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?
[Physics] Is the equilibrium temperature of a black body higher than other objects
equilibriumtemperaturethermal-radiationthermodynamics
Best Answer
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.