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QPages 98-102: Q15: HC Verma: Concepts of Physics - Volume 2 Heat Transfer Pages 98-102: Q15 Solution: Consider the situation of the previous problem. Assume that the …

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often incorrectly assume the ice thickness grows until the entire lake freezes or fails to recognize that steady state requires the heat flux through the ice layer to exactly equal the heat flux through the remaining water layer. You must set the conductive heat transfer rates equal ($K_{ice} \Delta T_{ice} / x = K_{water} \Delta T_{water} / (d-x)$) rather than simply summing temperature differences or assuming a fixed heat loss rate.

Consider the situation of the previous problem. Assume that the temperature of the water at the bottom of the lake remains constant at 4�C as the ice forms on the surface the heat required to maintain the temperature of the bottom layer may come from the bed of the lake). The depth of the lake is 1.0 m. Show that the thickness of the ice formed attains a steady state maximum value. Find this value. The thermal conductivity of water = 0.50 W m−1 �C−10.50 \text{ W m}^{-1} \text{ �C}^{-1}. Take other relevant data from the previous problem.

Step-by-Step Verified Solution

**Answer:** Maximum ice thickness = 0.89 m. **Step-by-Step Explanation:** For steady state: heat flow through ice = heat flow through water.
Kice×10x=Kwater×41−x\frac{K_{ice} \times 10}{x} = \frac{K_{water} \times 4}{1-x}
1.7×10x=0.5×41−x\frac{1.7 \times 10}{x} = \frac{0.5 \times 4}{1-x}
17(1−x)=2x⇒17=19x⇒x=0.89 m17(1-x) = 2x \Rightarrow 17 = 19x \Rightarrow x = 0.89 \text{ m}
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