# Would a black teapot midway between the temperatures of the tea and air cool down or heat up? If neither, then how do the 3 reach thermal equilibrium? [closed]

Say you had a matt black teapot with hot tea inside it but cool air around it. Originally, I’m guessing the matt black teapot should absorb more joules of IR energy than it emits per second. This is because originally the matt black teapot would be at room temperature like the air, so it would originally be cooler than what it was holding, so it would be trying to heat up to reach thermal equilibrium with the tea. However, when the teapot heats up, it will eventually get to a temperature where it is still cooler than the tea (say 50 degrees Celsius when the tea is at 80 degrees Celsius) but is hotter than the surrounding air (which would be at about 20 degrees Celsius). So what happens then? On the one hand, you might think the black teapot would carry on absorbing more IR radiation than it emits per second in order to come into thermal equilibrium with the tea, and so to carry on heating up. However, on the other hand, you might think the black teapot would start emitting more radiation than it absorbed, in order to cool down and come into thermal equilibrium with the air! So, would the black teapot just compromise by staying at a constant temperature at this point? If so, then it would need to come to thermal equilibrium by making the air hotter and the tea colder without changing its own temperature anymore. But how would it do this?

Say the black teapot was absorbing 20j per second and emitting 20j per second at this point: is the method by which it would still cause the air, itself and the tea to reach thermal equilibrium with each other because it would be absorbing almost all its 20j IR radiation per second (say 19j of it) from the tea (as the hot tea would have a net emission of IR radiation per second, unlike the cooler air) but the teapot would be emitting its 20j IR radiation back out in all directions, so meaning it might emit 9j of it back into the tea but 11j of it into the air per second? This would cause a net loss of 10j IR radiation from the tea in that second and a net gain of 10j IR radiation to the air (because 1j was taken from the air to make up the 20j absorbed, and then 11j were emitted into the air). This absorbing from one direction and emitting in all directions would allow the black teapot to essentially act like a conductor to help the air and tea get into thermal equilibrium with each other, whilst the teapot itself would stay at a constant temperature. So, is this how it would work in this situation? Or not?

• Hi, please edit your question so it's readable. As it stands, you have mixed the substance with a lot of wandering thought and further have not considered using paragraphs. May 26 at 12:56

Let’s first suppose that we have a very large teapot with a relatively small surface area, so that the tea stays at the same temperature for a long time (this is called a “heat reservoir”).

Heat flows from the tea to the teapot. This heats up the teapot. As the temperature of the teapot rises above the air temperature heat flows from the teapot to the air. Eventually the teapot reaches an intermediate temperature somewhere between the tea temperature and the air temperature at which the two heat flows are equal and opposite. As long as the temperatures of the tea and the air do not change, the teapot has now reached a stable temperature. The exact value of this stable temperature will depend on the thermal properties of the teapot.

Now drop the “heat reservoir” assumption and assume the tea cools down slowly as heat flows from it through the teapot to the air. The “stable” temperature for the teapot (which, remember, has to be between the tea and the air temperatures) also falls. So as the tea cools, the teapot also cools. Eventually the tea, teapot and air all reach equilibrium at the same temperature (although, in theory, it will take an infinite amount of time for them to reach exactly the same temperature).

So the short answer is the teapot initially heats up to an intermediate temperature, and then cools as the tea cools.

An important part of learning about heat transfer is understanding the concept of steady-state flux or dynamic equilibrium. Here, a material may be furiously transferring heat while remaining at a constant temperature $$T(x)$$:

The image shows a wall with a hotter fluid on one side and a cooler fluid on the other side, as a representation of a teapot wall separating hot tea from cold air. Fourier's law of conduction models the corresponding heat flux $$Q$$ (in units of power, or watts) as

$$Q=kA\frac{\Delta T}{L},$$

where $$k$$ is the teapot material thermal conductivity, $$A$$ is the surface area, and $$\Delta T$$ and $$L$$ are the temperature difference across and thickness of the teapot wall, respectively.

Over any small portion of time, the teapot wall maintains a near-constant temperature even while transferring this heat per unit time $$Q$$. Of course, this ultimately results in the tea cooling off and the air warming up. In any case, the point of confusion may be whether a material can transfer heat without changing temperature—the answer is yes.