Two identical spheres A and B are suspended in an air chamber which is maintained at a temperature of `50^@C`. Find the ratio of the heat lost per second from the surface of the spheres if
a. A and B are at temperatures `60^@C` and `55^@C`, respectively.
b. A and B are at temperatures `250^@C` and `200^@C`, respectively.

Explanation

Net heat loss per second per unit area `=esigma(T^4-T_0^4)` from stefan's law.
If `T-T_0` is small as compared to the temperature of surroundings, we have
Net heat loss per second per unit area`=("constant")xx(T-T_0)`
From newton's law of cooling
a. Here the temperature difference is small and hence we can use newton's law of cooling.
`E_Aprop(60-50)` and `E_Bprop(55-50)`
Hence, `(E_A)/(E_B)=(10)/(5)=2`
b. As the temperature difference is not negligible as compared to the temperature of surrounding, we use stefan's law for accurate answer.
`E_Aprop(250+273)^4-(50+273)^4`
`impliesE_Bprop(200+273)^44-(50+273)^4`
Hence, `(E_A)/(E_B)=(523^4-323^4)/(423^2-323^4)=1.632`