Question

Calculate the heat dissipation by radiation through a 0.2m^2 opening of a furnace at 1000K into an ambient at 300K. Assume that both the furnace and the ambient blackbodies.

Answer 1

Heat dissipation is 11.24 KW.

Step-by-step explanation:

Given that

Temperature of furnace= 1000 K

Temperature of surrounding = 300 K

Also given that both bodies are black bodies so ∈=1

We know heat radiation for black bodies is given as follows

`Q=\sigma A(T_1^4-T_2^4)`

Now by putting the values in the above formula

`Q=5.67* 10^(-8) * 0.2(1000^4-300^4)`

Q=11.24 KW

So heat dissipation is 11.24 KW.