The latent heat of vapourisation of H2O at body temperature (37°C) is 2.42x106 J Kg-1. To cool the body of a 75 Kg jogger (average specific heat capacity = 3500 J Kg-1-1) by 1.5 °C, how many Kilograms of water in the form of sweat have to be evaporated?

Q = mcΔT

I'm confused about what to do with the information regarding latent heat. Is it saying that this is how much energy you need to put into one kilogram of water before it will begin evaporating? and if it is, would I just add this value (per kilogram) onto the amount of energy require to raise x kilograms of water to 100°C?

I'd be really grateful if anyone can clear up any misunderstandings I have about what Latent heat is and how I would use it to solve a problem. I'd appreciate it if you left the actual question I quoted un-solved.

Thanks!

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gneill
Mentor
Hi Boas. Liquid water can evaporate directly to water vapor without raising the entire body of water to 100C first. The problem statement indicates that evaporating 1kg of water at 37C requires 2.42x106 Joules.

The energy to cause this evaporation comes from the body of the jogger. Your task is to determine how much water must be evaporated to drop the jogger's body temperature by 1.5C. I think you can assume that the latent heat of evaporation is essentially constant over the 1.5C change in temperature.

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So from the Q = mcΔT equation, I can work out how much energy the body must dump to drop it's temperature by 1.5°C and I know that the method it uses to get rid of the energy is sweating.

So I just need to work out how many kilograms of water can be evaporated by that amount of energy?

gneill
Mentor
So from the Q = mcΔT equation, I can work out how much energy the body must dump to drop it's temperature by 1.5°C and I know that the method it uses to get rid of the energy is sweating.

So I just need to work out how many kilograms of water can be evaporated by that amount of energy?
That's it, in a nutshell 