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(C) The rate of heat flow through the ice layer is given by $\frac{dQ}{dt} = \frac{KA(	heta_1 - 	heta_2)}{x}$.As the ice thickness increases by $dx$

Vedclass · Accepted Answer

$19.1$

Vedclass · Answer

(C) The rate of heat flow through the ice layer is given by $\frac{dQ}{dt} = \frac{KA(	heta_1 - 	heta_2)}{x}$.As the ice thickness increases by $dx$ in time $dt$,the heat released is $dQ = L \cdot dm = L \cdot ho \cdot A \cdot dx$.Equating the two,we get $\frac{KA(	heta_1 - 	heta_2)}{x} dt = L ho A dx$.Integrating from $x_1 = 5 \ cm$ to $x_2 = 10 \ cm$:$t = \frac{L ho}{K(	heta_1 - 	heta_2)} \int_{5}^{10} x \, dx = \frac{L ho}{K(	heta_1 - 	heta_2)} \left[ \frac{x^2}{2} ight]_{5}^{10}$.Substituting the values: $t = \frac{80 	imes 0.92}{0.004 	imes (0 - (-10))} 	imes \frac{100 - 25}{2} = \frac{73.6}{0.04} 	imes 37.5 = 1840 	imes 37.5 	ext{ seconds}$.Converting to hours: $t = \frac{1840 	imes 37.5}{3600} \approx 19.1 \ hours$.

$A$ $5 \ cm$ thick ice block is present on the surface of water in a lake. The temperature of the air is $-10^{\circ}C$. How much time will it take to double the thickness of the block? (Given: $L = 80 \ cal/g$,$K_{ice} = 0.004 \ cal/s \cdot cm \cdot ^{\circ}C$,$\rho_{ice} = 0.92 \ g/cm^3$)

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