When two identical metal blocks of mass m at temperatures T_1 and T_2 are brought into thermal contact in an isolated system, they reach an equilibrium temperature T_f. What is the total entropy change (Delta S) for this process?

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EEJ MAIN Chemistry QUESTION #7137

Question 1

When two identical metal blocks of mass $m$ at temperatures $T_1$ and $T_2$ are brought into thermal contact in an isolated system, they reach an equilibrium temperature $T_f$. What is the total entropy change ($\Delta S$) for this process?

$2Cp \ln(\frac{T_1+T_2}{2\sqrt{T_1T_2}})$✔️
$2Cp \ln(\frac{T_1+T_2}{2T_1T_2})$
$Cp \ln(\frac{(T_1+T_2)^2}{4T_1T_2})$
$2Cp \ln(\frac{T_1+T_2}{\sqrt{T_1T_2}})$

Correct Answer Explanation

The final temperature for two identical blocks is the arithmetic mean, $T_f = \frac{T_1+T_2}{2}$. The entropy change for the first block is $\Delta S_1 = Cp \ln(T_f/T_1)$ and for the second is $\Delta S_2 = Cp \ln(T_f/T_2)$. The total change $\Delta S = Cp \ln(\frac{T_f^2}{T_1T_2}) = 2Cp \ln(\frac{T_f}{\sqrt{T_1T_2}})$. Substituting $T_f$, we get $2Cp \ln(\frac{T_1+T_2}{2\sqrt{T_1T_2}})$.

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