Question

Two uniform copper solid spheres have radii $15cm$ and $20cm$ respectively and both have a temperature of $70^{\circ }C$ . If the surrounding temperature is $45^{\circ }C$ , If the ratio of rates of initial cooling of both the spheres $\frac{1.33}{x}$ then find $x$ .

Answer 1

Answer: 1

For two bodies made up of same material, rate of loss of heat will depend only on their surface area for same temperature change.
$\therefore \frac{{\left(\frac{dQ}{dt}\right)}_1}{{\left(\frac{dQ}{dt}\right)}_2}=\frac{A_1}{A_2}=\frac{r_1^2}{r_2^2}$
Ratio of initial rates of cooling $=\frac{{\left(\frac{d\theta }{dt}\right)}_1}{{\left(\frac{d\theta }{dt}\right)}_2}$
As, $\frac{dQ}{dt}=mc\frac{d\theta }{dt}$
$\therefore \frac{m_1{c}_1(d\theta /dt}{m_2{c}_2(d\theta /dt}=\frac{r_1^2}{r_2^2}\dots [$ from (i)]
since, both the spheres are made up of copper,
$c_1=c_2=c$
$\therefore \frac{{\left(\frac{d\theta }{dt}\right)}_1}{{\left(\frac{d\theta }{dt}\right)}_2}=\frac{r_1^2\times m_2}{r_2^2\times m_1}$
As, spheres have same densities,
$\begin{array}{l}\frac{m_1}{m_2}=\frac{V_1}{V_2}=\frac{r_1^3}{r_2^3}\\ \therefore \frac{(d\theta /dt}{(d\theta /dt}=\frac{r_1^2}{r_2^2}\times \frac{r_2^3}{r_1^3}=\frac{r_2}{r_1}=\frac{20}{15}=\frac{4}{3}=1.33\end{array}$