An iron rod of heat capacity C is heated to t | vedclass

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(D) Let the equilibrium temperature be $T$. Since the system is adiabatic,the total heat lost by the iron rod equals the total heat gained by the air

Vedclass · Accepted Answer

$\left( \frac{10C + 7R}{3C + 7R} \right) T_0$

Vedclass · Answer

(D) Let the equilibrium temperature be $T$. Since the system is adiabatic,the total heat lost by the iron rod equals the total heat gained by the air and the vessel-piston assembly.Heat lost by the iron rod = $C(8T_0 - T)$.Heat gained by the vessel and piston = $2C(T - T_0)$.Heat gained by $2$ moles of diatomic gas (at constant pressure,as the piston is movable and atmospheric pressure $P_0$ is constant) = $n C_p (T - T_0) = 2 	imes \frac{7R}{2} (T - T_0) = 7R(T - T_0)$.Equating heat lost and gained: $C(8T_0 - T) = 2C(T - T_0) + 7R(T - T_0)$.$8CT_0 - CT = 2CT - 2CT_0 + 7RT - 7RT_0$.$8CT_0 + 2CT_0 + 7RT_0 = CT + 2CT + 7RT$.$T_0(10C + 7R) = T(3C + 7R)$.$T = \left( \frac{10C + 7R}{3C + 7R} ight) T_0$.

List-$I$	List-$II$
$(A)$ Heat capacity of body	$(I)$ $J kg^{-1}$
$(B)$ Specific heat capacity of body	$(II)$ $J K^{-1}$
$(C)$ Latent heat	$(III)$ $J kg^{-1} K^{-1}$
$(D)$ Thermal conductivity	$(IV)$ $J m^{-1} K^{-1} s^{-1}$

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