JEE Advance - Physics (2021 - Paper 1 Online - No. 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 = 0, the temperature of the object is 200 K. The temperature of the object becomes 100 K at t = t1 and 50 K at t = t2. 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) is ____________.
Answer
9
Explanation
Ts = 0 K
Ti = 200 K, e = 1
$$ - Ms{{dT} \over {dt}} = {{dQ} \over {dt}} = \sigma eA{T^4}$$
$$ - {{dT} \over {dt}} = {{\sigma A{T^4}} \over {Ms}}$$
$${{\sigma A} \over {Ms}}\int\limits_{{t_i}}^{{t_f}} {dt = - \int\limits_{{T_i}}^{{T_f}} {{{dT} \over {{T^4}}}} } $$
$${{\sigma A} \over {Ms}}({t_f} - {t_i}) = {1 \over 3}\left( {{1 \over {T_f^3}} - {1 \over {T_i^3}}} \right)$$
$${{{{\sigma A} \over {Ms}}({t_1} - 0) = {1 \over 3}\left( {{1 \over {{{(100)}^3}}} - {1 \over {{{(200)}^3}}}} \right)} \over {{{\sigma A} \over {Ms}}({t_2} - 0) = {1 \over 3}\left( {{1 \over {{{(50)}^3}}} - {1 \over {{{(200)}^3}}}} \right)}}$$
$${{{t_1}} \over {{t_2}}} = {{{{{{(200)}^3} - {{(100)}^3}} \over {{{(100)}^3}{{(200)}^3}}}} \over {{{{{(200)}^3} - {{(50)}^3}} \over {{{(50)}^3}{{(200)}^3}}}}}$$
$$\therefore$$ $${{{t_2}} \over {{t_1}}} = {9 \over 1}$$
Ti = 200 K, e = 1
$$ - Ms{{dT} \over {dt}} = {{dQ} \over {dt}} = \sigma eA{T^4}$$
$$ - {{dT} \over {dt}} = {{\sigma A{T^4}} \over {Ms}}$$
$${{\sigma A} \over {Ms}}\int\limits_{{t_i}}^{{t_f}} {dt = - \int\limits_{{T_i}}^{{T_f}} {{{dT} \over {{T^4}}}} } $$
$${{\sigma A} \over {Ms}}({t_f} - {t_i}) = {1 \over 3}\left( {{1 \over {T_f^3}} - {1 \over {T_i^3}}} \right)$$
$${{{{\sigma A} \over {Ms}}({t_1} - 0) = {1 \over 3}\left( {{1 \over {{{(100)}^3}}} - {1 \over {{{(200)}^3}}}} \right)} \over {{{\sigma A} \over {Ms}}({t_2} - 0) = {1 \over 3}\left( {{1 \over {{{(50)}^3}}} - {1 \over {{{(200)}^3}}}} \right)}}$$
$${{{t_1}} \over {{t_2}}} = {{{{{{(200)}^3} - {{(100)}^3}} \over {{{(100)}^3}{{(200)}^3}}}} \over {{{{{(200)}^3} - {{(50)}^3}} \over {{{(50)}^3}{{(200)}^3}}}}}$$
$$\therefore$$ $${{{t_2}} \over {{t_1}}} = {9 \over 1}$$
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