Two identical systems, with heat capacity at constant volume that varies as Cv = bT3 (where b is a constant) are thermally isolated. Initially, one system is at a temperature 100 K and the other is at 200 K. The systems are then brought nto thermal contact and the combined system is allowed to reach thermal equilibrium. The final temperature (in K) of the combined system will be

Question

Two identical systems, with heat capacity at constant volume that varies as Cv = bT3 (where b is a constant) are thermally isolated. Initially, one system is at a temperature 100 K and the other is at 200 K. The systems are then brought nto thermal contact and the combined system is allowed to reach thermal equilibrium. The final temperature (in K) of the combined system will be (A) 171 (B) 141 (C) 150 (D) 180

Tardigrade Admin · Accepted Answer

Energy given by a system = Energy taken by another system

d Q_{1} = d Q_{2}

\Rightarrow = 100 \int T m c_{v} d T = - 200 \int T m c_{v} d T

\Rightarrow 100 \int T b T^{3} d T = - 200 \int T b T^{3} d T

\Rightarrow (T^{4})_{10}^{T} = - (T^{4})_{200}^{T}

\Rightarrow T^{4} - 10 0^{4} = - (T^{4} - 20 0^{4})

\Rightarrow 2 T^{4} = 17 \times 1 0^{8}

\Rightarrow T = 171 K

171

141

150

180

125

Energy given by a system = Energy taken by another system dQ1​=dQ2​ ⇒=100∫T​mcv​dT=−200∫T​mcv​dT ⇒100∫T​bT3dT=−200∫T​bT3dT ⇒(T4)10T​=−(T4)200T​ ⇒T4−1004=−(T4−2004) ⇒2T4=17×108 ⇒T=171K