não é realística: a máquina B forneceria cerca de 71{,}7\ \text{kW}, insuficientes para os 116{,}7\ \text{kW} demandados por 350 residências.

Claude Fable 5.1

Respondeu · bate com a resposta de referência dos cursinhos

a) T = 450 K; b) Q = 90 kW; c) Não é realístico: a potência exigida (~116,7 kW) supera o máximo possível (105 kW pela 1ª lei; ≤85 kW pela 2ª lei; ≤71,7 kW se ainda aquecer a água a 33,3 kW).

levou 19 s

Como pensou

For Company A, I convert 80 HP to 60 kW, giving an efficiency of 0.4 against a heat input of 150 kW. Since this is 80% of the Carnot efficiency, the Carnot efficiency works out to 0.5, which gives T = 450 K. The rejected heat is then 150 − 60 = 90 kW, and I'm starting to look at Company B next. For Company B, I check whether its claimed output of 350 units × 240 kWh/month is physically achievable: converting to a continuous power demand gives about 116.7 kW. Accounting for the water-heating load adds a required rejection of roughly 33.3 kW at 300 K, and testing the first law shows the work output would exceed the maximum 105 kW available — already impossible. Running the entropy inequality confirms this further, since even without water heating the work is capped near 85 kW, and with it, closer to 71.7 kW — nowhere near the claimed 116.7 kW, so I conclude the claim isn't physically realistic.

Dados iniciais (Empresa A)

Potência útil: W_A = 80 HP = 80 · (3/4) = 60 kW
Calor recebido: Q_H = 150 kW

a) Temperatura da fonte fria da empresa A

Rendimento real: η_A = W_A/Q_H = 60/150 = 0,40

Como η_A = 0,80 · η_Carnot:

η_Carnot = 0,40/0,80 = 0,50

η_Carnot = 1 − T/T_H ⟹ 1 − T/900 = 0,5 ⟹ T = 450 K

b) Taxa de calor rejeitada (empresa A)

Primeira lei para o ciclo: Q = Q_H − W_A = 150 − 60 = 90 kW

c) Análise da proposta da empresa B

Máquina B: recebe 150 kW a 900 K; rejeita Q/2 = 45 kW para a fonte a 450 K e Q' para a fonte a 300 K.

Demanda elétrica alegada:
Energia mensal: 350 × 240 = 84 000 kWh
Potência média: P_el = 84 000/720 ≈ 116,7 kW

Aquecimento de água alegado:
Vazão: 50 L/min = 50/60 kg/s
Q' = ṁ·c·ΔT = (50/60)(4000)(10) ≈ 33 333 W ≈ 33,3 kW (esse é o mínimo que precisa ir para a fonte de 300 K)

Primeira lei para a máquina B:
W_B = 150 − 45 − Q' = 105 − Q'

Mesmo com Q' = 0, W_B ≤ 105 kW < 116,7 kW. Com Q' = 33,3 kW (aquecimento da água), W_B = 71,7 kW. Para fornecer 116,7 kW seria necessário Q' = −11,7 kW (fonte de 300 K cedendo calor), o que é absurdo.

Segunda lei (Clausius) para a máquina B:
−150/900 + 45/450 + Q'/300 ≥ 0 ⟹ −1/6 + 1/10 + Q'/300 ≥ 0 ⟹ Q' ≥ 20 kW
Logo W_B ≤ 105 − 20 = 85 kW, no limite reversível. Incluindo o aquecimento da água (Q' ≥ 33,3 kW), W_B ≤ 71,7 kW.

Conclusão: a potência elétrica necessária (≈116,7 kW) excede o máximo permitido tanto pela 1ª lei (105 kW) quanto pela 2ª lei (85 kW, ou 71,7 kW com o aquecimento de água). As alegações do representante da empresa B não são realísticas (violariam as leis da termodinâmica); no máximo, com o aquecimento de água, atenderia cerca de 71,7·720/240 ≈ 215 residências.

Resposta final: a) T = 450 K; b) Q = 90 kW; c) Não é realístico: a potência exigida (~116,7 kW) supera o máximo possível (105 kW pela 1ª lei; ≤85 kW pela 2ª lei; ≤71,7 kW se ainda aquecer a água a 33,3 kW).

Claude Opus 5.5

Respondeu · bate com a resposta de referência dos cursinhos

a) T = 450 K; b) Q = 90 kW; c) Não são realísticos: seriam necessários ≈116,7 kW elétricos + 33,3 kW térmicos = 150 kW, mas a 1ª lei limita W + Q′ a 105 kW e a 2ª lei limita W a no máximo 85 kW.

levou 14 s

Como pensou

For Company A, with efficiency 0.4 being 80% of Carnot efficiency, I find the hot reservoir temperature comes out to 450 K, and the heat rejected is 90 kW. Now I'm working through Company B's setup, where it also draws 150 kW from the hot reservoir but rejects 45 kW at 450 K, then figuring out the remaining heat transfer to the 300 K reservoir needed to meet the heating demand of 350 residences at 240 kWh each. Checking the energy balance, the required work plus residual heat exceeds what's available, which violates the first law. Testing the second law via entropy generation, I find the maximum feasible work output is only about 85 kW, well below the 116.67 kW needed — confirming the claim of supplying all 350 residences is unrealistic, with only around 215 achievable instead.

Dados convertidos: potência útil de A: W_A = 80 HP · 0,75 = 60 kW; calor da fonte quente: Q_H = 150 kW a 900 K.

a) Temperatura da fonte fria de A

O rendimento real de A é
η_A = W_A/Q_H = 60/150 = 0,40.

Esse valor é 80 % do rendimento de Carnot:
η_Carnot = 0,40/0,80 = 0,50 = 1 − T/900.

Logo, T = 450 K.

b) Calor rejeitado por A

Pela 1ª lei:
Q = Q_H − W_A = 150 − 60 = 90 kW.

c) Análise da proposta B

Na máquina B, Q_H = 150 kW a 900 K. Ela rejeita Q/2 = 45 kW para a fonte a 450 K e Q′ para a fonte a 300 K.

Demandas anunciadas:

Residências: 350 · 240 kWh / 720 h = 350 · (1/3) kW ≈ 116,7 kW de potência elétrica.
Água: ṁ = 50 L/min = 5/6 kg/s, então Q′ = ṁ·c·ΔT = (5/6)·4000·10 ≈ 33,3 kW.

1ª lei (conservação de energia):
W + Q′ = 150 − 45 = 105 kW.

As demandas anunciadas exigiriam W + Q′ = 116,7 + 33,3 = 150 kW > 105 kW. A proposta viola a conservação de energia. Mesmo destinando 33,3 kW ao aquecimento da água, sobrariam só W = 71,7 kW, o que atende cerca de 215 residências.

2ª lei (entropia):
ΔS_univ = −150/900 + 45/450 + Q′/300 ≥ 0.

Isso exige Q′ ≥ 20 kW. Portanto W ≤ 85 kW, também insuficiente para os 116,7 kW necessários às 350 residências, ainda que se dispensasse o aquecimento da água.

Conclusão: as alegações da empresa B não são realistas. Pela 1ª lei, só há 105 kW disponíveis para W + Q′. Pela 2ª lei, a potência útil máxima é 85 kW, abaixo dos ≈116,7 kW requeridos, além dos 33,3 kW pedidos para a água.

Resposta final: a) T = 450 K; b) Q = 90 kW; c) Não são realísticos: seriam necessários ≈116,7 kW elétricos + 33,3 kW térmicos = 150 kW, mas a 1ª lei limita W + Q′ a 105 kW e a 2ª lei limita W a no máximo 85 kW.

