Question

A metallic cylindrical vessel whose inner and outer radii are $r_1$ and $r_2$ is filled with ice at $0^{\circ }\text{C}.$ The mass of the ice in the cylinder is $m$. Circular surfaces of the cylinder are sealed completely with adiabatic walls. The vessel is kept in air and the temperature of the air is ${50}^{\circ }\text{C}.$ How long will it take for the ice to melt completely?
[Thermal conductivity of the cylinder is $K$ and its length is $l$. Latent heat of fusion of ice is $L$].

• A. $\frac{mL}{50\pi Kl}\ln \left(\frac{r_1}{r_2}\right)$
• B. $\frac{mL}{200\pi Kl}\ln \left(\frac{r_1}{r_2}\right)$
• C. $\frac{mL\ln \left(\frac{r_2}{r_1}\right)}{100\pi Kl}$
• D. $\frac{mL\ln \left(\frac{r_2}{r_1}\right)}{50\pi Kl}$

Answer 1

The correct option is C $\frac{mL\ln \left(\frac{r_2}{r_1}\right)}{100\pi Kl}$

Consider a thin cylindrical shell of thickness $dr$ and radius $r$ from the centre.



We know that, rate of heat flow $H=\frac{dQ}{dt}=KA\frac{dT}{dr}=K\left(2\pi rl\right)\frac{dT}{dr}$
$\Rightarrow \frac{H}{2\pi Kl}{\int }_{r_1}^{r_2}\frac{dr}{r}={\int }_0^{50}dT$
$\Rightarrow \frac{H}{2\pi Kl}\ln \left(\frac{r_2}{r_1}\right)=50$
$\therefore H=\frac{100\pi Kl}{\ln \left(r_2/r_1\right)}$ is the rate of heat transfer across the cylinder.
$\because Q=mL$, we can say that $Ht=mL$ where $L$ is the latent heat of fusion of ice and $t$ is the time taken for ice to melt completely.
$\Rightarrow t=\frac{mL}{H}=\frac{mL\ln \left(r_2/r_1\right)}{100\pi Kl}$
Thus, option (c) is the correct answer.