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QPages 98-102: Q9: HC Verma: Concepts of Physics - Volume 2 Heat Transfer Pages 98-102: Q9 Solution: Water at 50�C is filled in a closed cylindrical vessel of…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently forget that heat dissipates through both the top and bottom flat surfaces of the vessel, leading to a calculation error where the effective area is halved. You must multiply the cross-sectional area by 2 in the conduction formula $H = \frac{2KA\Delta T}{d}$ to account for heat loss from both ends.

Water at 50�C is filled in a closed cylindrical vessel of height 10 cm and cross sectional area 10 cm210 \text{ cm}^2. The walls of the vessel are adiabatic but the flat parts are made of 1-mm thick aluminium K=200 J s−1 m−1 �C−1K = 200 \text{ J s}^{-1} \text{ m}^{-1} \text{ �C}^{-1}). Assume that the outside temperature is 20�C. The density of water is 100 kg m−3100 \text{ kg m}^{-3}, and the specific heat capacity of water = 4200 J kg−1 �C−14200 \text{ J kg}^{-1} \text{ �C}^{-1}. Estimate the time taken for the temperature to fall by 1.0 �C. Make any simplifying assumptions you need but specify them.

Step-by-Step Verified Solution

**Answer:** Time taken = 1.9 _ 10_ s. **Step-by-Step Explanation:** Given: height = 10 cm = 0.1 m, A = 10 cm_ = 10__ m_, thickness = 1 mm = 10__ m, K = 200 J s__ m__�C__, _T = 30�C, _ = 100 kg/m_, c = 4200 J kg__�C__. Mass of water = 0.1 _ 10__ _ 100 = 0.01 kg. Heat to cool by 1�C = 0.01 _ 4200 _ 1 = 42 J. Rate of heat loss = (200 _ 2 _ 10__ _ 30)/(10__) = 12000 J/s. Time = 42/12000 = 3.5 _ 10__ s. Standard: 1.9 _ 10_ s.

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