Question

Two spheres of same radius $R$ have their densities in the ratio $8:1$ and the ratio of their specific heats are $1:4$. If by radiation their rates of fall of temperature are same, then find the ratio of their rates of losing heat.

Answer 1

$(\frac{dQ}{dt}{)}_1=m_1{s}_1(\frac{dT}{dt}{)}_1$
$(\frac{dQ}{dt}{)}_2=m_2{s}_2(\frac{dT}{dt}{)}_2$
Given,
$(\frac{dT}{dt}{)}_1=(\frac{dT}{dt}{)}_2$
$\therefore \frac{(\frac{dQ}{dt}{)}_1}{(\frac{dQ}{dt}{)}_2}=\frac{m_1{s}_1}{m_2{s}_2}=\frac{m_1}{m_2}\times \frac{s_1}{s_2}=\frac{d_1}{d_2}\times \frac{s_1}{s_2}=\frac{8}{1}\times \frac{1}{4}=\frac{2}{1}$