Two bodies A and B have thermal emissivities of 0.01 and 0.81 respectively the outer surface area of the two bodies are the same. The twobodies emit total radiant spectral radiancy in the radiation from A and B respectively differ by 1.00 `mum`. If the temperature A is 5802 K, find a. the temperature of B `lamda_(B)`
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To solve the problem, we need to find the temperature of body B and the wavelength of body B given the emissivities and temperature of body A. Let's break it down step by step.
### Step 1: Understand the relationship between the powers emitted by the two bodies
The power emitted by a black body is given by the Stefan-Boltzmann law:
\[ P = \epsilon \sigma A T^4 \]
where:
- \( P \) is the power emitted,
- \( \epsilon \) is the emissivity,
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