JEE PYQ: Thermodynamics Question 35
Question 35 - 2019 (11 Jan Shift 1)
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:
(1) $C_p\ln\left[\dfrac{(T_1 + T_2)^2}{4T_1T_2}\right]$
(2) $2C_p\ln\left[\dfrac{(T_1 + T_2)^{\frac{1}{2}}}{T_1T_2}\right]$
(3) $2C_p\ln\left[\dfrac{T_1 + T_2}{4T_1T_2}\right]$
(4) $2C_p\ln\left[\dfrac{T_1 + T_2}{2T_1T_2}\right]$
Show Answer
Answer: (1)
Solution
Final temperature $= \dfrac{T_1 + T_2}{2}$, let $T_2 > T_1$. $dS = \dfrac{dq}{T} = \dfrac{C_p dT}{T}$. $\Delta S_{\text{total}} = C_p\ln\dfrac{T_f}{T_1} + C_p\ln\dfrac{T_f}{T_2} = C_p\ln\left[\dfrac{(T_1 + T_2)^2}{4T_1T_2}\right]$.
BETA
