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

For two bodies made up of same material, rate of loss of heat will depend only on their surface area for same temperature change.
$\therefore \frac{\left(\frac{d Q}{d t}\right)_{1}}{\left(\frac{d Q}{d t}\right)_{2}}=\frac{A_{1}}{A_{2}}=\frac{r_{1}^{2}}{r_{2}^{2}}$
Ratio of initial rates of cooling $=\frac{\left(\frac{d \theta}{d t}\right)_{1}}{\left(\frac{d \theta}{d t}\right)_{2}}$
As, $\frac{ dQ }{ dt }= mc \frac{ d \theta}{ dt }$
$\therefore \frac{ m _{1} c _{1}( d \theta / dt }{ m _{2} c _{2}(d \theta / dt }=\frac{r_{1}^{2}}{r_{2}^{2}} \ldots[$ from (i)]
since, both the spheres are made up of copper,
$ c_{1}=c_{2}=c $
$\therefore \frac{\left(\frac{d \theta}{d t}\right)_{1}}{\left(\frac{d \theta}{d t}\right)_{2}}=\frac{r_{1}^{2} \times m_{2}}{r_{2}^{2} \times m_{1}}$
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 / d t}{(d \theta / d t}=\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} $