Which cools faster?

Level 2

A solid cube and a solid sphere of the same material have equal surface area. Both are at the same temperature $70^{\circ} \ C$ . Then which of the following statements is true?

Assumptions :

Heat loss is solely due to radiation.
Temperature of surroundings remain relatively constant
Temperature difference between the bodies and the surroundings is relatively small.

The cube cools down faster than the sphere The one with greater mass cools down faster Both cool down at the same rate The sphere cools down faster than the cube

1 solution

Vignesh Rao
Dec 25, 2017

From Stefan–Boltzmann law, we get

$\dfrac{\text{d}Q}{\text{d}t} = \varepsilon A \sigma (T^4 - T_s^4)$

where $T_s$ is temperature of surroundings and $T$ is temperature of body.

Since temperature difference between surroundings and the body is small, we can write

$T = T_s + \Delta T$

Using the above equation we get:

$\dfrac{\text{d}Q}{\text{d}t} =4 \varepsilon A \sigma T_s^3 (T-T_s)$

$-\dfrac{\text{d}T}{\text{d}t}= \dfrac{4 \varepsilon A \sigma T_s^3 (T-T_s)}{ms}$

where $m$ is mass and $s$ is specific heat capacity.

$-\dfrac{\text{d}T}{\text{d}t}= \dfrac{4 \varepsilon A \sigma T_s^3 (T-T_s)}{\rho V s}$

where $\rho$ is density of material and $V$ is volume of the body.

$\Rightarrow - \dfrac{\text{d}T}{\text{d}t} \propto \dfrac{1}{V}$

For same surface area of cube and sphere, $V_{\text{sphere}}>V_{\text{cube}}$

Hence, the cube will cool faster than the sphere.

Which cools faster?

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