Question

A sphere, a cube, and a thin circular disc, all made of the same material, have the same mass. Their initial temperature is $ 3 \times {10^3}^\circ C $ . The body which cools rapidly is
(A) Sphere
(B) Cube
(C) Both sphere and cube
(D) Circular disc

Answer 1

Hint : The rate of heat loss is directly proportional to the exposed area of the hot body. Since they are made up of the same material (density) and mass, then they can be considered of the same volume.

Formula used: In this solution we will be using the following formulae;
 $ Q = mc\Delta T $ where $ Q $ is the total heat loss (or gain), $ m $ is the mass of the body, $ c $ is the specific heat capacity of the body, and $ \Delta T $ is the change in temperature for the particular heat lost.
 $ \dot Q = \varepsilon \sigma A\left( {{T_1}^4 - {T_2}^4} \right) $ where $ \dot Q $ is the instantaneous rate of heat loss through radiation, $ \varepsilon $ is the emissivity of the body, $ \sigma $ is the Stefan-Boltzmann constant, and $ A $ is the surface area exposed, and $ {T_1} $ and $ {T_2} $ are the temperature of the body and surrounding respectively.
 $ \dot Q = hA\Delta T $ where $ \dot Q $ is the instantaneous rate of heat loss due to convection $ h $ is convection coefficient.

Complete step by step answer
In the question, three bodies of different shapes but the same material and mass are losing heat.
Now since they are all of the same material, then they have the same density. They also have the same mass, hence they have the same volume.
Now, generally, heat must be lost due convection and radiation. For convectional heat loss, we have
 $ \dot Q = hA\Delta T $ where $ h $ is convectional coefficient and $ \Delta T $ is the temperature difference in body and environment.
For radiation heat loss,
 $ \dfrac{Q}{t} = \varepsilon \sigma A\left( {{T_1}^4 - {T_2}^4} \right) $ where $ \dot Q $ is the instantaneous rate of heat loss through radiation, $ \varepsilon $ is the emissivity of the body, $ \sigma $ is the Stefan-Boltzmann constant, and $ A $ is the surface area exposed, and $ {T_1} $ and $ {T_2} $ are the temperature of the body and surrounding respectively.
Since the materials are the same, hence all the constants do not create a difference in the shapes, hence we can say in general,
 $ \dot Q = kA $ where $ k $ is a constant encompassing all the constants.
Now, heat loss in general can be given as
 $ Q = mc\Delta T $ where $ Q $ is the total heat loss (or gain), $ m $ is the mass of the body, $ c $ is the specific heat capacity of the body, and $ \Delta T $ is the change in temperature for the particular heat lost.
Hence, by differentiating with time, we have
 $ \dot Q = mc\dfrac{{dT}}{{dt}} $
By substituting $ \dot Q = kA $ into this we have
 $ \dfrac{{dT}}{{dt}} = \dfrac{{kA}}{{mc}} $ which defines the rate of cooling, hence the larger surface area cools faster.
In our question, the larger surface area is the circular disc (for the same volume)
Hence, the one which cools rapidly is the disc.
Thus, the correct option is D.

Note
For clarity, the thin circular disc has a larger surface area because it means that for the same volume, it must have most of its body exposed since it is thin as compared to a solid cube or sphere where most of its body is under the outer layer.