Two blocks of the same metal having same mass | vedclass

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(A) When two blocks of the same mass and same heat capacity $C_p$ at temperatures $T_1$ and $T_2$ are brought into contact,the final equilibrium tempe

Vedclass · Accepted Answer

$2C_p \ln \left[ \frac{T_1 + T_2}{2\sqrt{T_1T_2}} \right]$

Vedclass · Answer

(A) When two blocks of the same mass and same heat capacity $C_p$ at temperatures $T_1$ and $T_2$ are brought into contact,the final equilibrium temperature $T_f$ is given by $T_f = \frac{T_1 + T_2}{2}$.The change in entropy for the first block is $\Delta S_1 = C_p \ln \left( \frac{T_f}{T_1} \right) = C_p \ln \left( \frac{T_1 + T_2}{2T_1} \right)$.The change in entropy for the second block is $\Delta S_2 = C_p \ln \left( \frac{T_f}{T_2} \right) = C_p \ln \left( \frac{T_1 + T_2}{2T_2} \right)$.The total entropy change is $\Delta S_{total} = \Delta S_1 + \Delta S_2 = C_p \ln \left( \frac{(T_1 + T_2)^2}{4T_1T_2} \right) = 2C_p \ln \left( \frac{T_1 + T_2}{2\sqrt{T_1T_2}} \right)$.

Two blocks of the same metal having same mass and at temperatures $T_1$ 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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