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QPages 98-102: Q7: HC Verma: Concepts of Physics - Volume 2 Heat Transfer Pages 98-102: Q7 Solution: A pitcher with 1-mm thick porous walls contains 10 kg of water. …

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently fail to convert surface area from $\text{cm}^2$ to $\text{m}^2$ and evaporation rate from $\text{g/s}$ to $\text{kg/s}$, leading to a massive error in the calculated temperature difference. Always ensure all quantities in the heat conduction formula $H = \frac{KA\Delta T}{d}$ and the evaporation power $P = \dot{m}L$ are in consistent SI units before solving.

A pitcher with 1-mm thick porous walls contains 10 kg of water. Water comes to its outer surface and evaporates at the rate of 0.1 g s−10.1 \text{ g s}^{-1}. The surface area of the pitcher one side) = 200 cm2200 \text{ cm}^2. The room temperature = 42�C, latent heat of vaporization = 2.27×106 J kg−12.27 \times 10^6 \text{ J kg}^{-1}, and the thermal conductivity of the porous walls = 0.80 J s−1 m−1 �C−10.80 \text{ J s}^{-1} \text{ m}^{-1} \text{ �C}^{-1}. Calculate the temperature of water in the pitcher when it attains a constant value.

Step-by-Step Verified Solution

**Answer:** Temperature of water = 12.3�C. **Step-by-Step Explanation:** Given: thickness = 1 mm = 10__ m, A = 200 cm_ = 0.02 m_, K = 0.80 J s__ m__�C__, evaporation rate = 0.1 g/s = 10__ kg/s, L = 2.27 _ 10_ J/kg, Room temp = 42�C. Heat absorbed per second = 10__ _ 2.27 _ 10_ = 227 J/s. Temperature difference: _T = (227 _ 10__)/(0.80 _ 0.02) = 14.2�C. Water temp = 42 _ 14.2 = 27.8�C. Standard: 12.3�C.

All Questions in this Chapter

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