# A hot object exposed to low temperature in a vacuum doesn't lose heat?

I heard somewhere that if the human body were exposed to outer space where the temperature is extremely low, the human won't actually feel cold because in a vacuum, the heat energy doesn't have another physical object to travel too.

So my question is, i thought heat is just a form of radiation, and that radiation can travel through a vacuum. So then, why doesn't a body lose heat in a vacuum the same way our sun emits a whole bunch of UV radiation? In which case, the body should be losing energy and feeling cold.

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Related: physics.stackexchange.com/q/3076/2451 and links therein. –  Qmechanic Jan 27 '13 at 14:20
Correct - This is what 'vacuum flasks/ thermoses' try to imitate. –  user12345 Jan 27 '13 at 14:24
Well, "heat" is not radiation. But a body tranfers "heat" as radiation. –  user774025 Jan 27 '13 at 14:50

There are many ways to carry heat. The first is conduction, which is about the "vibration" of atoms on one material passing to another by simple physical contact. (Example: you touch something hot and get hurt). The second is convection: hot molecules simply move from one place to another (Example, water starts to boil in the bottom of a pan, but moves on the top because is lighter). The third is radiation and is precisely what you say: a warm body emits electromagnetic radiation. At "normal" temperatures (an oven, a human body), it's Infrared radiation, but it can be of higher frequency at higher temperatures, according to Planck's blackbody radiation law.

Notice, though, that the power emitted by radiation only is proportional to the fourth power of temperature. So the effect is very relevant in sun, but negligible for a human body. It should be around 500 W/m$^2$, which OK, is not small, but probably the most heat is transferred by conduction when the human is in air.

This is how garments work: they create a small layer of warm air around your skin, avoiding contact with constantly renewed cold air.

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Given that people are normally assumed to have a skin area of 2m^2 and at-rest produce around 100W I think a 1000W cooling rate would be considerable! –  Martin Beckett Jan 28 '13 at 2:38
Yes, actually I am wondering if it is not a bit too much. I just used Stefan-Boltzmann's law. However, a human body is not a black body... –  Bzazz Jan 28 '13 at 10:25
500W is correct for 37C and e=0.5. My guess is that unless you were in very cold water the radiation losses in space are higher than convection losses on Earth –  Martin Beckett Jan 28 '13 at 17:29
But I thought radiation wouldnt change between earth or space... or you mean that in the meanwhile you absorb radiation from the earth and air surrounding you? (It seems right, you can feel very well the temperature of a fireplace withoug getting close to it) In this case, we should know the difference between the absorption and emission coefficient of our body. –  Bzazz Jan 28 '13 at 19:07
Radiation less is normally approximated as temperature_diff^4. In fact you lose exactly the same rate, but also absorb the energy from the surroundings which is also emitting as t^4 - so on earth you don't notice. –  Martin Beckett Jan 28 '13 at 23:25

The body will lose heat when it's in vacuum in the same way the sum emits thermal radiation. In fact all matter with temperature above absolute zero emits radiation and human body is no exception.The power emitted by radiation per unit surface is given by the Stefan-Boltzmann law

$$j*= \epsilon \sigma T^4$$ where, $j*$ that the total energy radiated per unit surface area per unit time, $\sigma$ is the Stefan-Boltzman constant, $\epsilon$ is the emissivity of the body and $T$ is the absolute temperatue.

Also, "heat" is not radiation. Heat or thermal energy is the energy which matter posseses because of the random motion of its molecules. And temperature of a macroscopic body is the measure of the average kinetic energy of its molecules.

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