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Suppose we place a black body and an object of lower emissivity, both of the same shape, into an evacuated glass contained, and we place that container in sun light. After time, if both reach thermal equilibrium (either effect the other), does the black body have a higher temperature at its equilibrium? Why or why not?

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  • $\begingroup$ When you expose an object to sun light, it is not in thermal equilibrium and will not be until the object's surroundings are at the same temperature as the sun. The assumption that the black body is in equilibrium is an error which leads to your confusion. $\endgroup$ – DanielSank Dec 5 '15 at 5:56
  • $\begingroup$ @DanielSank I guess I was assuming that the surrounding of the glass container also wouldn't effect the two object. If there is no air nor any other surrounding to the two object, then its thermal equilibrium will only relate to itself and the incoming radiations. $\endgroup$ – A. Jiao Dec 6 '15 at 4:05
  • $\begingroup$ You already accepted John Rennie's answer, so I guess you understand what I mean now :D $\endgroup$ – DanielSank Dec 6 '15 at 4:12
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Your objects will reach equilibrium when the energy absorbed from the sunlight is equal to the energy radiated by the body.

The energy radiated will be given by the Stefan-Boltzmann law:

$$ J_\text{out} = Ae\sigma T^4 $$

where $A$ is the surface area of the object and $e$ is the emissivity.

Now let $I$ be the energy per unit area in the sunlight incident on the object, so the total energy will be $W = aI$ where $a$ is the cross sectional area of the object. The proportion of the energy reflected is $rW$, where $r$ is the reflectivity, so the energy absorbed is:

$$ J_\text{in} = W - rW = (1-r)W = (1-r)aI $$

The reflectivity is related to the emissivity by $e+r=1$, which means $1-r=e$ and the equation simplifies to:

$$ J_\text{in} = eaI $$

At equilibrium we set $J_\text{in} = J_\text{out}$ to get:

$$ eaI = Ae\sigma T^4 $$

and rearranging gives:

$$ T = \left(\frac{aI}{A\sigma}\right)^\frac{1}{4} $$

So at equilibrium the temperature is not related to the emissivity. A higher emissivity means the body loses more energy due to black body radiation, but because higher emissivity means lower reflectivity it means the object absorbs more of the sunlight falling on it. The two effects balance out.

Note that although the emissivity does not affect the equilibrium temperature it does affect the speed that the object attains equilibrium. The higher the emissivity the faster the object will reach equilibrium. So while your two objects are warming up they will warm up at different rates and therefore will temporarily have different temperatures.

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  • $\begingroup$ Thank you @John Rennie. Your answer is very thorough and answers my question. There is a few details I am uncertain of. Is J(out) the power of its emission of radiation? And should [I] be the power per unit area of the sun's incident ray on the object? Thanks $\endgroup$ – A. Jiao Dec 6 '15 at 4:12
  • $\begingroup$ @A.Jiao: yes, $J_\text{out}$ is the total power emitted as black body radiation. $I$ is indeed the power per unit area of the sunlight. $\endgroup$ – John Rennie Dec 6 '15 at 6:25
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The temperature of blackbody could vary from 0 to $\infty$ (take our Sun and a much larger star as an example; you can assume both as a radiating blackbody but their temperatures are different). Based on this temperature, the spectrum they emit is different (see the figure). enter image description here

In your question the object's temperature has to depend on the source that is heating it up (here, the sun); the spectrum that this blackbody radiates is the same as the spectrum of sun, assuming a perfect blackbody.

Then if both your objects are exactly the same (mass, shape and material) their temperature also has to be the same. However, the difference in emissivity means difference in absorption as well, so, they have to be different in temperature and this does not depend on whether they are blackbody or not.

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