Question

A lake is covered with ice and the atmospheric temperature above the ice is $-{20}^{\circ }\text{C}$. Find the rate of formation of ice $\left(\text{in cm/hr}\right)$ when its thickness is $10\text{cm}$.
[Thermal conductivity of ice $=0.005\text{cgs}$ units, density of ice $=0.9\text{g/cc}$ and latent heat of fusion of ice $=80\text{cal/g}$].

• A. $50{\text{cm hr}}^{-1}$
• B. $25{\text{cm hr}}^{-1}$
• C. $0.5{\text{cm hr}}^{-1}$
• D. $0.25{\text{cm hr}}^{-1}$

Answer 1

The correct option is C $0.5{\text{cm hr}}^{-1}$

Consider a layer of ice that is $x\text{cm}$ thick. Just below the ice, temperature is $0^{\circ }\text{C}$ and above it, the temperature is $\theta =-{20}^{\circ }\text{C}$.
Let $A$ be the area of the ice layer.
Heat conducted through ice in next $dt$ seconds will be
$dQ=\frac{kA\theta }{x}dt$
If ice layer thickens by $dx$, the mass of ice frozen in time $dt$ will be $\rho Adx$
$\therefore dQ=mL=\left(\rho Adx\right)L$
$\Rightarrow \frac{kA\theta }{x}dt=\rho ALdx$
$\Rightarrow \frac{dx}{dt}=\frac{k\theta }{x\rho L}$
is the rate of formation of ice, when thickness is $x\text{cm}$.
From the data given in the question,
$\frac{dx}{dt}=\frac{\left(0.005{\text{cal cm}}^{-1}{\text{s}}^{-1}{}^{\circ }{\text{C}}^{-1}\right)\times \left({20}^{\circ }\text{C}\right)}{\left(10\text{cm}\right)\left(0.9{\text{g cm}}^{-3}\right)\left(80{\text{cal g}}^{-1}\right)}=0.00014{\text{cm s}}^{-1}$
$=0.00014\times 3600{\text{cm hr}}^{-1}\approx 0.5{\text{cm hr}}^{-1}$
Thus, option (c) is the correct answer.