How Does Latent Heat Affect Sweat Evaporation in Cooling the Body?

[BOAS]

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^-1C°^-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!

 

[gneill]

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.

 

[BOAS]

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]

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

 

[Nickie]

Hello,

Thank you for your question about latent heat and its application in solving a problem. Latent heat refers to the amount of energy required to change the state of a substance without changing its temperature. In the case of water, the latent heat of vaporization is the amount of energy required to change water from a liquid state to a gas state at a given temperature. In this case, it is at body temperature (37°C).

To solve the problem, we can use the formula Q = mcΔT, where Q represents the amount of energy, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature. In this case, we are trying to cool the body of a 75 kg jogger by 1.5°C. Therefore, using the given specific heat capacity of 3500 J Kg-1C°-1, we can calculate the amount of energy required as follows:

Q = (75 kg) (3500 J Kg-1C°-1) (1.5°C)
Q = 393,750 J

Now, we need to take into account the latent heat of vaporization of water. Since we are trying to cool the body by evaporating sweat, we need to calculate the amount of energy required to vaporize the sweat. This can be done by multiplying the mass of sweat by the latent heat of vaporization of water at body temperature (2.42x106 J Kg-1).

Therefore, the total amount of energy required to cool the body by 1.5°C is:

Total Q = 393,750 J + (m kg) (2.42x106 J Kg-1)

To solve for the mass of water (in kg), we can rearrange the equation as follows:

(m kg) = (Total Q - 393,750 J) / (2.42x106 J Kg-1)

Substituting the values, we get:

(m kg) = (393,750 J + (m kg) (2.42x106 J Kg-1)) / (2.42x106 J Kg-1)

Solving for m, we get:

m = 0.162 kg

Therefore, to cool the body of a 75 kg jogger by 1.5°C, approximately 0.162 kg of water in the form of sweat needs to be evaporated.

I hope this helps clarify