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Two identical finite bodies of constant heat capacity at temperatures T1 and T2 are available to do work in a heat engine. The final temperature Tf reached by the bodies on delivery of maximum work is

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  3. Tf = T1­ = T2

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Detailed Solution

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Explanation:

Assume T1 > T2

Q1 = cp (T1 - TF)

Q2 = cp (TF – T2)

∴ W = Q1 – Q2 = cp (T1 + T2 – 2TF)

For ‘w’ to be max, TF to be minimum

As,

Now,

For TF to be minimum

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