Question

# A lake is covered with ice $$2\ cm$$ thick. The temperature of ambient air is $$-15^{\circ}C$$. Find the rate of thickening of ice. For ice $$K = 4\times 10^{-4}k-cal\ m^{-1}s^{-1} (^{\circ}C)^{-1}$$, density $$= 0.9\times 10^{3} kg/ m^{3}$$ and latent heat of ice $$(L) = 80\ k\ cal./ kg$$.

Solution

## Heat energy flowing per sec is given by$$H = \dfrac {dQ}{dt} = KA \dfrac {\triangle \theta}{x} .....(i)$$If $$dm$$ is the mass of ice formed in time $$dt$$, then$$\dfrac {dm}{dt} = \dfrac {A\ dx \rho}{dt} = A.\rho. \dfrac {dX}{dt}$$Since, $$H = \left (\dfrac {dm}{dt}\right ) L$$$$\therefore H = A\rho \dfrac {dx}{dt} L ..... (ii)$$From eq. (i) and (ii)$$A\rho L \dfrac {dx}{dt} = -KA \dfrac {\triangle \theta}{x}$$Rate of thickening of ice $$= \dfrac {dx}{dt}$$$$\therefore \dfrac {dx}{dt} = -\dfrac {KA}{\rho AL}\dfrac {\triangle \theta}{x} = \dfrac {-K}{\rho L}\dfrac {\triangle \theta}{x} = \dfrac {4\times 10^{-4}}{0.9\times 10^{3} \times 80}\times \left (\dfrac {0 - (-15)}{2\times 10^{-2}}\right ) = 4.166\times 10^{-6} m/s$$$$= 1.45\ cm/ hour$$.Physics

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