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 $\mathrm{t}=0$, the temperature of the object is 200 K. The temperature of the object becomes 100 K at $\mathrm{t}=\mathrm{t}_{1}$ and 50 K at $\mathrm{t}=\mathrm{t}_{2}$. Assume the object and the container to be ideal black bodies. The heat capacity of the object does not depend on temperature. The ratio $\left(\mathrm{t}_{2}/\mathrm{t}_{1}\right)$ is ____.

JEE Advanced · 2021 · 03 Oct 2021 · Paper 1 · Heat & Thermodynamics

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 = 0

, the temperature of the object is 200 K. The temperature of the object becomes 100 K at

t = t_{1}

and 50 K at

t = 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

(t_{2} / t_{1})

is ____.

Answer: 9

The container is at 0 K, so the object only radiates. For an ideal black body of surface area

A

and heat capacity

C

(constant), Stefan–Boltzmann gives

C \frac{d T}{d t} = - σ A T^{4} \Rightarrow \frac{d T}{T ^{4}} = - k d t, k = \frac{σ A}{C} .

Integrating from the initial temperature

T_{0} = 200

K to a temperature

T

reached at time

t

\frac{1}{3} (\frac{1}{T ^{3}} - \frac{1}{T _{0}^{3}}) = k t \Rightarrow t \propto \frac{1}{T ^{3}} - \frac{1}{20 0 ^{3}} .

For

t_{1}

(

T = 100

K):

\frac{1}{10 0 ^{3}} - \frac{1}{20 0 ^{3}} = \frac{8 - 1}{20 0 ^{3}} = \frac{7}{20 0 ^{3}}

.

For

t_{2}

(

T = 50

K):

\frac{1}{5 0 ^{3}} - \frac{1}{20 0 ^{3}} = \frac{64 - 1}{20 0 ^{3}} = \frac{63}{20 0 ^{3}}

\frac{t _{2}}{t _{1}} = \frac{63}{7} = 9

Answer: 9

JEE Advanced · 2021 · 03 Oct 2021 · Paper 1 · Integer answer