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The thickness of ice in a lake is \( 10 cm \) and its temperature is \( -10^{0} C \). The thermal conductivity of ice \( =0.004 calcm ^{-1} \sec ^{-1} c^{-1} \) Density of ice \( =0.92 g / cc \), latent heat of ice \( =80 cal / g \). The time taken from thickness of ice to be doubled is \( \times \times 10^{5} sec \), find the value of \( x \) (two decimal place) |
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Answer» Given thickness of ice is x1 = 10 cm Temperature T = +10°C Thermal conductivity k = 0.004 cal cm-1 sec-1 c-1 Density of ice ρ = 0.9 g/cc, Latent heat of ice = 80 cal/g Thickness double, then time t = ? x2 = 2x1 ⇒ 2 x 10 ⇒ 20 cm Using formula, Δt = \(\frac{\rho L}{2kT}(x_2^2-x_1^2)\) Δt = \(\frac{92\times 10^{-2}\times 80}{2\times 4 \times 10^{-3}\times 10}\)[(20)2 - (10)2] Δt = 92 x 10 [400 - 100] Δt = 92 x 10 x [300] Δt = 276000 sec Δt = 2.76 x 105 sec then value of x = 2.76 |
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