Question

An icebox made of $ 1.5\text{ cm} $ thick Styrofoam has dimensions $ 60\text{ cm }\times \text{ 60 cm }\times \text{30 cm} $ . It contains ice at $ 0{}^\circ \text{C} $ and is kept in a room at $ 40{}^\circ \text{C} $ . Find the rate at which ice is melting. Latent heat of fusion of ice $ =3.36\times {{10}^{6}}J/kg $ $ $
And thermal conductivity of Styrofoam is $ =0.04W/m{{-}^{\circ }}C $ .

Answer 1

Hint First calculate the rate of heat flow into the box, and then the rate at which ice melts can be calculated by dividing rate of heat flow by latent heat of fusion of ice.
Use the following formulas while deriving the answer.
 $ \text{Total surface area of a cuboid = }2\left( lb+bh+hl \right) $
(Where, l, b, h are length, breadth and height respectively).
Rate of heat flow is given by,
Where, $ x $ : thickness, $ \dfrac{\Delta Q}{\Delta t}=\dfrac{KA({{\theta }_{1}}-{{\theta }_{2}})}{x} $
K: Thermal conductivity of material
A: surface area
 $ {{\theta }_{1}}-{{\theta }_{2}} $ : Temperature difference
Rate at which ice changes to liquid $ =\dfrac{1}{L}\left( \dfrac{\Delta Q}{\Delta t} \right) $
Where, L is the latent heat of fusion.

Complete step by step solution:
 $ \begin{align}
  & \text{Total surface area of walls of icebox = 2}\left( lb+bh+hl \right) \\
 & =2\left( 60\times 60+60\times 30+30\times 60 \right)\times {{10}^{-4}}{{m}^{2}} \\
 & =1.44{{m}^{2}}
\end{align} $
Given, thickness of the walls of the container $ =1.5\times {{10}^{-2}}m=0.015m $
Rate of heat flow into the box , $ \dfrac{\Delta Q}{\Delta t}=\dfrac{KA({{\theta }_{1}}-{{\theta }_{2}})}{x} $
$ \begin{align}
  & =\dfrac{0.04\times 1.44\times (40-0)}{0.015} \\
 & =154W
\end{align} $
$ \begin{align}
& \dfrac{0.04\times 1.44\times (40-0)}{0.015}=154W \\
& \\
\end{align} $
Where $ {{\theta }_{1}} $ is the temperature of ice $ \left( 0{}^\circ \text{C} \right) $
And $ {{\theta }_{2}} $ is the room temperature $ \left( 40{}^\circ \text{C} \right) $
Therefore, rate at which ice melts is $ =\dfrac{1}{L}\left( \dfrac{\Delta Q}{\Delta t} \right)=154/\text{Latent heat of fusion} $
$ \begin{align}
& =\dfrac{154}{3.36\times {{10}^{5}}\times {{10}^{-3}}} \\
& =0.46g/s \\
\end{align} $

Note:
The enthalpy of fusion of a substance, also known as (latent) heat of fusion is the change in its enthalpy resulting from providing energy, typically heat, to a specific quantity of the substance to change its state from a solid to a liquid, at constant pressure.
Do calculations involved carefully, especially while converting one unit to another. Learn the formulas involved in the question, to use them directly in problems. Surface area of various geometric figures like cube, cuboid, cylinder and sphere should be known to students.
 $ \dfrac{154}{3.36\times {{10}^{5}}\times {{10}^{-3}}}=0.46g/s $