Question:

Two blocks of the same metal having same mass and at temperature \(T_1\) and \(T_2\), respectively, are brought in contact with each other and allowed to attain thermal equilibrium at constant pressure. The change in entropy, \(\Delta S\), for this process is :

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Final temperature of two equal blocks is \(T_f = (T_1+T_2)/2\). Add \(\Delta S = C_p\ln(T_f/T_i)\) for each block.
Updated On: Nov 5, 2026
  • \(2C_p \ln\left[\frac{(T_1+T_2)^{1/2}}{T_1T_2}\right]\)
  • \(2C_p \ln\left[\frac{(T_1+T_2)}{2T_1T_2}\right]\)
  • \(C_p \ln\left[\frac{(T_1+T_2)^2}{4T_1T_2}\right]\)
  • \(2C_p \ln\left[\frac{(T_1+T_2)}{4T_1T_2}\right]\)
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The Correct Option is C

Solution and Explanation

Equal masses of the same metal have equal heat capacity \(C_p\). Final temperature: \(T_f = \frac{T_1+T_2}{2}\).
Block 1: \(\Delta S_1 = C_p\ln\frac{T_f}{T_1}\); block 2: \(\Delta S_2 = C_p\ln\frac{T_f}{T_2}\).
\(\Delta S = C_p\ln\frac{T_f^2}{T_1T_2} = C_p\ln\left[\frac{(T_1+T_2)^2}{4T_1T_2}\right]\).
(This is positive because \((T_1+T_2)^2 \ge 4T_1T_2\).)
Answer: (3) \(C_p \ln\left[\frac{(T_1+T_2)^2}{4T_1T_2}\right]\)
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