Q29 / 160 · NTA Abhyas (JEE Main) 2020 · Physics · Thermodynamics

Two identical systems, with heat capacity at a 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 200K . The systems are then brought to thermal contact and the combined system is allowed to reach thermal equilibrium. If T04=n×108 K then what will be the value of n, where T0 is the final temperature in kelvin?

Correct answer: 8.5

Solution

Energy given by a system = Energy taken by another system

dQ1=dQ2

⇒  ∫100TmcvdT=-∫200TmcvdT

⇒ ∫100TbT3dT=-∫200TbT3dT⇒T410T=-T4200T

⇒   T4-1004=-T4-2004

⇒   2T4=17×108⇒T=171 K

⇒ T4=8.5×108 K

Hence, the correct answer is 8.5.

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