Claude Sonnet 5.5

Respondeu · bate com a resposta de referência dos cursinhos

a) T_F = 450 K; b) Q = 90 kW; c) Não são realistas: o ciclo B fornece ≈71,7 kW (máx. ideal 85 kW), contra ≈116,7 kW necessários para 350 residências com aquecimento de água de 33,3 kW.

levou 12 s

Como pensou

For company A, with W=60 kW and QH=150 kW, the efficiency is 0.4, which equals 0.8 times Carnot efficiency, giving ηCarnot=0.5 and Tc=450 K. The rejected heat is 90 kW. Then for company B I'm setting up QH=150 kW split so half (45 kW) goes to the 450 K reservoir and the rest to 300 K. Checking the 350 residences demand: each needs 240 kWh over 720 hours, about 0.333 kW, totaling 116.67 kW required. For water heating, 50 L/min works out to 33.33 kW rejected at 300 K, leaving only W=150−45−33.33=71.67 kW of work — well short of the 116.67 kW needed. I verify the second law isn't violated here, but even solving for the minimum Q3 satisfying entropy balance (20 kW) caps the maximum possible work at 85 kW, still insufficient. Overall, meeting both demands simultaneously would require 195 kW of input, exceeding the 150 kW available, so the claim violates energy conservation and isn't physically realistic.

Dados úteis: W_A = 80 HP = 80 · 0,75 = 60 kW; Q_H = 150 kW; T_H = 900 K.

a) Temperatura da fonte fria da empresa A

Eficiência real: η = W/Q_H = 60/150 = 0,40.

Como η = 0,8 · η_Carnot, temos η_Carnot = 0,40/0,80 = 0,50.

η_Carnot = 1 − T_F/T_H, logo T_F = 900 · (1 − 0,5) = 450 K.

b) Taxa de calor rejeitada no projeto A

Pela 1ª lei, Q = Q_H − W = 150 − 60 = 90 kW.

c) Análise da proposta da empresa B

Demanda das residências. O consumo por residência é 240 kWh / 720 h = 1/3 kW. Para 350 residências, a potência necessária é P = 350/3 ≈ 116,7 kW.

Aquecimento da água. A vazão é 50 L/min = 50 kg/60 s ≈ 0,833 kg/s. A taxa de calor para a fonte de 300 K é:

Q₃ = ṁ c ΔT = 0,833 · 4000 · 10 ≈ 33,3 kW.

Balanço de energia (1ª lei) em B. Para a fonte de 450 K, Q/2 = 45 kW. Então:

W_B = 150 − 45 − 33,3 ≈ 71,7 kW.

Esse valor é menor que os 116,7 kW exigidos. Atender à demanda e ao aquecimento exigiria 116,7 + 45 + 33,3 ≈ 195 kW de entrada, mais que os 150 kW disponíveis. Isso viola a 1ª lei. Com 71,7 kW, só seriam atendidas cerca de 215 residências.

Verificação pela 2ª lei. A produção de entropia é:

S_ger = 45/450 + 33,3/300 − 150/900 = 0,100 + 0,111 − 0,167 ≈ 0,044 kW/K > 0.

O ciclo é termodinamicamente possível, mas não atende ao que foi prometido.

Melhor caso possível. Com ciclo reversível (S_ger = 0), o calor mínimo rejeitado a 300 K seria Q₃ = 300 · (150/900 − 45/450) = 20 kW. O trabalho máximo seria então 150 − 45 − 20 = 85 kW, ainda menor que 116,7 kW. Portanto, nem mesmo um ciclo ideal cumpriria a promessa.

Conclusão: os aspectos mencionados pelo representante da empresa B não são realistas. A potência necessária (≈116,7 kW) excede o trabalho que o ciclo pode fornecer (≈71,7 kW, no máximo 85 kW).

Resposta final: a) T_F = 450 K; b) Q = 90 kW; c) Não são realistas: o ciclo B fornece ≈71,7 kW (máx. ideal 85 kW), contra ≈116,7 kW necessários para 350 residências com aquecimento de água de 33,3 kW.

Grok 4.7

Respondeu · bate com a resposta de referência dos cursinhos

a) 450 K; b) 90 kW; c) não são realísticos

levou 1 min 17 s

Como pensou

The problem is from IME 2027, Physics, second phase. Empresa A has 80 HP useful power and extracts 150 kW from a hot source at 900 K. Empresa B uses a multifonte cycle with two cold sources: one at the same temperature as A's, receiving half of A's heat rejection, and another at 300 K. Both companies share the same hot source temperature and heat input. Empresa B claims to supply 350 residential units and heat water at 50 L/min as a byproduct of rejecting heat to the 300 K source. The data includes 1 HP = 3/4 kW, A's cycle efficiency at 80% of maximum admissible, monthly consumption of 240 kWh per unit, specific heat of water 4000 J/(kg·K), temperature variation +10 K, and 720 hours in a month. The tasks are to determine the cold source temperature for A and evaluate its heat rejection rate. Empresa A's useful power is 60 kW. Its efficiency is 40%, which equals 80% of the Carnot maximum. Solving for the cold source temperature yields T = 450 K. The heat rejection rate for Empresa A is 90 kW. For Empresa B, half of that heat, or 45 kW, would be rejected to the 450 K source. The remaining heat is rejected to the 300 K source. The work output of Empresa B is 105 kW minus the heat rejected to the 300 K source. The second law requires that Q_H / T_H ≤ Q_450 / 450 + Q_300 / 300. This inequality determines whether the cycle is thermodynamically possible. The minimum heat rejection to the 300 K source is 20 kW for a reversible cycle. The representative's two claims are now being evaluated: supplying electrical energy to 350 residential units and heating water at 50 L/min as a byproduct of rejecting heat to that cold source. The water heating requirement is being quantified. A flow rate of 50 L/min with a 10 K temperature increase demands roughly 33 kW. This value is being compared against the heat rejected to the 300 K source. The 33 kW heating requirement is greater than the minimum 20 kW required by the second law for a cycle to be possible. Therefore, the byproduct claim is thermodynamically feasible. The maximum possible work output for Empresa B is 85 kW under a reversible cycle. This value must be compared against the total electrical demand of 350 residential units to determine if the proposal is feasible. With 33.3 kW rejected to the 300 K source, the entropy balance is satisfied and the engine produces about 71.7 kW of work. The next step is to check whether this output can cover the monthly electrical demand of 350 residential units. The average power demand per residential unit is about 0.33 kW. For 350 units, the total monthly requirement reaches roughly 116.7 kW. The engine's maximum possible work output is only 85 kW, so it cannot supply that many units even under ideal conditions. The representative's claim about supplying electrical energy to 350 residential units is therefore thermodynamically impossible. The "additional asset" is the water heating byproduct. The question now asks for a thermodynamic analysis of this aspect. The final task is to conclusively judge whether the representative's claims are realistic. Both the 350-unit supply and the water-heating byproduct are impossible simultaneously. The maximum work output of 85 kW remains below the 116.7 kW required for 350 units, and the fixed heat rejection to the 450 K source cannot be reduced further. The calculations are being re-verified. The engine can supply only about 215 residential units. This is well below the 350 claimed. The fixed heat rejection to the 450 K source and the exact half-heat split are being re-examined. The impossibility is now being established through energy conservation alone. The total heat input of 150 kW must equal the work output plus the two heat rejections. The representative's claims would require a combined demand of 150 kW, but the engine can produce at most 105 kW after rejecting the fixed 45 kW to the 450 K source. This energy balance makes the proposal impossible without invoking the second law.

A potência útil da máquina da empresa A é

P_A=80\times\frac{3}{4}=60\,\mathrm{kW}.

Com a taxa de calor recebido da fonte quente \dot{Q}_H=150\,\mathrm{kW}, o rendimento vale

\eta_A=\frac{60}{150}=0{,}4.

Esse rendimento corresponde a 80\,\% do máximo admissível (rendimento de Carnot entre 900\,\mathrm{K} e a temperatura T da fonte fria):

0{,}4=0{,}8\left(1-\frac{T}{900}\right)\implies T=450\,\mathrm{K}.

Pela primeira lei aplicada ao ciclo,

\dot{Q}_F=\dot{Q}_H-P_A=150-60=90\,\mathrm{kW}.

No projeto da empresa B a fonte quente e o calor recebido são os mesmos, enquanto a rejeição para a fonte a 450\,\mathrm{K} é a metade da rejeição da empresa A:

\dot{Q}_{450}=45\,\mathrm{kW}.

