Question

A $100cm$ long cylindrical flask with inner and outer radius $r_1=2cm$ and $r_2=4cm$ respectively, is completely filled with ice at $0{}^{\circ }C$ as shown in the figure. The constant temperature outside (surrounding) the flask is ${40}^{\circ }C$. Assume heat exchange occurs only through the curved surface of the flask. (Thermal conductivity of the flask is $\ln 2\frac{W}{m^{\circ }C},L_{ice}=80cal/g$ and $1cal=4.2J$)

• A. Rate of heat flow from surrounding to the flask is $80\pi J/s$.
• B. The rate at which ice melts is $\frac{\pi }{4200}kg/s$.
• C. The rate at which ice melts is $100\pi kg/s$.
• D. Rate of heat flow from surrounding to the flask is $40\pi J/s$.

Answer 1

Answer: (A), (B)

The correct options are
A Rate of heat flow from surrounding to the flask is $80\pi J/s$.
B The rate at which ice melts is $\frac{\pi }{4200}kg/s$.

Heat will flow from higher temperature to lower temperature.
Here, $H$ is rate of flow of heat $\left(dQ/dt\right)$

$H=\frac{KAd\theta }{dr}=K\left(2\pi rl\right)\frac{d\theta }{dr}$

$\therefore H{\int }_{r_1}^{r_2}\frac{dr}{r}=2\pi Kl({\theta }_2-{\theta }_1)$
$\because l=1m$
$\Rightarrow H(\ln 4-\ln 2)=2\pi \ln 2(40-0)$
$\Rightarrow H\ln 2=80\pi \ln 2$
$\Rightarrow H=80\pi J/s$
The heat from flask will get transferred to ice and they will melt by taking latent heat from flask. Now,we know that,
$L=Q/m\Rightarrow mL=Q\Rightarrow \frac{dm}{dt}L=\frac{dQ}{dt}=H$
$\Rightarrow \frac{dm}{dt}L=80\pi$

$\Rightarrow \frac{dm}{dt}=\frac{80\pi }{L}=\frac{80\pi }{80\times 4.2\times 1000}=\frac{\pi }{4200}kg/s$