Question

The only possibility of heat flow in a thermos flask is through its cork which is $75c{m^2}$ in the area and $5cm$ thick. Its thermal conductivity is $0.0075\dfrac{{cal}}{{cm - s - {C^ \circ }}}$ . The outside temperature is ${40^ \circ }C$ and the latent heat of ice is $80\dfrac{{cal}}{g}$ . Time is taken by $500g$ of ice at ${0^ \circ }C$ in the flask to melt into the water at ${0^ \circ }C$ is
(A) $2.47h$
(B) $4.27h$
(C) $7.42h$
(D) $4.72h$

Answer 1

First find the heat flow through the thermos flask. This heat flow will be equal to the latent heat of ice. By doing that we will find the Time taken by $500g$ of ice at ${0^ \circ }C$ in the flask to melt into the water at ${0^ \circ }C$ .

Complete answer:
A thermos flask also known as vacuum flask is an insulating storage vessel that extends the duration that the contents of the flask remain hotter or cooler than the surroundings.
Given the cork's surface area $A = 75c{m^2}$
The cork's thickness $\Delta x = 5cm$
It has a high heat conductivity $K = 0.0075\dfrac{{cal}}{{cm - s - {C^ \circ }}}$
The temperature in the room ${T_a} = {40^ \circ }C$
The ice's temperature ${T_{ice}} = {0^ \circ }C$
The ice's mass $m = 500g$
The temperature of the ice's latent heat $L = 80\dfrac{{cal}}{g}$
Let the time it takes for the ice ${0^ \circ }C$ to melt into water ${0^ \circ }C$ be $t$
For a period of time, heat was transferred into the bottles equal to the ice's latent heat
$\Rightarrow KA\dfrac{{{T_a} - {T_{ice}}}}{{\Delta x}} \times t = mL$
$\Rightarrow 75 \times {10^{ - 4}} \times 75 \times \dfrac{{40 - 0}}{5} \times t = 500 \times 80$
$\Rightarrow t = 8888\sec$
$\Rightarrow t = 2.47hr$
Therefore, Time is taken by $500g$ of ice at ${0^ \circ }C$ in the flask to melt into the water at ${0^ \circ }C$ is $t = 2.47hr$
Hence, option A) $2.47hr$ is the correct option.

Note:
The heat or energy absorbed or released during a phase shift of a substance is known as latent heat. It could be a transition from a gas to a liquid, or from a liquid to a solid, or from ice to liquid, and vice versa. In the above question there won’t be any unit conversion as area, thickness and thermal conductivity are all in $cm$ .