A primeira lei fornece então

P_B+\dot{Q}_{300}=105\,\mathrm{kW}.

O consumo médio de cada residência é 240/720=1/3\,\mathrm{kW}, de modo que 350 residências exigem

P_\mathrm{nec}=\frac{350}{3}\approx116{,}7\,\mathrm{kW}>105\,\mathrm{kW}\ge P_B.

Logo já é impossível atender 350 unidades. O aquecimento de água a 50\,\mathrm{L/min} com \Delta T=10\,\mathrm{K} demanda ainda

\dot{Q}_\mathrm{água}=\frac{5}{6}\times4000\times10=\frac{100}{3}\approx33{,}3\,\mathrm{kW},

o que tornaria a soma P_B+\dot{Q}_{300} ainda maior que 105\,\mathrm{kW}, violando a conservação da energia. (O limite imposto pela segunda lei, P_B\le85\,\mathrm{kW}, apenas reforça a impossibilidade.) Os aspectos alegados pelo representante da empresa B não são, portanto, realísticos.

Resposta final: a) 450 K; b) 90 kW; c) não são realísticos

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a) 450\text{ K}; b) 90\text{ kW}; c) Não são realistas, pois a primeira lei limita P_B\le 105\text{ kW}, inferior aos 116{,}7\text{ kW} necessários, e o aquecimento de água reduziria ainda mais a potência disponível.

