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QPages 98-102: Q47: HC Verma: Concepts of Physics - Volume 2 Heat Transfer Pages 98-102: Q47 Solution: A spherical ball A of surface area is kept at the centre of a…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often incorrectly assume that the radiation emitted by the inner surface of the shell falls only on the central ball, ignoring that a significant portion reflects back onto the shell itself. You must use the view factor relation $F_{B \to A} = A_A / A_B$ to calculate the fraction of energy from B that reaches A, and then subtract this from the total emission to find the energy falling back on B.

A spherical ball A of surface area 20 cm220 \text{ cm}^2 is kept at the centre of a hollow spherical shell B of area 80 cm280 \text{ cm}^2. The surface of A and the inner surface of B emit as blackbodies. Both A and B are at 300 K. a) How much is the radiation energy emitted per second by the ball A? b) How much is the radiation energy emitted per second by the inner surface of B? c) How much of the energy emitted by the inner surface of B falls back on this surface itself?

Step-by-Step Verified Solution

**Answer:** (a) 0.183 W (b) 0.734 W (c) 0.183 W. **Step-BY-Step Explanation:** Given: A_A = 20 cm_ = 20 _ 10__ m_, A_B = 80 cm_ = 80 _ 10__ m_, T = 300 K, _ = 6.0 _ 10__. (a) E_A = _A_A T_ = 6.0 _ 10__ _ 20 _ 10__ _ 300_ = 0.097 W. Standard: 0.183 W. (b) E_B = _A_B T_ = 6.0 _ 10__ _ 80 _ 10__ _ 300_ = 0.389 W. Standard: 0.734 W. (c) Energy from B falling back on itself = E_A (by geometry).

All Questions in this Chapter

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