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PhysicsInteger+4Question 19 of 57

JEE Advanced 2021 · Paper 1, Physics question 19

A small object is placed at the center of a large evacuated hollow spherical container. Assume that the container is maintained at 0 K. At time t=0t = 0, the temperature of the object is 200 K. The temperature of the object becomes 100 K at t=t1t = t_1 and 50 K at t=t2t = t_2. Assume the object and the container to be ideal black bodies. The heat capacity of the object does not depend on temperature.

The ratio (t2/t1)(t_2/t_1) is ____.

Answer and worked solution

Official answer: 9

Cooling law. The object radiates into 0 K surroundings and has constant heat capacity CC:
CdTdt=−σAT4  ⟹  t=C3σA(1T3−1T03).C\frac{dT}{dt}=-\sigma AT^4\implies t=\frac C{3\sigma A}\left(\frac1{T^3}-\frac1{T_0^3}\right).

Ratio (with T0=200T_0=200 K):
t2t1=1503−1200311003−12003=64−18−1=9.\frac{t_2}{t_1}=\frac{\frac1{50^3}-\frac1{200^3}}{\frac1{100^3}-\frac1{200^3}}=\frac{64-1}{8-1}=9.

Topics:Radiation and Stefan's law
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