# Please help me with this doubt about radiant heat transfer

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1. Feb 1, 2016

### ajaysabarish

My textbook says that net rate of heat transfer due to radiation is εσA(T^4-To^4) but i couldn't understand it.
Rate of emission is εσAT^4 and rate of absorption is aσATo^4 so net rate of heat transfer must be εσAT^4-aσATo^4(where T is the temperature of body and To is the temperature of surroundings).it is possible only if a=e which means body is in thermal equilbrium.But that is not always the case.please help

Last edited: Feb 1, 2016
2. Feb 1, 2016

### Staff: Mentor

One "T" is the temperature of the object and the other "T" is the temperature of the surroundings.
http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/stefan.html

3. Feb 1, 2016

### SteamKing

Staff Emeritus
A body is in thermal equilibrium if and only if T = To. The factor ε (and probably a) is called the emissivity of the object.

https://en.wikipedia.org/wiki/Stefan–Boltzmann_law

The emissivity of a body is dependent on, among other things, the wavelength of the radiation emitted.

4. Feb 1, 2016

5. Feb 1, 2016

### ajaysabarish

yes sir,but what is the net rate of heat transfer?

6. Feb 1, 2016

### SteamKing

Staff Emeritus
7. Feb 1, 2016

8. Feb 1, 2016

### SteamKing

Staff Emeritus
Well, what exactly do you need explained?

9. Feb 1, 2016

### ajaysabarish

sir,net power of radiation is given to be εσA(T^4-To^4),but shouldn't it be power emitted - power absorbed?
power emitted is equal to εσAT^4 and power absorbed is equal to aσATo^4,so net power transferred must be equal to εσAT^4-aσATo^4,but why is it εσAT^4-εσATo^4(a is changed to ε).
where ε is the emissivity,a is the absorption coefficient,T is the temperature of body and To is the temperature of surroundings.
i need this to be explained sir.

10. Feb 1, 2016

### SteamKing

Staff Emeritus
All the Stefan-Boltzmann law governs is the heat lost due to radiation from one body which is hotter than its surroundings.

If you want to find out about heat absorbed by other bodies, then you have to measure the temperature of each of these absorbing bodies and estimate their emissivities and areas of absorption.

There is a corollary to the Stefan-Boltzmann law called Kirchhoff's Law, which governs radiative heat transfer by "real" bodies.

Kirchhoff's Law implies that for certain bodies, ε = α, or a body which radiates heat readily can also absorb heat just as easily. Of course, there are certain conditions which must be met, as described here:

http://web.mit.edu/16.unified/www/FALL/thermodynamics/notes/node135.html

The temperatures used in the Stefan-Boltzmann law are absolute temperatures.

If the temperature of the surroundings of a body were at absolute zero, then P = εσAT4, but the surroundings are usually at some other temperature To, so the net heat loss by radiation is
P = εσA(T4 - To4).

If To should happen to be greater than T, then there will be a net gain of heat by the body, and P will be negative.

http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/stefan.html

11. Feb 2, 2016

### ajaysabarish

doubt cleared thank you very much sir

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