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JEE Advanced 2021 Paper 1 · Physics · Work, Energy and Power · hard

Two identical copper blocks, each of mass 2.0 kg and specific heat 400 J kg^-1K^-1 (independent of temperature), are initially at 300 K and 500 K and…

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Two identical copper blocks, each of mass 2.0 kg2.0\ \mathrm{kg} and specific heat 400 J kg−1K−1400\ \mathrm{J\,kg^{-1}K^{-1}} (independent of temperature), are initially at 300 K300\ \mathrm{K} and 500 K500\ \mathrm{K} and are isolated from the surroundings. Choose the correct statement(s).
  1. A. If they are placed in direct thermal contact, the common final temperature is 400 K400\ \mathrm{K}.
  2. B. In this irreversible process the total entropy of the blocks increases by about 51.6 J K−151.6\ \mathrm{J\,K^{-1}}.
  3. C. If a reversible engine is run between the two blocks until they reach a common temperature, the maximum work obtained is about 20.3 kJ20.3\ \mathrm{kJ}.
  4. D. In the reversible-engine case the common final temperature is also 400 K400\ \mathrm{K}.
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Answer: A, B, C

(A) Energy conservation for equal heat capacities gives Tf=(300+500)/2=400T_f=(300+500)/2=400 K: correct. (B) For direct contact, ΔS=mcln⁡Tf300+mcln⁡Tf500=800ln⁡4002300×500=800×0.06454=51.6 J K−1\Delta S=mc\ln\frac{T_f}{300}+mc\ln\frac{T_f}{500}=800\ln\frac{400^2}{300\times500}=800\times0.06454=51.6\ \mathrm{J\,K^{-1}}, which is positive as the process is irreversible: correct. (C) For a reversible engine, total entropy is constant: ln⁡Tf300+ln⁡Tf500=0\ln\frac{T_f}{300}+\ln\frac{T_f}{500}=0, so Tf=300×500=387.3T_f=\sqrt{300\times500}=387.3 K. Work equals heat drawn from the hot block minus heat dumped into the cold block: W=800[(500−387.3)−(387.3−300)]=800×25.4=20.3W=800[(500-387.3)-(387.3-300)]=800\times25.4=20.3 kJ: correct. (D) The reversible final temperature is 387.3387.3 K, lower than 400400 K because work is extracted, so (D) is incorrect.
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