levou 32 s

Como pensou

We need answer in Portuguese, solve. Need analyze thermodynamics. Let's restate. We have two companies. Empresa A: turbine/heat engine MT1. Heat from hot source at 900 K at rate 150 kW. Useful power output 80 HP. Rejects heat Q to cold sink at unspecified temperature T. Efficiency is 80% of maximum admissible (Carnot) between 900 K and T. Need determine T (cold source temp), and heat rejection rate. Dados: 1 HP = 3/4 kW. So 80 HP = 80 * 0.75 = 60 kW. Useful power W_dot = 60 kW. Heat input Q_H_dot = 150 kW. Efficiency eta_A = W_dot / Q_H_dot = 60/150 = 0.4 = 40%. It says efficiency of cycle of A: 80% of maximum admissible. Maximum admissible Carnot = 1 - T_L/T_H = 1 - T/900. So eta_A = 0.8 * eta_Carnot = 0.8*(1 - T/900) = 0.4. Thus 1 - T/900 = 0.5 => T/900 = 0.5 => T = 450 K. Part a: 450 K. Part b: heat rejected rate = Q_H_dot - W_dot = 150 - 60 = 90 kW (or Q_C = Q_H*(1-eta)=150*0.6=90 kW). Part b: 90 kW. Empresa B: "multifonte" cycle. Same hot source temp and heat received: T_H=900 K, Q_H_dot = 150 kW same? The text: "a temperatura de fonte quente e o calor recebido por seu ciclo termodinâmico são os mesmos do concorrente." So Q_H=150 kW, T_H=900K. It rejects heat to two cold sources: one has same temperature as competitor receiving, however half of heat associated with corresponding of company A. Wait: "calor é rejeitado a taxas distintas, para duas fontes frias: uma delas possui a mesma temperatura da concorrente recebendo, no entanto, metade do calor associado ao correspondente da empresa A. Já a segunda fonte encontra-se à temperatura de 300 K, recebendo certa quantidade de calor inicialmente não especificado." Need parse. "uma delas possui a mesma temperatura da concorrente recebendo" = one cold source has same temperature as the competitor (A) receives? The competitor receives? Actually "fonte fria da concorrente" maybe T=450K? "recebendo, no entanto, metade do calor associado ao correspondente da empresa A." So the first cold source at T=450K receives Q/2? Or "metade do calor associado ao correspondente da empresa A" = half of the heat rejected by A (90 kW)? Let's parse: It rejects heat at different rates to two cold reservoirs: one has same temperature as the competitor [A's cold source] receiving, however, half of the heat associated with the corresponding of company A. Could mean the heat rejected to the 450 K reservoir is half of A's rejected heat, i.e., Q1 = Q_A/2 = 45 kW. The second source at 300 K receives certain amount initially unspecified. Also representative B argues product can supply 350 residential units at same time, and heat water for medical post at 50 L/min as byproduct of heat rejection to 300K source. Need assess realistic? Not enough; need use thermodynamics (first/second law? entropy? Maybe determine if claimed hot water heating possible from rejection to 300K plus power output? Need compute. Part c asks argument conclusively if aspects mentioned by representative B are realistic. Need evaluate claims: It can supply 350 units of energy AND heat water at 50 L/min to ΔT=10K as byproduct of heat rejection to 300 K, with same heat input and hot source. It maybe violates first law or second law if too low heat rejection to 300K? We need compute required heat for water heating and energy for 350 units, compare with available heat rejected to 300K and total heat rejected, and maybe check cycle efficiency/entropy. Let's identify all data. Empresa A efficiency 40%, W=60 kW, Q_H=150 kW, Q_C=90 kW to T=450 K. Empresa B: Same Q_H=150 kW, T_H=900 K. Rejects Q1 to 450 K = half of A's Q_C = 45 kW? It says "metade do calor associado ao correspondente da empresa A" = half of heat associated with A's corresponding cold source = 90/2 = 45 kW. So Q2 to 300 K = Q_H - W_B - Q1. But W_B? The text doesn't explicitly state useful power of B? It says representative argues product capable of supplying needs ... maybe we can compute W_B from first law: W_B = Q_H - Q1 - Q2. Need determine Q2 maybe from claim/realism? Or maybe there is maximum efficiency? Need inspect cycle. "produto que opere em um ciclo termodinâmico 'multifonte'. Nessa opção, calor é rejeitado a taxas distintas, para duas fontes frias...". This is a heat engine rejecting heat to two reservoirs at different temperatures. The cycle may be reversible? It says "multifonte" maybe to improve efficiency? Need analyze. Need evaluate if representative B's claims realistic: "seu produto é capaz de suprir as necessidades de energia elétrica residencial de 350 unidades ao mesmo tempo, sendo ainda possível aquecer água para um posto de atendimento médico próximo, a uma taxa de 50 L por minuto, como subproduto da rejeição de calor para a fonte de 300 K." That is: Useful power output enough for 350 residential units; and it can heat water at 50 L/min by heat rejected to 300K source. Need check if thermodynamic constraints allow both. We need compute energy consumption of 350 units: "consumo mensal no município por residência: 240 kW·h por unidade; número de horas em um mês: 720 h." So average power per unit = 240 kWh / 720 h = 1/3 kW? 240/720 = 1/3 kW ≈ 0.333 kW. For 350 units, total power = 350 * 1/3 = 116.666... kW = 350/3 kW ≈ 116.7 kW. But B's hot source input is 150 kW. If output power to electricity is ~? Let's compute: To supply 116.7 kW, W_B must be at least 116.7 kW. But Q_H=150 kW, so efficiency at least 116.7/150 = 0.777... = 77.8%. That's very high for 900K and cold reservoirs 450K/300K? Maximum possible efficiency for a heat engine with two sinks? If it uses only 300K sink, Carnot efficiency = 1 - 300/900 = 66.7%, so impossible. But it rejects heat to 450K and 300K; using both sinks could maybe increase efficiency? Actually rejecting heat at higher temperature reduces Carnot efficiency? Wait, for a heat engine with multiple heat rejection reservoirs, the maximum work output for given heat input and given sink temperatures might be obtained by rejecting heat reversibly at a range of temperatures? The maximum efficiency is still 1 - T0/T_H if all heat rejected at lowest temperature T0=300K? Actually if heat is rejected at 450K and 300K, reversible engine would reject all heat at the lowest available temperature? Let's recall: For a heat engine operating between hot reservoir T_H and cold reservoir T_C, maximum efficiency = 1 - T_C/T_H. If there are multiple cold reservoirs, the maximum work for fixed Q_H is when all heat is rejected to the lowest temperature? But here first cold source at 450K receives fixed 45 kW? Maybe the cycle is irreversible? Need infer. Wait representative B says "calor é rejeitado a taxas distintas, para duas fontes frias: uma delas possui a mesma temperatura da concorrente recebendo, no entanto, metade do calor associado ao correspondente da empresa A." This might mean in B's cycle, the heat rejection to the 450 K source is fixed at 45 kW (half of A's 90 kW). The second source at 300 K receives unspecified. So W_B = 150 - 45 - Q2 = 105 - Q2 kW. To supply 116.7 kW, need Q2 negative, impossible. Therefore not realistic. But maybe "suprir as necessidades" refers to 350 units over month? Let's check: 350 units * 240 kWh/month = 84,000 kWh/month. B's engine operates at 150 kW heat input. In 720 h/month, if electric output maybe 150 kW? max output <150 kW, energy < 108,000 kWh; could be enough if output > 84,000/720 = 116.7 kW. Still need W>116.7 kW, impossible because Q_H 150 kW and heat rejection fixed at 45 kW to 450K, so W <=105 kW (if Q2>=0). Thus not enough. But maybe B's engine output? Let's parse carefully: "O representante da empresa B argumenta que seu produto é capaz de suprir as necessidades de energia elétrica residencial de 350 unidades ao mesmo tempo..." That may mean output power 350 * (240 kWh/month / 720 h) = 116.7 kW. That is impossible due first law: if Q_H=150 kW and Q1=45 kW, max W=105 kW. So unrealistic. But maybe the water heating is extra byproduct: heat rejected to 300K can heat water: Q_water = m_dot c ΔT = 50 L/min * 1 kg/L * 4000 J/(kg·K) * 10 K = 50*4000*10 = 2,000,000 J/min = 33,333.33 J/s = 33.33 kW. This heat is "subproduto da rejeição de calor para a fonte de 300 K"; i.e., the heat Q2 to 300K is used to heat water. If Q2 = 33.33 kW, then B's electric output W = 150 - 45 - 33.33 = 71.67 kW, not 116.7. If also need supply 350 units, impossible. But maybe water heating load could be provided by Q2 plus additional? It says "como subproduto da rejeição de calor para a fonte de 300 K", so Q2 must be at least 33.33 kW to heat water; then W max 71.67 kW, much less than 116.7. So claim impossible. But need be rigorous: maybe B can reject more heat to 300K and use only part for water? But total Q_H fixed 150, Q1 fixed 45, so Q2 + W = 105. To produce W≥116.7, Q2 negative, impossible. To heat water Q2≥33.33, so W≤71.67. So cannot both. Conclusion not realistic. However, must consider possible thermodynamic limits: For B's "multifonte" cycle, maybe it can produce work while rejecting heat to both reservoirs, and maybe there is a maximum efficiency depending on fraction rejected to 450K? Let's derive maybe from second law / entropy generation? If the cycle is reversible? It could have multiple heat rejections at different temperatures; maybe to maximize work, it could use the 300K sink for all heat rejection, but if it rejects Q1=45 kW to 450K, then it's irreversible? But if it's a "multifonte" maybe it uses a Carnot-like with finite heat capacity? Let's not need; first law enough. But let's check other parts: a) T=450 K; b) Q_rej=90 kW. c) Need analyze realistic? We must "conduza uma análise termodinâmica da situação descrita e: a) determine a temperatura da fonte fria do ciclo termodinâmico referente à empresa A; b) avalie a taxa de calor rejeitada para a fonte fria no projeto da empresa A; c) argumente conclusivamente se os aspectos mencionados pelo representante da empresa B são realistas." So final answer should include calculations. Need maybe mention first law and impossibility. But maybe there is nuance: They may ask to assess "aspectos mencionados" including "suprir as necessidades de energia elétrica residencial de 350 unidades" and "aquecer água a 50 L/min". Need compute. Let's do full solution. First convert units: 1 HP = 3/4 kW. 80 HP = 60 kW. Q_H_dot = 150 kW. η_A = W/Q_H = 60/150 = 0.4 = 40%. η_Carnot = 1 - T_F/T_Q = 1 - T/900. Since η_A = 80% of η_Carnot: 0.4 = 0.8 (1 - T/900) => 0.5 = 1 - T/900 => T/900 = 0.5 => T = 450 K. Second: heat rejected rate: Q_C_dot = Q_H_dot - W_dot = 150 - 60 = 90 kW. Alternatively from definition: Q_C = Q_H(1-η) = 150*(0.6)=90 kW. Then B: Q_H=150 kW, T_H=900K. First cold source at same T=450K receives "metade do calor associado ao correspondente da empresa A." Could be "metade do calor associado ao correspondente da empresa A" = half of heat A rejects to its cold source = Q1 = 90/2 = 45 kW. So Q1=45 kW. Second cold source at 300 K receives Q2 unspecified. By first law for B's cycle: W_B = Q_H - Q1 - Q2 = 150 - 45 - Q2 = 105 - Q2 (kW). To supply 350 residential units: average power per unit = 240 kWh/month / 720 h/month = 1/3 kW. Thus P_res = 350 / 3 ≈ 116.67 kW. (Actually 350 * (240/720) = 116.67 kW.) For B to supply that, need W_B ≥ 116.67 kW. But from first law, W_B = 105 - Q2 ≤ 105 kW because Q2 ≥ 0. Contradiction. So it cannot supply 350 units. Even before considering water heating. Additionally, water heating: Q_dot_agua = m_dot c ΔT, with m_dot = 50 L/min = 50 kg/min = 50/60 kg/s = 5/6 kg/s. c=4000 J/(kg·K). ΔT=10 K. So Q = (5/6)*4000*10 = 200000/6? Wait 5/6*4000=3333.33; *10=33333.33 J/s = 33.33 kW. Yes. This heat must come from Q2 (rejection to 300K). If Q2 = 33.33 kW, then W_B ≤ 71.67 kW. Still below 116.67. If B's output is less, cannot meet residential demand. Conversely if W_B ≥116.67 then Q2 would need to be negative, impossible. Also if Q2 is at least 33.33 kW, max W=71.67 kW. But wait, maybe the water heating is "como subproduto da rejeição de calor para a fonte de 300 K." It could be that Q2 is not only used for water; it rejects to 300K, and part of that heat goes to water. But if Q2 >33.33, more heat to 300K, less work. If Q2 = 33.33, max work 71.67; if Q2 larger, less work. To supply 350 units need W≥116.67, so Q2 = 105 - W ≤ -11.67? No, W=105 - Q2. For W≥116.67, Q2 ≤ -11.67, impossible. So no. Maybe there is a second-law maximum? Let's consider if B's cycle is claimed "multifonte" maybe it can output W while rejecting heat to 450 K and 300 K. The first law alone shows W ≤ 105 kW because Q1 fixed 45 kW and Q2 nonnegative. So sufficient. But maybe the "calor rejeitado a taxas distintas" means Q1 is not half of A's Q_C? Let's verify. Original text: "uma delas possui a mesma temperatura da concorrente recebendo, no entanto, metade do calor associado ao correspondente da empresa A." The phrase: "metade do calor associado ao correspondente da empresa A" could mean "half of the heat [rejected] associated with the corresponding one of company A"? I.e., the first cold source receives half of the heat that the corresponding source of company A receives. Company A rejects Q=90 kW to its cold source. So yes Q1=45 kW. Good. But "a segunda fonte encontra-se à temperatura de 300 K, recebendo certa quantidade de calor inicialmente não especificado." It doesn't say Q2 fixed; we can choose. But for water heating, Q2 must be enough. Need maybe "aspectos mencionados pelo representante da empresa B são realistas?" Could also include the claim "produto é capaz de suprir as necessidades de energia elétrica residencial de 350 unidades ao mesmo tempo, sendo ainda possível aquecer água..." So both. Let's solve in a formal way. Potential issue: Efficiency of A: "80% do máximo admissível" maybe maximum admissible = Carnot. Yes likely. Potential issue: 1 HP = 3/4 kW; 80 HP = 60 kW. If they maybe use metric HP? But given 1HP=3/4 kW. Let's produce final with item by item. Need ensure "b) avalie a taxa de calor rejeitada para a fonte fria no projeto da empresa A": 90 kW = 90 kJ/s. Could also compute as Q_out = 150 - 60 = 90 kW. Part c: We need "argumente conclusivamente se os aspectos mencionados... são realistas." Need maybe mention that from first law, B's work output maximum if second cold source receives zero heat: W_max = Q_H - Q1 = 150 - 45 = 105 kW. But required residential power = 116.7 kW, so impossible. Also water heating requires at least 33.3 kW of heat rejection to 300 K; then W_B ≤150 - 45 - 33.3 = 71.7 kW, even more insufficient. Therefore claims are not realistic. Maybe also note second law? Let's see. Maybe there is additional second law constraint: If B were a reversible engine operating between 900K and 300K and 450K? Let's compute maximum possible work with given Q_H and given Q1? But first law already. Maybe someone could argue B could produce 116.7 kW by rejecting less to 450K? But Q1 is fixed by "metade do calor associado..."? Actually Q1 is specified by representative's own statement, not necessarily required by cycle. But we take statement as claim; if they claim Q1=45 kW, then W max is 105. So impossible. But could "metade do calor associado ao correspondente da empresa A" mean maybe "half of Q_H"? No, "calor associado ao correspondente" likely rejected heat. We assumed. Let's check alternative interpretations: The first cold source receives "metade do calor associado ao correspondente da empresa A" could mean half of A's heat rejection "Q" = 45 kW. The second at 300K receives unspecified. The representative says product can supply 350 units and heat water. If Q1=45 and water heating needs 33.33, then Q2=33.33, W=71.67. If not heating water, maybe Q2=0, W=105. Still less than 116.67. So no. Maybe "consumo mensal no município por residência: 240 kWh por unidade; número de horas em um mês: 720 h" maybe the 350 units' consumption is 350 * 240 = 84000 kWh per month. The engine's maximum energy output in a month = 105 kW * 720 h = 75600 kWh, less. So also energy shortfall. Good to mention. Let's also maybe check "energia elétrica residencial de 350 unidades ao mesmo tempo" means instantaneous power? "ao mesmo tempo" may mean simultaneously; yes average power 116.7 kW. Maybe there is need to compute "suprir as necessidades" maybe using monthly energy: Required monthly = 350 * 240 = 84000 kWh. Available from B if W=105 kW over 720h = 75600 kWh, short by 8400 kWh (10%). If W=71.67, much short. Good. But wait, B's cycle heat input 150 kW is "taxa" maybe continuous; yes. Let's think if there is any possibility of using the rejected heat to 300 K for water heating AND also generating electricity? The rejected heat is after generating work, so cannot generate work from same heat. It can heat water. The claim says "como subproduto da rejeição de calor para a fonte de 300 K", so yes. Let's craft final response in Portuguese, with equations. Potential final: a) $P_u = 80\,HP \times \frac{3}{4}\,kW/HP = 60\,kW$. $Q_q = 150\,kW$. $\eta_A = P_u/Q_q = 60/150 = 0{,}40 = 40\%$. Sendo $\eta_A = 0{,}8\,\eta_{Carnot}=0{,}8(1 - T_f/900)$, temos $0{,}40 = 0{,}8(1 - T_f/900) \Rightarrow 1 - T_f/900 = 0{,}5 \Rightarrow T_f = 450\,K$. b) $Q_f = Q_q - P_u = 150 - 60 = 90\,kW$. (ou 90 kJ/s) c) Empresa B: $Q_q=150\,kW$, $T_q=900\,K$. Fonte fria 1: $T=450\,K$, recebe $Q_1 = Q_{f,A}/2 = 45\,kW$. Fonte fria 2: $T=300\,K$, recebe $Q_2$. Primeira lei: $P_B = 150 - 45 - Q_2 = 105 - Q_2$ kW. Necessidade das 350 residências: $P_{res} = 350 \times \frac{240\,kWh/mês}{720\,h/mês} = 350 \times \frac{1}{3}\,kW = 116{,}7\,kW$. Para atender, seria preciso $P_B \ge 116{,}7\,kW$, mas $P_B \le 105\,kW$, pois $Q_2 \ge 0$. Logo, nem sem aquecimento de água é possível. Além disso, o aquecimento de água a 50 L/min com $\Delta T=10K$ exige $\dot Q_{água} = 50\,kg/min \times 4000\,J/(kgK) \times 10K = 2\,000\,000\,J/min = 33{,}3\,kW$. Esse calor tem de sair de $Q_2$. Se $Q_2 \ge 33{,}3\,kW$, então $P_B \le 150 - 45 - 33{,}3 = 71{,}7\,kW$, muito menor que 116,7 kW. Portanto, as alegações não são realistas; violam a primeira lei da termodinâmica (e a limitação de Carnot para a operação com as fontes disponíveis). Should I mention second law? Maybe not necessary; first law suffices. But "análise termodinâmica" maybe include entropy? Could add second law: The proposed operation would be impossible even before considering irreversibilities because first law. If they try to reject Q2=0 and output 116.7, then Q_H=150, Q1=45, W=116.7 implies Q2=-11.7, impossible. Good. Let's verify numbers: 350 units * 240 kWh/month / 720 h = 350*(1/3) = 116.666... kW. Yes. Maybe they want "taxa de calor rejeitada para a fonte fria no projeto A" could be computed as "Q = 90 kW". Need item b. Let's final with "Resposta final: a) 450 K; b) 90 kW; c) não são realistas..." But instruction says "Termine com uma linha neste formato: Resposta final: <o resultado de cada item...>". So final line. Need include development in Portuguese. The final answer can be moderate. Include equations. Since final should be in Portuguese. Let's craft final with proper formatting. Potential nuance: In item c, the text says "argumente conclusivamente se os aspectos mencionados pelo representante da empresa B são realistas." Maybe they expect "Não, pois o trabalho máximo possível é 105 kW e a necessidade é 116,7 kW; além disso, o aquecimento de água exigiria 33,3 kW, reduzindo o trabalho máximo a 71,7 kW." Good. Let's finalize.

