- #1

- 57

- 0

If an area of 1 m

Q/t = e(5.67E-8)(A)T

54E-3 = 1(5.67E-8)T

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.

^{2}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)T

^{4}54E-3 = 1(5.67E-8)T

^{4}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.

Last edited: