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Two bodies A and B have emissivities 0.01 and 0.81 respectively, with equal outer surface areas. They radiate energy at the same rate. The wavelength of maximum spectral radiancy of B differs from that of A by 1.00 micrometre. If A is at 5802 K, find the temperature of B.

1. the temperature of B is 1934 K
2. the temperature of B is 11604 K
3. the temperature of B is 2901 K
4. the temperature of B is 934 K

Correct answer: the temperature of B is 1934 K

Solution

Equal power gives T_B = T_A (e_A/e_B)^(1/4) = 5802 * (1/81)^(1/4) = 5802/3 = 1934 K.

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