a) Cálculo da temperatura da fonte fria da empresa A:

A potência útil é

P_u=80\text{ HP}=80\cdot\frac34\text{ kW}=60\text{ kW}

e o calor recebido da fonte quente é

Q_q=150\text{ kW}.

Assim, o rendimento da máquina A é

\eta_A=\frac{P_u}{Q_q}=\frac{60}{150}=0{,}40=40\%.

Como esse rendimento é 80% do máximo admissível (Carnot) entre 900 K e T_f:

0{,}40=0{,}8\left(1-\frac{T_f}{900}\right)

1-\frac{T_f}{900}=0{,}5

\frac{T_f}{900}=0{,}5

T_f=450\text{ K}.

b) Taxa de calor rejeitada pela empresa A:

Q_f=Q_q-P_u=150-60=90\text{ kW}.

c) Análise da empresa B:

A empresa B também recebe

Q_q=150\text{ kW},\quad T_q=900\text{ K}.

Ela rejeita para uma fonte fria a 450 K metade do calor rejeitado por A:

Q_1=\frac{90}{2}=45\text{ kW}.

Para a fonte de 300 K, seja Q_2 a taxa de calor rejeitada. Pela primeira lei:

P_B=Q_q-Q_1-Q_2=150-45-Q_2=105-Q_2.

Como Q_2\ge 0,

P_B\le 105\text{ kW}.

A necessidade elétrica das 350 unidades residenciais é:

P_{res}=350\cdot\frac{240\text{ kWh/mês}}{720\text{ h/mês}} =350\cdot\frac13\text{ kW} \approx 116{,}7\text{ kW}.

Logo, mesmo sem aquecimento de água, seria necessário P_B\ge 116{,}7\text{ kW}, mas a primeira lei permite no máximo 105 kW. Portanto, a alegação já é impossível.

