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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.)
954614_044f8a358fe54ced89669f6f13d363fd.png


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