# Cooling faster in space, or in Antarctica?

1. Nov 30, 2005

### Blahness

Here's the situation:

A heated object is in space, with no close proximity to any star. It is simply in one position, motionless.

Another object of equal heat is in Antarctica, in the dead of night, laying half-inside the snow. Temperature at -100 Degrees Celsius.

Which would lose heat faster?

2. Nov 30, 2005

### siddharth

3. Nov 30, 2005

### Blahness

But does space have a temperature, since it has so little matter in it?

I'm thinking a vaccum, like how a Thermos works. Is space like a Thermos?

4. Nov 30, 2005

### alfredblase

space is not like a thermos flask. And space does have a temperature, (around 3 Kelvin I think)

Perhaps this makes me a poor experimental physicist, (true) but I've never broken a thermos flask to see how it works. However I think a thermos flask works in two ways, cuts down the heat loss in the body due to convection and conduction via the glass vacuum casing, and reduces heat loss due to radiation via reflection of infrared by the reflective material which the casing employs. Space cuts down on convection and conduction almost completely, I think, but it hardly provides any impediment to radiation loss. So comparing a perfect thermos to a perfect vacuum.. well the thermos is an infinitely better insulator.

As far as the original question goes, (antarctica vs. space), Stefan's law will help you compare radiation heat loss in the two cases, but you would still need to find a way of at least estimating the heat loss in antarctica due to convection and conduction. I can't think how to do this, but I strongly suspect I'd freeze faster in deep space than in antartica deep winter time... heh

Last edited: Nov 30, 2005
5. Nov 30, 2005

### Staff: Mentor

I'd tend to agree, but the answer really depends on how hot this object is and what it's geometry is.

The hotter the object, Blahness, the bigger role radiation plays.

6. Nov 30, 2005

### pervect

Staff Emeritus

The object will lose heat due to conduction, convection, and radiation. The radiation law has already been given which will model the heat loss due to radiation. The object half-buried in the snow will only be able to radiate over half the sphere, however.

We now need to know the heat loss of the buried object due to conduction and convection. To compute the heat loss due to conduction we need to know the temperature of the object, the temperature of the snow, and some information about the thermal resistance of the object as calculated from its geometry and conductivity

i.e. heat flow = (thermal conductivity) * (area / length) * (delta-T)

To calculate the heat loss of the buried object due to convection, we need similar information to apply "Newton's law of cooling"

heat flow = (area) * (constant) * (delta-T)

7. Dec 1, 2005

### Blahness

So the hotter the object is, the more that radiation plays a role in it?

Got it.

8. Dec 1, 2005

### Staff: Mentor

I should probably be a little more specific (thorough): The Sefan-Boltzman Equation shows that heat transfer via radiation increases by a power of four of temperature (T^4, ie, double the temperature difference, and heat transfer increases by a factor of 16), whereas heat transfer via conduction or convection is a simple direct proportion (double the temperature difference and heat transfer doubles).

9. Dec 1, 2005

### alfredblase

Hmm the question still hasnt been answered properly I dont think. Pervect gave us big clues, without giving any indication as to how his formulas could be applied. The one for conduction he gave I think can only be directly aplied to a case such as two bodies conected by a wire. In the case of a sphere half buried in snow, the "wire" is actually the boundary between the sphere and the snow. Also pervect didnt give the constant needed for the convection formula. In fact determining the constant I think is rather tricky (http://en.wikipedia.org/wiki/Convection).

Neverheless in the case of conduction the solution can easily be approximated, I think, by a bit of dimensional analysis:

The heat flow at the instant we half bury a sphere in thermal equilibrium in snow is a function of the temperature difference between the sphere and the ice, the thermal conductivity of the boundary and the radius of the sphere. Since we are dealing with only half the surface area buried we can assume it is best to stick a 2pi infront of the function:

Q = 2pi f(deltaT, r, thermal conductivity of boundary)

Now I'm not completely sure but my intuition tells me that the thermal conductivity of the boundary would be the average of the sphere and the snow conductivity as they both contribute equally to the boudary. So Blahness gives us the material and size of the sphere and we can estimate the heat flow due to conduction.

As far as the convection, the problem seems very yucky and unatractive to me, and I'd rather not tackle it for the moment. Maybe later if people continue to show interest in the thread I would give it a bash.