Question

When the air temperature is $-{\theta }^{\circ }C$, the ice on a large shallow lake increases by $xm$ in $t$ seconds. The thermal conductivity of ice is:
(Density of ice $=\rho kg{m}^{-3}$ and $L=$ Latent heat in $Jk$).

• A. $\frac{\rho L{x}^2}{2\theta t}$
• B. $\frac{\rho L{x}^2}{4\theta t}$
• C. $\frac{\rho L{x}^2}{\theta t}$
• D. $\frac{2\rho L{x}^2}{\theta t}$

Answer 1

The correct option is A $\frac{\rho L{x}^2}{2\theta t}$

Let the temperature of air be $-{\theta }^{o}C$. At a certain time let the thickness of ice be $x$ and let it increase by a further thickness$dx$ in a time $dt$.

Quantity of heat lost to increase $dx=\rho LAdx$

where $\rho$ - density of ice, $L$ - specific heat of fusion of water, $A$ - area of surface, $x$ - thickness of ice after time $t$

Rate of loss of heat $=\frac{LA\rho dx}{dt}=\frac{k\theta A}{x}\Rightarrow \frac{dx}{dt}=\frac{k\theta }{L\rho x}$

(On Integrating)

$\Rightarrow x={\left(\frac{2k\theta t}{L\rho }\right)}^{1/2}\Rightarrow x^2=\frac{2k\theta t}{L\rho }\Rightarrow k=\frac{L\rho x^2}{2\theta t}$