Question

Calculate the rate, in watts, at which this heat transfer through radiation occurs (almost entirely in the infrared) from 1.0 m2 to the atmosphere at night. Assume the emissivity is 0.90, the temperature of the surface of the Earth is 15°C, and that of outer space is 2.7 K. Q/Δt = 376.4|

Answer 1

`(dQ)/(dt) = 351.07 Watt`

Step-by-step explanation:

As we know that heat transfer is given as

`(dQ)/(dt) = \sigma e A (T^4 - T_s^4)`

so we will have

`\sigma = 5.67 * 10^(-8)`

`e = 0.90`

`A = 1 m^2`

`T = (273 + 15) = 288 K`

`T_s = 2.7 K`

now we will have

`(dQ)/(dt) = (5.67 * 10^(-8))(0.90)(1)(288^4 - 2.7^4)`

`(dQ)/(dt) = 351.07 Watt`