Question

# A lake surface is exposed to an atmosphere where the temperature is less than $$0^oC$$. If the thickness of the ice layer formed on the surface grows from $$2cm$$ to $$4cm$$ in $$1$$ hour, the atmosphere temperature will be (Thermal conductivity of ice,$$K=4\times 10^{-3}\,cal\,\,cm^{-1}s^{-1}\,^oC^{-1}$$, density of ice $$=0.9\,\,g\,\,cc^{-1}$$. Latent heat of fusion of ice $$=80\,cal\,\,g^{-1}$$ Neglect the change of density during the state change. Assume that the water below the ice has $$0^oC$$ temperature every where.)

A
20oC
B
0oC
C
30oC
D
15oC

Solution

## The correct option is C $$-30^o C$$The rate of heat flow through the layer of the ice formed ice $$-\theta <0^oC$$$$i_H=\dfrac{0-(-\theta)}{(y/KA)}=\dfrac{\theta KA}{y}=\dfrac{dQ}{dt}$$         .........(i)Also, rate of heat gain during fusion of ice,$$\dfrac{dQ}{dt}=L\dfrac{dm}{dt}=L\rho .A\dfrac{dy}{dt}$$           ........(ii)From equations (i) and (ii), we get $$\dfrac{KA\theta}{y}=\rho AL\dfrac{dy}{dt}$$$$\int^4_2 ydy=\int^{3600}_0 \left(\dfrac{K\theta}{\rho L}\right)dt$$$$\left[\dfrac{y^2}{2}\right]^{4}_{2}=\dfrac{K\theta}{\rho L}[t]_0^{3600}$$$$\Rightarrow \dfrac{1}{2}\times [16-4]=\dfrac{4\times 10^{-3}\times \theta \times (3600-0)}{0.9\times 80}$$$$\Rightarrow \theta =\dfrac{1}{2}\times \dfrac{12\times 0.9 \times 80}{4\times 3600 \times 10{-3}}=30^o C$$$$\therefore$$ the atmospheric temperature $$=-\theta =-30^o C$$Physics

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