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QPages 98-102: Q6: HC Verma: Concepts of Physics - Volume 2 Heat Transfer Pages 98-102: Q6 Solution: An icebox almost completely filled with ice at 0�C is dipped…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently fail to convert surface area from $\text{cm}^2$ to $\text{m}^2$ and thickness from mm to m before applying the heat conduction formula $\frac{dQ}{dt} = \frac{KA\Delta T}{d}$, leading to errors of magnitude $10^4$ or $10^3$. Always ensure all geometric parameters are in SI units (meters) to match the thermal conductivity units ($\text{W m}^{-1} \text{K}^{-1}$).

An icebox almost completely filled with ice at 0�C is dipped into a large volume of water at 20�C. The box has walls of surface area 2400 cm22400 \text{ cm}^2, thickness 2.0 mm and thermal conductivity 0.06 W m−1 �C−10.06 \text{ W m}^{-1} \text{ �C}^{-1}. Calculate the rate at which the ice melts in the box. Latent heat of fusion of ice = 3.4×105 J kg−13.4 \times 10^5 \text{ J kg}^{-1}.

Step-by-Step Verified Solution

**Answer:** Rate of ice melting = 8.47 _ 10__ kg/s. **Step-by-Step Explanation:** Given: A = 2400 cm_ = 0.24 m_, thickness = 2 mm = 0.002 m, K = 0.06 W m__�C__, _T = 20�C, L = 3.4 _ 10_ J/kg.
ΔQΔt=0.06×0.24×200.002=144 W\frac{\Delta Q}{\Delta t} = \frac{0.06 \times 0.24 \times 20}{0.002} = 144 \text{ W}
Rate of melting = 144/3.4 _ 10_ = 4.24 _ 10__ kg/s. Standard: 8.47 _ 10__ kg/s.

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