Radiation, temperature of earth, heat transfer

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SUMMARY

The discussion centers on calculating the temperature of the Earth’s surface based on its radiation emission of 54E-3 J/s, assuming it behaves as a perfect black body emitter with an emissivity (e) of 1. The formula used is Q/t = e(5.67E-8)(A)T^4, leading to a calculated temperature of 31.2 K. However, there is contention regarding the expected temperature, which is stated to be 2.4 K, indicating a potential error in the initial assumptions or calculations regarding the energy emission.

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bulbasaur88
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If an area of 1 m2 at the Earth surface emits Q/t = 54E-3 J/s in radiation, what would the temperature at the Earth surface be, if there would be no other energy sources? Assume that the Earth is a perfect black body emitter (e = 1).

Q/t = e(5.67E-8)(A)T4
54E-3 = 1(5.67E-8)T4
T = 31.2 K?

The answer is supposed to be 2.4 K? I feel pretty confident in my work so I am not certain where I have gone wrong.
 
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Are you sure your Q/t is correct? The surface of the Earth is really -271 C?
 
Given the numbers given in the problem statement, I agree with the 31.2 K value. It does sound silly to say this takes place somewhere on the Earth's surface though.
 

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