Além disso, o aquecimento de água a 50\text{ L/min} com \Delta T=10\text{ K} exige:

\dot Q_{\text{água}}=\dot m c\Delta T =\left(\frac{50}{60}\text{ kg/s}\right)(4000\text{ J/(kg·K)})(10\text{ K}) \approx 33{,}3\text{ kW}.

Esse calor teria de vir da rejeição Q_2 para a fonte de 300 K. Se Q_2\ge 33{,}3\text{ kW}, então

P_B\le 150-45-33{,}3=71{,}7\text{ kW},

muito abaixo dos 116,7 kW necessários. Assim, os aspectos mencionados pelo representante da empresa B não são realistas termodinamicamente.

Resposta final: a) 450\text{ K}; b) 90\text{ kW}; c) Não são realistas, pois a primeira lei limita P_B\le 105\text{ kW}, inferior aos 116{,}7\text{ kW} necessários, e o aquecimento de água reduziria ainda mais a potência disponível.

Gemini 3.8 Flash

Respondeu · bate com a resposta de referência dos cursinhos

a) 450 K; b) 90 kW; c) Não são realísticos, pois violam a 1ª e a 2ª Leis da Termodinâmica.

levou 49 s

Como pensou

**Analyzing the Prompt** I am now focusing on understanding the core of the IME exam problem. My current thought process involves carefully dissecting the provided text and any associated images to grasp the full context and constraints of the question at hand. **Evaluating Turbine Options** I am currently comparing two distinct turbine proposals. One, from Company A, involves an 80 HP unit with a heat input of 150 kW from a 900 K source. Company B offers a "multi-source" cycle, matching Company A's hot source temperature and heat input but rejecting heat to two cold sources, one at the same temperature as A's but receiving half the heat, and another at 300 K. **Calculating Heat Rejection** I am now calculating the heat rejected by Company A's turbine to its cold source. This involves using the given efficiencies and heat inputs, as well as the conversion factor for horsepower to kilowatts. I'm also determining the temperature of this cold source based on the provided thermodynamic cycle efficiency. **Verifying Company B's Claims** I'm now critically assessing the feasibility of Company B's multi-source cycle proposal. My current focus is on mathematically substantiating whether their proposed heat rejection to two distinct cold sources, while maintaining the same hot source and heat input, aligns with thermodynamic principles and their claimed performance. **Calculating Heat Rejection** I've determined the cold source temperature for Company A's turbine to be 450 K. I am now calculating the heat rejected by Company A's turbine to its cold source, which is 90 kW. **Assessing Company B's Cycle** I am now examining Company B's multi-source cycle. I've confirmed their hot source and heat input match Company A. My current focus is on their proposed heat rejection to two cold sources, one at 450 K receiving 45 kW. **Calculating Water Heating Needs** I'm now calculating the minimum heat required to warm the specified volume of water. My focus is on determining the heat rejection rate to the 300 K source, accounting for flow rate, specific heat, and a 10 K temperature rise. This calculation is crucial for validating Company B's proposal. **Validating Cycle Efficiency** I'm now cross-referencing the electrical power output claims with the energy balance for Company B. The calculations confirm that their proposed electrical generation of approximately 116.67 kW is achievable within the energy constraints derived from the First Law of Thermodynamics. **Detecting Thermodynamic Violation** I have identified a critical flaw in Company B's proposal. Their claimed power output and heat rejection to the secondary cold source, when combined with the heat rejected to the primary cold source, exceed the total heat input. This directly violates the First Law of Thermodynamics, rendering their design infeasible. **Proving the Flaw** I am now rigorously applying the Second Law of Thermodynamics, specifically Clausius' Inequality, to Company B's proposed cycle. My calculations demonstrate that their claimed heat rejection to the secondary cold source is insufficient, leading to a maximum possible work output far below their stated requirement. This definitively proves their design is thermodynamically impossible. **Refuting Company B's Design** I have definitively proven that Company B's proposed cycle violates the First Law of Thermodynamics. Their claimed power output and required water heating exceed the total heat input, confirming their design's impossibility. **Proving Thermodynamic Impossibility** I've rigorously proven that Company B's design is thermodynamically impossible. My analysis, utilizing both the First and Second Laws, demonstrates that their proposed power output and heat rejection to the secondary cold source, when summed with the heat rejected to the primary source, exceed the total heat input. This definitively refutes their claims. **Refuting Company B's Cycle** I've now confirmed that Company B's multi-source cycle violates the First Law of Thermodynamics. Their claimed power output and total heat rejection exceed the total heat input, definitively proving their design is impossible. **Proving Impossibility** I've established that Company B's design is thermodynamically impossible. My calculations, using both the First and Second Laws, show their claimed power output and required heat rejection exceed the total heat input, definitively refuting their proposal. **Refuting Final Claim** My analysis confirms Company B's proposal violates the First Law. Their claimed power output, combined with the minimum heat rejection required for water heating, exceeds the total heat input. This definitively proves their design is impossible. **Confirming Impossibility** My analysis definitively shows Company B's design violates the First Law. Their claimed power and required heat rejection exceed the input heat. The Second Law also demonstrates their claimed heat rejection to the secondary source is insufficient.

Para resolver a questão, faremos uma análise termodinâmica detalhada das propostas apresentadas por ambas as empresas.

a) Determinação da temperatura da fonte fria da Empresa A

A potência mecânica/útil fornecida pela turbina da empresa A é:

\dot{W}_A = 80\text{ HP} = 80 \times \frac{3}{4}\text{ kW} = 60\text{ kW}

A taxa de calor extraída da fonte quente (T_Q = 900\text{ K}) é \dot{Q}_Q = 150\text{ kW}. Assim, o rendimento real do ciclo da máquina da empresa A é dado por:

\eta_A = \frac{\dot{W}_A}{\dot{Q}_Q} = \frac{60\text{ kW}}{150\text{ kW}} = 0,40 \quad (40\%)

Sabe-se que essa eficiência corresponde a 80\% do máximo admissível (rendimento de Carnot entre a fonte quente T_Q e a fonte fria T):

\eta_{\text{Carnot}} = 1 - \frac{T}{T_Q} = 1 - \frac{T}{900}

\eta_A = 0,80 \cdot \eta_{\text{Carnot}} \implies 0,40 = 0,80 \left(1 - \frac{T}{900}\right)

Resolvendo para T:

1 - \frac{T}{900} = \frac{0,40}{0,80} = 0,50 \implies \frac{T}{900} = 0,50 \implies T = 450\text{ K}

b) Avaliação da taxa de calor rejeitada para a fonte fria na Empresa A

Pela Primeira Lei da Termodinâmica aplicada a um ciclo térmico em regime permanente:

\dot{Q}_Q = \dot{W}_A + \dot{Q}_A

onde \dot{Q}_A é a taxa de calor rejeitada para a fonte fria de temperatura T = 450\text{ K}:

\dot{Q}_A = \dot{Q}_Q - \dot{W}_A = 150\text{ kW} - 60\text{ kW} = 90\text{ kW}

c) Análise da proposta da Empresa B

No projeto da empresa B, o ciclo opera com três reservatórios térmicos:

Fonte quente: T_Q = 900\text{ K}, com \dot{Q}_{Q,B} = 150\text{ kW};
Primeira fonte fria: T_1 = 450\text{ K}, recebendo calor a uma taxa \dot{Q}_{B,1} = \frac{\dot{Q}_A}{2} = \frac{90}{2} = 45\text{ kW};
Segunda fonte fria: T_2 = 300\text{ K}, recebendo calor a uma taxa \dot{Q}_{B,2}.

