Question

A lake surface is exposed to an atmosphere where the temperature is $<0^o$C. If the thickness of the ice layer formed on the surface grows from $2$cm to $4$cm in $1$ hour. Find the atmospheric temperature, $T_a$.

(Thermal conductivity of ice $K=4\times {10}^{-3}$ cal/cm/s/${}^{o}C$, density of ice $=0.9$ gm/cc. Latent heat of fusion of ice$=80$ cal/gm. Neglect the change of density during the state change. Assume that the water below the ice has $0^o$ temperature everywhere).

Answer 1

A lake surface is exposed to an atmosphere where the temperature $<0°C$

Thickness of ice layer formed $=2cmto4cm$

Time $t=1h$

Atmospheric temperature $=T_a$

Thermal conductivity of ice $K=4\times {10}^{-3}cal/cm/s°C$

Density of the ice $=0.9gm/cc$

Latent heat of fusion of ice $=80cal/gm$

Neglect the change of density during the state change

Assume that the water below the ice has $0°C$ temperature everywhere.

Solution 'We shall use the following relation which determines the rate of change of the thickness of ice

$\frac{dx}{dt}=K\theta {/}_x\frac{dx}{dt}=\frac{4N{O}^{-3}\times \theta }{0.9\times 80\times 2}4-2=\frac{4\times {10}^{-3}\times 10}{90\times 80\times 2}\frac{2\times 4\times 9}{{10}^{-3}\times 1}=\theta \theta =8\times 9\times {10}^3=72\times {10}^3°C=72000°C$