Question

A layer of ice $0.15\text{m}$ thick has formed on the surface of a deep pond. If the temperature of upper surface of ice is constant and equal to that of the air which is $-{12}^{\circ }\text{C}$ , the time taken to increase the thickness of ice layer by $0.2\text{mm}$ is?
$[$ Given : latent heat of ice = $80\text{cal /g}$
density of ice = $0.91\frac{g}{c{m}^3}$
thermal conductivity of ice = $0.5\frac{cal}{smK}]$

• A. $182.68\text{s}$
• B. $364.24\text{s}$
• C. $546.35\text{s}$
• D. $300.55\text{s}$

Answer 1

The correct option is B $364.24\text{s}$

Let A be the area of the upper face of the ice layer.
increases in thickness = $0.2\text{mm}=0.02\text{cm}$
So mass of ice to be frozen is
$m=\rho \times$volume
$m=(0.91\times 0.02\times A)\text{g}$
Now latent heat of ice is $L=80\text{cal/g}$
$\Rightarrow \Delta Q=mL$
$\Delta Q=A\times 0.02\times 0.91\times 80\text{cal}$
The average thickness through which heat has to pass is
$\Delta x=\frac{15+(15+0.02)}{2}=15.01\text{cm}$
The thermal conductivity is given as
$k=0.5\frac{cal}{smK}$
$k=0.5\times {10}^{-2}\frac{cal}{scmK}$
Now we use $\frac{\Delta Q}{\Delta t}=KA\frac{\Delta T}{\Delta x}$
$\Rightarrow \frac{A\times 0.02\times 0.91\times 80}{\Delta t}=0.5\times {10}^{-2}A\times \frac{12}{15.01}$
$\Rightarrow \Delta t=\frac{15.01\times 0.02\times 0.91\times 80}{0.5\times 12\times {10}^{-2}}$
$\Delta t=364.24\text{s}$