Question

A solid copper cube and sphere, both of same mass & emissivity are heated to same initial temperature and kept under identical conditions. What is the ratio of their initial rate of fall of temperature?

Answer 1

If a is the length of cube, surface area: $S_{cube}=6a^2$
If r is the radius of sphere surface area: $S_{sphere}=4\pi r^2$
Both the cube and sphere have the same volume as given.
$\therefore a^3=\frac{4}{3}\pi r^3$
$\Rightarrow \frac{a}{r}=(\frac{4\pi }{3}{)}^{\frac{1}{3}}$
$\frac{{Area}_{cube}}{{Area}_{sphere}}=\frac{6a^2}{4\pi r^2}=\frac{6}{4\pi }(\frac{a}{r}{)}^2=\frac{6}{4\pi }\ast (\frac{4\pi }{3}{)}^{\frac{2}{3}}=(\frac{6}{\pi }{)}^{\frac{1}{3}}$
Rate of fall in temp $=$ Rate of heat radiated(Power)
$\therefore$ Rate of fall in temp $=(\frac{6}{\pi }{)}^{\frac{1}{3}}$