Vamos avaliar as demandas declaradas pelo representante da empresa B:

Demanda de energia elétrica residencial:
- A potência média contínua consumida por unidade residencial é:
  
  P_{\text{unidade}} = \frac{240\text{ kW}\cdot\text{h}}{720\text{ h}} = \frac{1}{3}\text{ kW}
- Para atender a 350 unidades residenciais simultaneamente, a potência elétrica exigida é:
  
  \dot{W}_B \ge 350 \times \frac{1}{3}\text{ kW} = \frac{350}{3}\text{ kW} \approx 116,67\text{ kW}
Taxa de calor para aquecimento de água:
- A vazão mássica de água é:
  
  \dot{m} = \rho \cdot \dot{V} = 1\text{ kg/L} \times \frac{50\text{ L}}{60\text{ s}} = \frac{5}{6}\text{ kg/s}
- A taxa de calor mínima transferida para a água aquecida (\Delta T = 10\text{ K}) é:
  
  \dot{Q}_{\text{água}} = \dot{m} \cdot c \cdot \Delta T = \frac{5}{6}\text{ kg/s} \times 4000\text{ J/(kg}\cdot\text{K)} \times 10\text{ K} = \frac{200\,000}{6}\text{ W} = \frac{100}{3}\text{ kW} \approx 33,33\text{ kW}
- Como o aquecimento da água é subproduto da rejeição para a fonte de 300\text{ K}, devemos ter:
  
  \dot{Q}_{B,2} \ge \frac{100}{3}\text{ kW}
Verificação da 1ª Lei da Termodinâmica:
Pela conservação de energia na máquina térmica da empresa B:

\dot{Q}_{Q,B} = \dot{W}_B + \dot{Q}_{B,1} + \dot{Q}_{B,2}

150\text{ kW} = \dot{W}_B + 45\text{ kW} + \dot{Q}_{B,2} \implies \dot{W}_B + \dot{Q}_{B,2} = 105\text{ kW}

Porém, somando as demandas alegadas pelo representante:

\dot{W}_{B,\text{requerido}} + \dot{Q}_{B,2,\text{requerido}} = \frac{350}{3}\text{ kW} + \frac{100}{3}\text{ kW} = \frac{450}{3}\text{ kW} = 150\text{ kW}

Portanto, a taxa total de energia demandada saindo da turbina seria:

\dot{E}_{\text{saída}} = \dot{W}_B + \dot{Q}_{B,1} + \dot{Q}_{B,2} = 150\text{ kW} + 45\text{ kW} = 195\text{ kW} > 150\text{ kW} = \dot{Q}_{Q,B}

O que viola frontalmente o Princípio da Conservação da Energia (Primeira Lei da Termodinâmica).
Verificação da 2ª Lei da Termodinâmica:
Pela desigualdade de Clausius:

\frac{\dot{Q}_{Q,B}}{T_Q} - \frac{\dot{Q}_{B,1}}{T_1} - \frac{\dot{Q}_{B,2}}{T_2} \le 0 \implies \frac{150}{900} - \frac{45}{450} - \frac{\dot{Q}_{B,2}}{300} \le 0

\frac{1}{6} - \frac{1}{10} \le \frac{\dot{Q}_{B,2}}{300} \implies \frac{1}{15} \le \frac{\dot{Q}_{B,2}}{300} \implies \dot{Q}_{B,2} \ge 20\text{ kW}

Assim, a potência teórica máxima admissível (caso reversível) seria:

\dot{W}_{B,\text{máx}} = \dot{Q}_{Q,B} - \dot{Q}_{B,1} - \dot{Q}_{B,2,\text{mín}} = 150 - 45 - 20 = 85\text{ kW}

Como a potência demandada para suprir as residências é de \frac{350}{3}\text{ kW} \approx 116,67\text{ kW} > 85\text{ kW}, a proposta viola também a Segunda Lei da Termodinâmica.

Conclusão: Os aspectos mencionados pelo representante da empresa B não são realísticos, pois violam simultaneamente a Primeira e a Segunda Leis da Termodinâmica.

Resposta final:
a) 450 K;
b) 90 kW;
c) Não são realísticos, pois violam a 1ª e a 2ª Leis da Termodinâmica.

Gemini 4 Argon

Respondeu · bate com a resposta de referência dos cursinhos

a) 450 K; b) 90 kW; c) Não são realísticos, pois violam a Primeira Lei da Termodinâmica (a soma das taxas de energia de saída, 195 kW, excede a taxa de calor fornecida pela fonte quente, 150 kW) e excedem a potência máxima admissível pela Segunda Lei da Termodinâmica.

levou 1 min 30 s

Como pensou

**Analyzing Physics Problem** I am currently processing the initial details of a physics problem from the IME 2027 exam, focusing on understanding the parameters of "Empresa A" and its hot reservoir temperature. **Processing Engine Parameters** I'm now analyzing the specifications for Empresa A's turbine, noting its $80\text{ HP}$ useful power output and the heat absorption rate of $150\text{ kW}$ from a $900\text{ K}$ hot reservoir. I'm also starting to map out the components of Empresa B's proposed system. **Evaluating Thermodynamic Cycles** My focus has shifted to analyzing Empresa B's multi-source thermodynamic cycle. I'm detailing its heat rejection to two cold reservoirs, one at the same temperature as Empresa A's but receiving half the heat, and another at $300\text{ K}$. I'm also considering the implications of its ability to simultaneously provide electrical power and hot water. **Determining Cold Reservoir Temperature** I'm now calculating the cold reservoir temperature for Empresa A, using its efficiency relative to the maximum possible and its power output. This involves applying thermodynamic principles and the provided conversion factor from horsepower to kilowatts. **Confirming Cold Reservoir Temperature** I've confirmed the cold reservoir temperature for Empresa A is $450\text{ K}$ by relating its actual efficiency to the Carnot efficiency. The calculation was remarkably straightforward, directly yielding the result. **Verifying Empresa B's Claims** I'm now evaluating the realism of Empresa B's assertions. This involves calculating the heat rejected to its two cold reservoirs and the electrical power it can provide, then applying thermodynamic principles to determine if these claims are feasible. **Calculating Heat Rejection** I am now calculating the heat rejected by Empresa B to its second cold reservoir at $300\text{ K}$. This involves determining the mass flow rate and specific heat required to heat water at $50\text{ L/min}$ by $10\text{ K}$. **Checking Energy Balances** I am now cross-referencing the energy balance for Empresa B, specifically focusing on the discrepancy between the sum of rejected heat and work output, and the total heat input. It appears there's a potential issue with the first law not being fully satisfied when accounting for all heat flows. **Investigating Energy Discrepancies** I'm currently examining the energy balance for Empresa B, specifically a discrepancy where the calculated energy output exceeds the input. This suggests a potential violation of the First Law of Thermodynamics and requires further investigation into the provided figures. **Revising Energy Balance** I'm now re-evaluating Empresa B's energy balance. It appears the claimed power output and heat rejection to both reservoirs exceed the heat input from the hot reservoir, suggesting a potential violation of the First Law. I'm recalculating to find the maximum possible work output under these conditions. **Calculating Maximum Work** I am now calculating the maximum possible work output for Empresa B, given its heat input and rejection to the first cold reservoir, to see if its claimed performance is thermodynamically feasible. This involves applying the Second Law to find the upper limit on work. **Applying Second Law Principles** I've successfully applied the Clausius inequality to determine the minimum heat rejection ($\dot{Q}_2$) to the second cold reservoir for Empresa B. This yields a maximum theoretical work output, which I'm now comparing against the claimed performance to assess feasibility. **Verifying Heat for Water** I am now checking if the calculated heat rejection for Empresa B is sufficient to heat the water as specified. This involves ensuring the heat transferred to the $300\text{ K}$ reservoir meets the requirements for the hot water production. **Calculating Final Work Output** I'm now calculating the final maximum work output for Empresa B, considering the actual heat required for hot water production. This step confirms the thermodynamic limitations on its performance under the given conditions. **Analyzing Heat Water Requirement** I am now focusing on the thermal power needed to heat the water, noting that heating water at $300\text{ K}$ by $10\text{ K}$ using a cold reservoir at $300\text{ K}$ implies the reservoir is the sink for heat transfer, not the source for heating. The calculations for this thermal requirement are underway. **Resolving Thermal Inconsistencies** I've identified that the proposed energy output for Empresa B significantly exceeds its heat input, a clear violation of the First Law of Thermodynamics. My current focus is on recalculating its maximum potential work output to rigorously assess the feasibility of its claims.

