Question

A sphere and a cube, both made of copper, have equal volumes and are blackened. These are allowed to cool at same temperature and in same atmosphere. The ratio of the rates of loss of heat will be :

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

Answer 1

Answer: (B)

$(\pi /6{)}^{1/3}:1$

Given: sphere (of radius $r$) and cube (of side $a$) of same material, volume and cooled at same temperature in the same atmosphere

Rate of heat loss is

$R=\frac{dQ}{dt}=hA(\theta -{\theta }_0)$

we have $volum{e}_{sphere}=volum{e}_{cube}\frac{4}{3}\pi r^3=a^3\frac{r}{a}=(\frac{3}{4\pi }{)}^{\frac{1}{3}}r=a(\frac{3}{4\pi }{)}^{\frac{1}{3}}$

now $R_{sphere}:R_{cube}=h{A}_{sphere}(\theta -{\theta }_0):h{A}_{cube}(\theta -{\theta }_0)=A_{sphere}:A_{cube}=4\pi (r{)}^2:6(a{)}^2=4\pi \left(a\right(\frac{3}{4\pi }{)}^{\frac{1}{3}}{)}^2:6(a{)}^2=(\frac{\pi }{6}{)}^{\frac{1}{3}}:1$