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QPages 98-102: Q40: HC Verma: Concepts of Physics - Volume 2 Heat Transfer Pages 98-102: Q40 Solution: Calculate the amount of heat radiated per second by a body of…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often apply $P=\varepsilon\sigma A T^4$ directly and report the emitted power, forgetting that in thermal equilibrium the body also absorbs $\varepsilon\sigma A T_{room}^4$, so the net radiated power is $\varepsilon\sigma A (T_{body}^4- T_{room}^4)$ which is zero when $T_{body}=T_{room}$. Always subtract the ambient term for net heat transfer.

Calculate the amount of heat radiated per second by a body of surface area 12 cm212 \text{ cm}^2 kept in thermal equilibrium in a room at temperature 20�C. The emissivity of the surface = 0.80 and σ=6.0×10−8 W m−2 K−4\sigma = 6.0 \times 10^{-8} \text{ W m}^{-2} \text{ K}^{-4}.

Step-by-Step Verified Solution

**Answer:** Heat radiated per second = 0.83 W. **Step-by-Step Explanation:** Given: A = 12 cm_ = 12 _ 10__ m_, T = 20�C = 293 K, e = 0.80, _ = 6.0 _ 10__.
E=eσAT4=0.80×6.0×10−8×12×10−4×(293)4=0.83 WE = e\sigma A T^4 = 0.80 \times 6.0 \times 10^{-8} \times 12 \times 10^{-4} \times (293)^4 = 0.83 \text{ W}

All Questions in this Chapter

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