Question

A sphere and a cube, both of copper have equal volumes and are black. They are allowed to cool at same temperature and in same atmosphere. The ratio of their rate of loss of heat will be

• A. 1:1
• B. $(\frac{\pi}{6})^{\frac{2}{3}}$
• C. $(\frac{\pi}{6})^{\frac{1}{3}}$
• D. $\frac{4\pi}{3}:1$

Hint:

Physics Tip: For a given volume, a sphere has the minimum surface area. Therefore, it will always have a lower rate of heat loss compared to any other shape of the same volume.

Answer 1

The Correct Option is C

Concept: Physics (Thermal Properties) - Stefan's Law and Rate of Radiation.

Step 1: Relate the volumes of the sphere and cube. Given that the volumes are equal ($V_s = V_c$: $$ a^3 = \frac{4}{3}\pi r^3 \text{ } $$ $$ a = \left( \frac{4}{3}\pi r^3 \right)^{\frac{1}{3}} \text{ } $$

Step 2: Establish the relationship for rate of heat loss. The rate of loss of heat ($Q$) by radiation is directly proportional to the surface area ($A$) of the object. $$ Q \propto A \text{} $$

Step 3: Calculate the ratio of the surface areas. $$ \frac{Q_{sphere}}{Q_{cube}} = \frac{A_{sphere}}{A_{cube}} = \frac{4\pi r^2}{6a^2} $$

Step 4: Substitute for 'a' and simplify. $$ \frac{Q_{sphere}}{Q_{cube}} = \frac{4\pi r^2}{6 \left( \frac{4}{3}\pi r^3 \right)^{\frac{2}{3}}} \text{ } $$ After algebraic simplification, the ratio becomes: $$ \frac{Q_{sphere}}{Q_{cube}} = \left( \frac{\pi}{6} \right)^{\frac{1}{3}} \text{} $$ $$ \therefore \text{The ratio of their rate of loss of heat is } \left( \frac{\pi}{6} \right)^{\frac{1}{3}}. $$