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On average a human consumes around $2000$ kilocalories per day. This converts to roughly $2000000$ calories / $86400$ seconds or around $100$ joules / second giving roughly $100$ watts.

But if you use human's body temperature of 310 Kelvin, the Stefan-Boltzmann law $$P = e \sigma A T^4$$ says the power radiated by a human with a surface area of $2 \, \text{m}^2$ and emissitivity $1$ is 1000 watts.

What's wrong here?

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  • $\begingroup$ It is likely in your conversion to joules, you need to multiply by 4.18 in the conversion of calories to joules $\endgroup$ – Triatticus Sep 29 '18 at 19:24
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    $\begingroup$ This is exactly why I wear a sweater when it is cold outside. $\endgroup$ – safesphere Sep 29 '18 at 19:27
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    $\begingroup$ Notes: the unit kelvin has no degree associated; and it's watt not Watt. $\endgroup$ – Massimo Ortolano Sep 29 '18 at 22:17
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    $\begingroup$ Most of the skin area is colder than 37°C, especially if not shielded by clothes. Because of the ^4 exponent, this makes a quite big difference. $\endgroup$ – jpa Sep 30 '18 at 10:37
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    $\begingroup$ @jpa: It doesn't make much difference at low temperatures. At 32°C, the human skin would still radiate 960W. To only radiate 100W, the skin would need to be at 173K, that is, -100°C. $\endgroup$ – Eric Duminil Sep 30 '18 at 11:27
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You are forgetting that you also absorb radiation from the environment. The formula you want is

$$P_\text{net} = \epsilon A\sigma\left(T_\text{skin}^4 - T_\text{env}^4\right)$$

You can find more info on Hyperphysics: http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/bodrad.html

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    $\begingroup$ The question makes the implicit assumption that all radiated heat must be generated within the human. But if some heat is absorbed, not as much needs to be generated. A person in deep space would be insulated by the spacesuit so that the inner surface radiates some heat back, and the outer surface is much cooler and radiates less heat to space. $\endgroup$ – jpa Sep 30 '18 at 10:31
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    $\begingroup$ If you put a human in deep space without any insulation, wouldn't, they freeze to death? $\endgroup$ – Peter Shor Sep 30 '18 at 11:02
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    $\begingroup$ @coniferous_smellerULPBG-W8ZgjR: As far as I can tell, this answer is correct and addresses the question. The body only needs to "pay" for what's missing to achieve thermal equilibrium. When there's a deficit (i.e. the surrounding is colder than the human skin), the body needs to burn more food in order for the body temperature to stay constant. In a warm bath, you need to consume less energy than when laying in the snow. And naked in space, you'd need to eat 20 000 kilocalories/day. $\endgroup$ – Eric Duminil Sep 30 '18 at 11:43
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    $\begingroup$ If you use 32°C and 20°C, you get around 160W, which is the correct order of magnitude. Since we're (usually) not walking around naked, the body needs to burn less. $\endgroup$ – Eric Duminil Sep 30 '18 at 11:48
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    $\begingroup$ @coniferous_smellerULPBG-W8ZgjR So you are just talking about one part of the heat flow, not the net heat flow? $\endgroup$ – Aaron Stevens Sep 30 '18 at 11:53
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Short answer:

You are going wrong in assuming that the calorie intake accounts for all of the radiation of the human body. A human body really emits a power of roughly $1\ \mathrm {kW}$.

Long answer: As you've shown, any black body near room temperature with a surface of $\sim 2 \ \mathrm {m^2}$ emits in the ballpark of $1 \ \mathrm {kW}$. This means any object with that surface area and whose emissivity is near 1, emits such a power at room temperature (so even a dead body!). The extra $100\ \mathrm W$ due to calorie intake a human can make use of is usually mostly used in heating, because body temperature is usually at a higher temperature than room temperature. Nevertheless, if the room temperature is at a higher than 37°C (human body temperature), then this extra 100W due to calorie intake will be used to keep the body temperature near 37°C, for example by sweating.

In short, the extra calorie intake usually translate in emitting $\sim 1 \ \mathrm {kW}$ + $100 \ \mathrm W$. Thus we see that the calorie intake only accounts for about 10% of the total power radiated. That's what you were missing.

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  • $\begingroup$ @user27542 I've completely edited my answer, focusing on the important point(s) only. $\endgroup$ – thermomagnetic condensed boson Sep 30 '18 at 13:55
  • $\begingroup$ A dead body is warmer than surroundings for only a short time - it cannot keep emitting 1kW unless there is energy from food coming in. Eventually it has to be balanced, with energy coming in equaling energy going out. $\endgroup$ – jpa Sep 30 '18 at 14:09
  • $\begingroup$ @jpa it sure can. All it takes is a room temperature near 30°C. It's not about net power, it's about the power radiated. And it needs not to be warmer than the surroundings. You've missed point 1 of my answer. $\endgroup$ – thermomagnetic condensed boson Sep 30 '18 at 14:15
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    $\begingroup$ This is obviously not what the OP had in mind with his question. He was confused about how you could emit more energy than you take in, which is impossible. The reason that even a dead body emits energy is because it also absorbs energy from the environment. $\endgroup$ – HiddenBabel Sep 30 '18 at 14:54
  • $\begingroup$ I say, let the asker decides what he thinks answers best his question. I still believe I've done a good job in answering it. Cheers and thanks for all the comments. $\endgroup$ – thermomagnetic condensed boson Sep 30 '18 at 16:22
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What is wrong, is that you are using a formula applicable to an isolated system and applying it to a non-isolated system.

As it has been pointed out, a black body with the given parameters, radiates about 1,000 watts (as an isolated system). However, in an environment of the same temperature, it also absorbs 1,000 watts. So, the net radiation, is zero! Only when the environment is colder than the skin temperature, would the person need to use some of the 100 watts available from the food intake.

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protected by ACuriousMind Sep 30 '18 at 10:08

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