Question

# The thickness of ice in a lake is $$5 cm$$ and its atmospheric temperature is $$-10^{o}C$$. Calculate the time required for the thickness of ice to grow to $$7 cm.$$ Thermal conductivity of ice = $$4 \times 10^{-3} cal/cm^{-1}s^{-1}C^{-10}$$; density of ice $$0.92 g/cm^3$$ and specific latent heat of fusion for ice $$= 80 cal/g.$$

Solution

## Here, $$x_1 = 5 cm, x_2 = 7 cm; T = 10^{o}C, k = 4 \times 10^{-3} cal/cm^{-1}s^{-1}C^{-10}; L = 80 cal/g,\rho = 0.92 g/cm^2$$Using                   $$\Delta t = \dfrac{\rho L}{2KT} (x_2^2 - x_1^2)$$We have the required time $$\Delta t$$ as$$\Delta t = \dfrac{92 \times 10^{-2}} {2 \times 4 \times 10^{-3} cal/cm^{-1}s^{-1}C^{-10}}\dfrac{g}{cm^{3}} \times \dfrac{80}{10^{o}C} \dfrac{cal}{g} (7^2 - 5^2) cm$$  $$= \dfrac{92 \times 80 \times 24}{8}s = 22080 s = 6.13 h$$Physics

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