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A black sphere of diameter L is enclosed (without touching) inside a black cube of side L, with vacuum in between. The sphere is at temperature T1 and the cube at T2. What is the net rate of radiative heat exchange between them?
- sigma*pi*L²*(T1⁴ - T2⁴)
- sigma*L²*(T1⁴ - T2⁴)
- 6*sigma*L²*(T1⁴ - T2⁴)
- 4*sigma*pi*L²*(T1⁴ - T2⁴)
Correct answer: sigma*pi*L²*(T1⁴ - T2⁴)
Solution
Net exchange = sigma * A_sphere * (T1⁴ - T2⁴), and the sphere's area is pi*L², giving sigma*pi*L²*(T1⁴ - T2⁴).
Related JEE Advanced Physics questions
- Match the temperature of a blackbody listed in Group-I to the corresponding statement in Group-II, and select the correct answer.
[Given: Wien’s constant = 2.9 × 10⁻³ m-K and hc/e = 1.24 × 10⁻⁶ V-m]
Group-I:
(P) 2000 K
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(R) 5000 K
(S) 10000 K
Group-II:
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(2) The peak wavelength of emitted radiation falls within the visible spectrum.
(3) The peak wavelength of emitted radiation produces the broadest central diffraction maximum in a single-slit setup.
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(5) The peak wavelength of emitted radiation is suitable for imaging human bones.
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