Question

A solid copper sphere (density$\rho$ and specific capacity C) of radius r at an initial temperature of $200{\text{ }}K$ is suspended inside a chamber whose walls are at almost $0{\text{ }}K$ . The time required for the temperature of the sphere to drop to $100{\text{ }}K$ is
A. $\dfrac{{7r\rho C}}{{72 \times {{10}^6}\sigma }}$
B. $\dfrac{{r\rho C}}{{6\sigma }}$
C. $\dfrac{{r\rho C}}{\sigma }$
D. None of the above.

Answer 1

:In the given situation, the heat loss by the sphere is equal to the heat loss by the radiation. Equate both the equations and integrate within the suitable limits. The heat loss interests in the ventilation of hot processes which is known as convectional heat loss. The total heat loss of the object involves losses occurring by radiation, convection, and conduction.

Complete step by step answer:
Let $\dfrac{{dT}}{{dt}}$ be the rate of change of temperature. Let m be the mass of the sphere and A be its surface area.
Heat loss from the sphere = $mC\dfrac{{dT}}{{dt}}$ , where C is the specific heat capacity of the sphere.
Volume of the sphere of radius $r = \dfrac{4}{3}\pi {r^3}$
It is given that density of the sphere = $\rho$
$m = \left( {\dfrac{4}{3}\pi {r^3}} \right) \times \rho$
Therefore, heat loss from the sphere = $\left( {\dfrac{4}{3}\pi {r^3}} \right) \times \rho C\dfrac{{dT}}{{dt}}$

Heat loss by the sphere due to radiation =$\sigma A{T^4}$ ,
where A is the surface area of the sphere and is the Stefan’s constant.
Here, A = $4\pi {r^2}$
Now, Heat loss from the sphere = Heat loss due to radiation
$\left( {\dfrac{4}{3}\pi {r^3}} \right) \times \rho C\dfrac{{dT}}{{dt}} = \sigma \left( {4\pi {r^2}} \right){T^4}$
$\Rightarrow\dfrac{r}{3} \times \rho C\dfrac{{dT}}{{dt}} = \sigma {T^4} \Rightarrow\dfrac{{r\rho C}}{{3\sigma {T^4}}}dT = dt$

Let t be time during which the temperature of the sphere falls from 200 K to 100 K. Then integrating both sides within the suitable limits, we get
$\dfrac{{r\rho C}}{{3\sigma }}\int\limits_{100}^{200} {\dfrac{{dT}}{{{T^4}}}} = \int\limits_0^t {dt}$
$\Rightarrow\dfrac{{r\rho C}}{{3\sigma }}\int\limits_{100}^{200} {\dfrac{{dT}}{{{T^4}}}} = \int\limits_0^t {dt}$
$\Rightarrow\dfrac{{r\rho C}}{{9\sigma \times {{10}^6}}}\left( {1 - \dfrac{1}{8}} \right) = t$
$\therefore t = \dfrac{{7r\rho C}}{{72\sigma \times {{10}^6}}}$

Hence, the correct option is A.

Note: The initial temperature of the body was 200 K and the final temperature was 100 K. As the temperature dropped so the lower limit will be 100 and the upper limit of the integral will be 200. If temperature rises then the limits will be from lower temperature to the raised temperature. If the limits are not taken correctly, the value of the integral will come out incorrectly. So choose limits accordingly.