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Approximately how quickly will the temperate increase per second with an adult at rest inside a perfectly insulated room of say 100 ft3? I get about 1.6 degrees per minute, others get about 2 degrees per minute, but at those rates the person inside would be boiling water in about an hour with only his body temperature, are those rates roughly right?

This is the formula I'm using: ΔT = energy/((mass)(specific_heat))

Using about 100W for the person's energy output * 60 seconds. The aproximate mass of air in 100ft3 is 3661.1 grams. The specifiv heat of air is 1.01 J/g C.

ΔT = 100*60/(3661.1*1.01) = 1.622 deg/min

If that's right then the temperature would rise from 0 to 97 degrees celsius in about an hour, is that possible?

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closed as off-topic by AccidentalFourierTransform, stafusa, sammy gerbil, John Rennie, David Z Dec 2 '17 at 8:35

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    $\begingroup$ That's more of a box than a room. 100 cubic feet would be a cube which is around 4.6 feet per side. $\endgroup$ – J. Murray Dec 1 '17 at 21:00
  • $\begingroup$ Yeah, the room (cube) size is just to make calcs easier. But say the person generates 100W of heat energy, we could find some constant temperature increase rate even if in reality it's not that uniform. $\endgroup$ – Haalef Dec 1 '17 at 21:26
  • $\begingroup$ What is your difficulty in doing this calculation yourself? $\endgroup$ – sammy gerbil Dec 2 '17 at 4:05
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The heat capacity at constant pressure of nitrogen is about 7/2 R = 29.1 J/K per mole. Very roughly, the molar volume is a cubic foot. So the heat capacity of the air in the box is about 3 kJ/K.

Heat production is about 100 watt = 100 J/s = 6 kJ/min. So the temperature will rise about 2 degrees per minute.

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  • $\begingroup$ 2 degrees per minute sounds like a little bit too much, the guy would be boiling water in less than an hour using only his body temperature. Solving for temperature in the formula (q = mcΔT taken from thoughtco.com/heat-capacity-final-temperature-problem-609496) I get 0.0023 degrees per minute wich sounds more realistic. I get about 2 degrees per minute with that formula if I don't convert the mass of air to grams. $\endgroup$ – Haalef Dec 2 '17 at 0:00
  • $\begingroup$ @Haalef The mass of air is about 100 moles or about 3 kg. Of course the guy will never boil water. The estimate of 100 watts is a bit high but after about 10 or 15 minutes he will feel severe discomfort, when humidity and temperature make it difficult to dispose of his body heat. We have a 1 m$^3$ styrofoam box in which we put student volunteers for about 5 minutes. $\endgroup$ – Pieter Dec 2 '17 at 0:26
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    $\begingroup$ Yeah, I get what you mean. I've made a mistake with the specific heat of air, it's about 1.01 J/gC and the mass in 100 ft3 is about 3661.1 grams. So now I get 1.62 degrees Celsius per minute. I still think it's too much, but I'll go with that for now. $\endgroup$ – Haalef Dec 2 '17 at 0:54

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