# I Time of cooling to equilibrium in space plasma gas

1. Apr 14, 2017

### Albertgauss

Hi all,

If I have a hot object in space (not a star but say an oven or just a hot gas as would be on Earth < 10,000 degrees Kelvin) glowing at a temperature T and I want to know long it takes to come to equilibrium with the vacuum of space around it, how can I calculate such a time? I could find the Stephen Boltzmann Law for power radiated but am not sure how to get the time to completely cool off. Seems like it should be easy but it eludes me somehow.

Oh also, slight mistake in the title. Its not a plasma I want but either a gas or an object just at some general temperature T.

2. Apr 14, 2017

### CWatters

The answer is an infinite amount of time...

The rate of heat loss depends on the temperature of the object so the rate of heat loss reduces as the temperature reduces. The result is a curve that is asymptotic to the temperature of the surroundings. At best you can work out how long it takes to get within some value of its surroundings.

3. Apr 16, 2017

### Albertgauss

What about an example? Suppose the background temperature of space is 3 Kelvin. What if I have an object that starts at 1000 Kelvin and goes to 10 Kelvin, would there be a formula to calculate how long the object cools to 10 Kelvin? Ignore phase changes.

I know there must be something more concrete than an infinite "amount of time." For example, I know that car engines operate at around 300 degrees Fahrenheit and cool off to room temperature ~75 degrees Fahrenheit. I've heard of people calculating this time of cooling (~ tens of minutes I think) but I can't find a formula they used.

4. Apr 17, 2017

### CWatters

For an object in space the heat loss is all by radiation (eg not conduction or convection) so try..
http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/cootime.html

Its based on Stefan-Boltzmann law and assumes that the thermal conductivity of the object is very good. If the thermal conductivity is poor then the outer layers of the object effectively insulate the inner layers and slow down the heat loss process.

5. Apr 17, 2017

### Albertgauss

Yes, that did the trick! Just what I was looking for. Pretty easy to play around with the numbers. All done here.