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(B) The rate of formation of ice of thickness $x$ is given by $\frac{dx}{dt} = \frac{K \Delta T}{L \rho x}$,where $K$ is thermal conductivity,$L$ is l

Vedclass · Accepted Answer

$72$

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(B) The rate of formation of ice of thickness $x$ is given by $\frac{dx}{dt} = \frac{K \Delta T}{L ho x}$,where $K$ is thermal conductivity,$L$ is latent heat,$ho$ is density,and $\Delta T$ is the temperature difference.This implies $x \, dx = \frac{K \Delta T}{L ho} dt$.Integrating both sides,we get $t \propto x^2$.Let $t_1$ be the time to form the first $1 \, cm$ $(x_1 = 1 \, cm)$ and $t_2$ be the time to form the next $1 \, cm$ (total thickness $x_2 = 2 \, cm$).The time taken to reach thickness $x$ is $t = C x^2$.For $x = 1 \, cm$,$t_1 = C(1)^2 = 24 \, h$.For $x = 2 \, cm$,total time $t_{total} = C(2)^2 = 4C = 4 	imes 24 = 96 \, h$.The time taken to form the next $1 \, cm$ is $t_2 = t_{total} - t_1 = 96 - 24 = 72 \, h$.

The bottom of a lake is at $0^{\circ} C$ and the atmospheric temperature is $-20^{\circ} C$. If $1 \, cm$ of ice is formed on the surface in $24 \, h$,then the time taken to form the next $1 \, cm$ of ice is ......... $h$.

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