a) Para o ciclo termodinâmico da empresa A (máquina térmica \text{MT 1}), a potência útil fornecida é:

\dot{W}_A = 80\text{ HP} = 80 \times \frac{3}{4}\text{ kW} = 60\text{ kW}

Como a taxa de calor extraída da fonte quente (T_Q = 900\text{ K}) é \dot{Q}_Q = 150\text{ kW}, o rendimento efetivo (\eta_A) do ciclo da empresa A vale:

\eta_A = \frac{\dot{W}_A}{\dot{Q}_Q} = \frac{60\text{ kW}}{150\text{ kW}} = 0,40 = 40\%

Sabendo que esse rendimento corresponde a 80\% do rendimento máximo admissível (rendimento de Carnot, \eta_{\text{máx}} = 1 - \frac{T}{T_Q}), tem-se:

\eta_A = 0,80 \cdot \left(1 - \frac{T}{T_Q}\right) \implies 0,40 = 0,80 \cdot \left(1 - \frac{T}{900}\right)

1 - \frac{T}{900} = 0,50 \implies \frac{T}{900} = 0,50 \implies T = 450\text{ K}

b) Pela Primeira Lei da Termodinâmica aplicada ao ciclo da empresa A em regime permanente, a taxa de calor rejeitada para a fonte fria (\dot{Q}) é dada por:

\dot{Q}_Q = \dot{W}_A + \dot{Q} \implies \dot{Q} = \dot{Q}_Q - \dot{W}_A = 150\text{ kW} - 60\text{ kW} \implies \dot{Q} = 90\text{ kW}

c) Para avaliar se a proposta da empresa B é realística, calculam-se inicialmente as taxas de transferência de energia mencionadas pelo representante:

Taxa de calor recebida da fonte quente (T_Q = 900\text{ K}):

\dot{Q}_Q = 150\text{ kW}
Taxa de calor rejeitada para a primeira fonte fria (T_1 = T = 450\text{ K}):

\dot{Q}_{F1} = \frac{\dot{Q}}{2} = \frac{90\text{ kW}}{2} = 45\text{ kW}
Potência elétrica necessária para suprir N = 350 residências simultaneamente (\dot{W}_B):
Como cada residência consome E_{\text{mês}} = 240\text{ kW}\cdot\text{h} em um mês de \Delta t = 720\text{ h}, a potência média demandada por residência é:

P_{\text{res}} = \frac{240\text{ kW}\cdot\text{h}}{720\text{ h}} = \frac{1}{3}\text{ kW}

Para as 350 unidades residenciais:

\dot{W}_B = 350 \times \frac{1}{3}\text{ kW} = \frac{350}{3}\text{ kW} \approx 116,67\text{ kW}
Taxa de calor necessária para aquecer a água na segunda fonte fria (T_2 = 300\text{ K}):
A vazão mássica de água é \dot{m} = \rho \cdot \dot{V} = 1\text{ kg/L} \times 50\text{ L/min} = 50\text{ kg/min} = \frac{5}{6}\text{ kg/s}.
Para elevar sua temperatura de \Delta T_{\text{água}} = +10\text{ K}, a taxa de calor mínima rejeitada para a fonte de 300\text{ K} deve ser:

\dot{Q}_{F2} = \dot{m} c \Delta T_{\text{água}} = \frac{5}{6}\text{ kg/s} \times 4000\text{ J/(kg}\cdot\text{K)} \times 10\text{ K} = \frac{100}{3}\text{ kW} \approx 33,33\text{ kW}

Análise pelas Leis da Termodinâmica:

Primeira Lei da Termodinâmica (Conservação da Energia):
Somando as taxas de energia que saem do ciclo proposto pela empresa B, obtém-se:

\dot{E}_{\text{saída}} = \dot{W}_B + \dot{Q}_{F1} + \dot{Q}_{F2} = \frac{350}{3}\text{ kW} + 45\text{ kW} + \frac{100}{3}\text{ kW} = 150\text{ kW} + 45\text{ kW} = 195\text{ kW}

Como a potência total que sai do ciclo (195\text{ kW}) supera a taxa de calor fornecida pela fonte quente (\dot{Q}_Q = 150\text{ kW}), a proposta viola a Primeira Lei da Termodinâmica (mesmo desconsiderando o aquecimento de água, \dot{W}_B + \dot{Q}_{F1} = 161,67\text{ kW} > 150\text{ kW}).
Segunda Lei da Termodinâmica (Desigualdade de Clausius):
Pela desigualdade de Clausius (\sum \frac{\dot{Q}_i}{T_i} \le 0):

\frac{\dot{Q}_Q}{T_Q} - \frac{\dot{Q}_{F1}}{T_1} - \frac{\dot{Q}_{F2}}{T_2} \le 0 \implies \frac{150}{900} - \frac{45}{450} \le \frac{\dot{Q}_{F2}}{300} \implies \frac{1}{6} - \frac{1}{10} \le \frac{\dot{Q}_{F2}}{300} \implies \dot{Q}_{F2} \ge 20\text{ kW}

Portanto, o limite máximo absoluto de potência útil para essa máquina reversível seria \dot{W}_{B,\text{máx}} = 150 - 45 - 20 = 85\text{ kW} (capaz de atender no máximo 255 residências). Caso se rejeite \dot{Q}_{F2} = \frac{100}{3}\text{ kW} \ge 20\text{ kW} para aquecer a água, a potência elétrica restante fica limitada pela Primeira Lei a no máximo \dot{W}_B = 150 - 45 - \frac{100}{3} = \frac{215}{3}\text{ kW} \approx 71,67\text{ kW} (suficiente para apenas 215 residências).

Conclui-se, portanto, que os aspectos mencionados pelo representante da empresa B não são realísticos, pois violam as leis da Termodinâmica.

Resposta final: a) 450 K; b) 90 kW; c) Não são realísticos, pois violam a Primeira Lei da Termodinâmica (a soma das taxas de energia de saída, 195 kW, excede a taxa de calor fornecida pela fonte quente, 150 kW) e excedem a potência máxima admissível pela Segunda Lei da Termodinâmica.

GPT-6.1 Terra

Respondeu · bate com a resposta de referência dos cursinhos

a) 450\ \text{K}; b) 90\ \text{kW}; c) não é realista: no máximo 215 residências com o aquecimento indicado.

levou 16 s

Como pensou

a) A potência útil da máquina A é

P_A=80\cdot\frac34=60\ \text{kW}.

Sua eficiência é \eta_A=60/150=0{,}40. Como ela corresponde a 80\% da eficiência máxima de Carnot,

0{,}40=0{,}80\left(1-\frac{T}{900}\right),

de onde

\boxed{T=450\ \text{K}}.

b) Pela conservação da energia no ciclo,

\dot Q=150-60=\boxed{90\ \text{kW}}.

c) A máquina B rejeita para a fonte a 450\ \text{K} a taxa

\frac{\dot Q}{2}=45\ \text{kW}.

Para aquecer a água, a potência térmica necessária é, no mínimo,

\dot Q_{300} =\frac{50}{60}\cdot4000\cdot10 =\frac{100}{3}\ \text{kW}.

A potência elétrica média necessária às 350 residências é

P_{\rm res}=\frac{350\cdot240}{720} =\frac{350}{3}\ \text{kW}.

Mesmo supondo aproveitamento integral do calor na água, a potência máxima disponível para as residências seria, pela primeira lei,

P_B\leq 150-45-\frac{100}{3} =\frac{215}{3}\ \text{kW},

inferior aos 350/3\ \text{kW} requeridos. Portanto, a proposta não é realista: as duas demandas não podem ser atendidas simultaneamente. Esse limite de potência corresponderia ao consumo médio de, no máximo, 215 residências.

Resposta final: a) 450\ \text{K}; b) 90\ \text{kW}; c) não é realista: no máximo 215 residências com o aquecimento indicado.