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QPages 98-102: Q14: HC Verma: Concepts of Physics - Volume 2 Heat Transfer Pages 98-102: Q14 Solution: On a winter day when the atmospheric temperature drops to…

⚠️ Common Exam Traps & Conceptual Pitfalls

When integrating the growth rate $\frac{dx}{dt}=\frac{K\Delta T}{\rho L x}$, students often treat the thickness $x$ as a constant and obtain a linear time dependence. The correct integration yields $t=\frac{\rho L}{2K\Delta T}x^2$, so the time grows with the square of the ice thickness. Remember to keep $x$ in the denominator during integration and square it in the final expression.

On a winter day when the atmospheric temperature drops to _10�C, ice forms on the surface of a lake. a) Calculate the rate of increase of thickness of the ice when 10 cm of the ice is already formed. b) Calculate the total time taken in forming 10 cm of ice. Assume that the temperature of the entire water reaches 0�C before the ice starts forming. Density of water = 1000 kg m−31000 \text{ kg m}^{-3}, latent heat of fusion of ice = 3.36×105 J kg−13.36 \times 10^5 \text{ J kg}^{-1} and thermal conductivity of ice = 1.7 W m−1 �C−11.7 \text{ W m}^{-1} \text{ �C}^{-1}. Neglect the expansion of water on freezing.

Step-by-Step Verified Solution

**Answer:** (a) Rate = 5 _ 10__ m/s (b) Time = 27.45 hours. **Step-by-Step Explanation:** Given: K = 1.7 W/m�C, _ = 1000 kg/m_, L = 3.36 _ 10_ J/kg, _T = 10�C. (a) For ice thickness x = 10 cm: rate = K _T / (x _ L) = 1.7 _ 10 / (0.1 _ 1000 _ 3.36 _ 10_) = 5.06 _ 10__ m/s. (b) Time = _L/(2K_T) _ x_ = (1000 _ 3.36 _ 10_ _ 0.01)/(2 _ 1.7 _ 10) = 9.88 _ 10_ s = 27.45 hours [citation:14].

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