# Thermal energy in vacuum

Hi guys.

I was wondering, what happens to the temperature of a object if it's put into vacuum?

Say you have a box of steal (something that transfers thermal energy fast between the inside and outside).
Inside the box you have a rock. The rock is held in the center of the box using something that does not transfer any kind of thermal energy.

Outside the box there is air with constant temperature.
The box itself has the same temperature as the air outside.
The rock and the air inside the box also has the same temperature.

Now you suck the air out of the box, so the rock is left in vacuum. What will happen to the temperature of the rock?

## Answers and Replies

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Drakkith
Staff Emeritus
Thank you for the reply.

If I get you right the temperature doesn't drop when the air is sucked out?

When air is compressed it's getting warmer. I would expect the opposite to happen when air is sucked out, and this would make the rock colder.

K^2
It will cool while you are pumping the air out, yes. As the pressure drops, the gas in the box will, indeed, cool down and cool the rock. But this will be temporary. Once you've reached sufficient vacuum and stopped pumping, temperature between the rock and the walls will equalize due to thermal radiation.

Ah. I see.
Thank you

Chestermiller
Mentor
It will cool while you are pumping the air out, yes. As the pressure drops, the gas in the box will, indeed, cool down and cool the rock. But this will be temporary. Once you've reached sufficient vacuum and stopped pumping, temperature between the rock and the walls will equalize due to thermal radiation.
And, if the mass of the rock is much greater than the mass of the original air, the rock will experience very little cooling by the air. If the air were somehow removed instantaneously (or at least very rapidly), the rock temperature would virtually not change at all.