Question

Consider two identical, finite, isolated systems of constant heat capacity \( C \) at temperatures \( T_1 \) and \( T_2 \) (\( T_1>T_2 \)). An engine works between them until their temperatures become equal. Taking into account that the work performed by the engine will be maximum (\( W_{\text{max}} \)) if the process is reversible (equivalently, the entropy change of the entire system is zero), the value of \( W_{\text{max}} \) is:

• A. \( C(T_1 - T_2) \)
• B. \( C(T_1 - T_2)/2 \)
• C. \( C(T_1 + T_2 - \sqrt{T_1 T_2}) \)
• D. \( C(\sqrt{T_1} - \sqrt{T_2})^2 \)

Hint:

The maximum work done in an entropy-conserving process can be derived using the temperatures of the two systems and their heat capacities.

Answer 1

The Correct Option is C

Step 1: Understanding the maximum work.
The maximum work done in a thermodynamic process where the entropy change of the system is zero occurs when the temperatures become equal. The work done is given by \( W_{\text{max}} = C(T_1 + T_2 - \sqrt{T_1 T_2}) \), based on the first and second laws of thermodynamics.
Step 2: Conclusion.
Therefore, the correct answer is option (C).