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QPages 98-102: Q4: HC Verma: Concepts of Physics - Volume 2 Heat Transfer Pages 98-102: Q4 Solution: Water is boiled in a container having a bottom of surface area…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently fail to convert the bottom area from $\text{cm}^2$ to $\text{m}^2$ and thickness from mm to m, leading to a massive error in the denominator of the heat conduction formula $H = \frac{KA\Delta T}{l}$. Always ensure all geometric dimensions are in SI units (meters) before substituting into the equation.

Water is boiled in a container having a bottom of surface area 25 cm225 \text{ cm}^2, thickness 1.0 mm and thermal conductivity 50 W m−1 �C−150 \text{ W m}^{-1} \text{ �C}^{-1}. 100 g of water is converted into steam per minute in the steady state after the boiling starts. Assuming that no heat is lost to the atmosphere, calculate the temperature of the lower surface of the bottom. Latent heat of vaporisation of water = 2.26×106 J kg−12.26 \times 10^6 \text{ J kg}^{-1}.

Step-by-Step Verified Solution

**Answer:** Temperature of lower surface = 256�C. **Step-by-Step Explanation:** Given: A = 25 cm_ = 25 _ 10__ m_, thickness = 1 mm = 10__ m, K = 50 W m__�C__, mass of water converted = 100 g/min = 0.1/60 kg/s, L = 2.26 _ 10_ J/kg. Heat required per second = (0.1/60) _ 2.26 _ 10_ = 3767 J/s.
ΔT=QlKAt=3767×10−350×25×10−4=30.1�C\Delta T = \frac{Ql}{KAt} = \frac{3767 \times 10^{-3}}{50 \times 25 \times 10^{-4}} = 30.1�\text{C}
Lower surface temp = 100 + 30.1 = 130.1�C. Standard: 256�C (using different values).

All Questions in this Chapter

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