# Emissivity and Final Temperature of a Black and White object

Objects can be categorized as blackbodies (emissivity $\epsilon = 1$), grey bodies (emissivity $\epsilon < 1$) and white bodies (emissivity $\epsilon = 0$).

If we placed two objects (identical shape), one black and one white, in sunlight for a very long time (say 15 hrs), what would happen?

1) I think both objects would absorb radiation from the sun and their temperatures would increase until they reach their final temperatures.

2) The final temperature for the white object is lower than for the black object.

3) The white object will take longer than the black object to reach its final temperature.

4) The black object behaves like a blackbody. Blackbodies are said to be perfect emitters and perfect absorbers. But even a white object reflects (in theory) all the incident energy. So what is the difference? That reflection does not contribute to raising the T of the object?

5) In some cases, two different objects, left in the same room, reach the same temperature as the room. Why doesn't that happen when the same two object are left under the sun? What if they were left in the shadow? Would they eventually have the same temperature?

1) I think both objects would absorb radiation from the sun and their temperatures would increase until they reach their final temperatures.

A (perfect) white body, by definition, reflects all incident radiation and thus would not absorb any radiation from the sun.

2) The final temperature for the white object is lower than for the black object.

If only considering radiative heating, yes. The white body will not heat up at all (see also 5).

3) The white object will take longer than the black object to reach its final temperature.

No, see 2.

4) The black object behaves like a blackbody. Blackbodies are said to be perfect emitters and perfect absorbers. But even a white object reflects (in theory) all the incident energy. So what is the difference? That reflection does not contribute to raising the T of the object?

They will produce different spectra. The white body reflects all the incident radiation, so it's spectrum will be the same as that of the incoming sunlight. The black body, however, absorbs all incoming sunlight and emits blackbody radiation.

5) In some cases, two different objects, left in the same room, reach the same temperature as the room. Why doesn't that happen when the same two object are left under the sun? What if they were left in the shadow? Would they eventually have the same temperature?

In general, there are more ways for heat transfer to occur than through radiation. Two other ways are conduction and convection. These are the main mechanisms by which objects reach thermal equilibrium with their surroundings. If you were to place two objects outside in the shadow, they would both eventually reach thermal equilibrium with the air (mainly through conduction and convection).

Sorry, but you've got it backwards.

Let's start with the exception - a perfect white body. This is an exception because, since reflectivity equals 1 minus the emissivity, it is a perfect reflector. In this case, your statement that "both objects would absorb radiation" is clearly wrong. A PERFECT reflector will not absorb energy, although it's important to keep in mind that making a perfect reflector at all wavelengths (not just visible) is easier said than done. So let's eliminate the perfect white body, and consider objects whose emissivity may be close to zero, but not exactly.

In the short run, as you've stated, the white body will heat up more slowly than the black body since, as you realize, its increased reflectivity will reduce the amount of energy which it absorbs. However, it will hold on to its absorbed energy much more tightly than a black body, and the retention of energy will more than compensate for the small energy input.

At equilibrium, the absorbed energy must equal the radiated energy, with the radiated energy being expressed as $$P =\sigma kT^4$$ where $\sigma$ is nal to emissivity. So we can write $$P_{in} (1-\sigma) = \sigma kT^4$$ Then $$T^4 =\frac{1-\sigma}{\sigma} \frac{P_{in}}{k}$$ The smaller $\sigma$, the greater the final temperature, and a nearly perfect reflector can have a very high temperature.

Of course, for a good reflector, it can take a very long time to heat up, and for objects heated by sunlight the sun may well set before significant heating occurs. In practical construction, the lower ambient temperature associated with night-time allows convective cooling before morning.

• Clearly, at the same wavelength, if reflectivity is high, emissivity is low. But real bodies usually absorb and reflect in a certain spectral band and emit in a different band. So the emissivity should be evaluated over a different spectral region than reflectivity.... – Brett Cooper May 29 '18 at 16:24