What is the net energy radiated per unit time by a black body at temperature $T$ with emissivity $e$ and absorptivity $a$, if it’s placed in an environment of temperature $T_o$.Assume that $T$$\gt$$T_o$

Here’s my calculation:

I used the Stefan-Boltzmann law to calculate the $Q$ radiated by object due to its temperature which is $$(Q_{rad})_{obj}=\sigma eAT^4 $$ Where $\sigma$ is Stefan-Boltzmann constant.

The energy is being radiated by the surroundings also so,$$(Q_{rad})_{surr}=\sigma AT_o^4$$So the object absorbs and also emits the energy due the radiation from surroundings which is given by$$Q_{abs}=\sigma aAT_o^4$$ $$Q_{em}=\sigma eAT_o^4$$ Therefore the net energy emitted is given by $$Q_{net}=(Q_{rad})_{obj}-Q_{abs}+Q_{em}$$ which equals to $$Q_{net}=\sigma eAT^4-\sigma aAT_o^4+\sigma eAT_o^4$$

Now if this all is correct, how to simplify it further, because the answer is $Q_{net}=\sigma eA(T^4-T_o^4)$ which I had an attempt to prove.So please guide me.


Assume first that the temperature of the blackbody environment and the object are the same. Hence the object will radiate the same spectrum as the bb env but reduced by emissivity. From the outside you see this emission plus the reflection. Since e+r=1, this will be the same intensity as the radiation falling on it, consistent with thermal equilibrium.

If we raise the temperature of the object it will radiate more energy than it receives by an amount of $e\sigma (T^4 - T_0^4)$. It will still absorb and reflect the same amount of energy, as this is determined solely by the blackbody environment.

| cite | improve this answer | |
  • $\begingroup$ The temperature aren’t same $\endgroup$ – P-S-S May 10 '19 at 13:18
  • $\begingroup$ Correct, I edited my answer accordingly. $\endgroup$ – my2cts May 11 '19 at 11:53
  • $\begingroup$ I’m not getting you can you calculate please!kind of ok but unable to do it! $\endgroup$ – P-S-S May 11 '19 at 13:54
  • $\begingroup$ calculate it please,I’m not able to $\endgroup$ – P-S-S May 12 '19 at 4:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.