# Question about Phase Change & Latent Heat

1. Feb 14, 2019 at 5:00 AM

### IshaanM8

1. The problem statement, all variables and given/known data

When running, some (heat) energy is lost by the evaporation of sweat. If a runner loses 1.0 MJ( Mega Joules) of energy via sweating, calculate an approximate mass of sweat evaporated from the runner.

2. Relevant equations
Q = McΔT
Q (Joules) = M (Mass) c (Maximum Heat Capacity) ΔT (Change in Temperature)

Evaporation (Phase Change - Latent Heat of Vaporization)
22.6 x 105

Maximum Heat capacity of Human body is 3500 Joules

3. The attempt at a solution

I'm sorry, I don't have an attempt here because I think we're missing something.

We have Q & c
We don't have the mass of the human body or the temperature change. Don't we need 2 equations minimum to solve 2 unknowns?

Is this question even solvable? I mean, how? The answer given was 0.4 kg. But how though...

2. Feb 14, 2019 at 5:06 AM

### stockzahn

Ahoihoi @PF!

- Water (sweat) is an azeotropic fluid. What does that mean for the temperature during evaporation?
- The mass $M$ doesn't correspond to the mass of the human body - assuming a sensible heat transfer, what would $M$ stand for?

3. Feb 14, 2019 at 6:32 AM

### IshaanM8

I'm sorry but I don't know either of those terms. It's only first 3 weeks of year 11 physics.

4. Feb 14, 2019 at 6:41 AM

### Staff: Mentor

That's the wrong equation. This is for a change in heat due to a difference in temperature. There is a different equation when considering phase transformations.

5. Feb 14, 2019 at 6:55 AM

### stockzahn

Azeotropic fluids are "pure" fluids. If they are vaporised, the temperature does not increase (given a constant pressure, which can be assumend when sweating). I just wanted to point out that a temperature difference isn't necessary (or even possible) in your scenario, so you have to find another equation (what @DrClaude wrote directly). You need what you called the latent heat of evaporation - it is called latent (=hidden), since during the heat transfer the temperature does not change (compared to the sensible heat transfer with temperature change). The latent heat of evaporation (or specific evaporation enthalpy) is what the specific heat capacity is for the sensible heat transfer - for the first you just don't need a temperature difference, because it doesn't exist.

Do you know the units of the specific heat capacity $c$ and the specific evaporation enthalpy $r$? If you write them down, the difference will become obvious.

And still: What do you think the symbols $M$ stands for?

6. Feb 14, 2019 at 7:34 AM

### IshaanM8

Q = mL ?
Q is Energy
m is mass
L is latent heat

7. Feb 14, 2019 at 7:41 AM

### IshaanM8

Well, I think that units for the specific heat capacity for the human body is 3500 Joules / KgK and for water is 4200 Joules / KgK. Latent Heat of Evaporation would be 22.6 x 105Joules / Kg.
The term $M$ would be for the mass of water in the body? In kilograms?

8. Feb 14, 2019 at 7:49 AM

### stockzahn

Well, then you can see the difference, one of the properties has the Temperature below the fraction bar, the other not.

Suppose you have a coolant extracting a certain amount of heat from an object, e.g. 1 MJ. Does it matter what's the mass of the cooled object if you already know the heat you have to deal with?

9. Feb 14, 2019 at 7:52 AM

### IshaanM8

Now I look at it that way, no, you don't.

10. Feb 14, 2019 at 8:00 AM

### stockzahn

Then let's take your equation for the sensible heat transfer: $Q = mc\Delta T$. As you've already stated yourself, for a latent heat transfer (evaporation), this equation changes to: $Q = mr$. You know the heat and the evaporation enthalpy, therefore $m=\frac{Q}{r}$. It says, the larger the heat ($r=const.$), the larger the mass $m$. Now just think about sweating: If you have to extract mort heat from your body, which mass has to increase?

11. Feb 14, 2019 at 8:02 AM

### IshaanM8

The water content of sweat, right?

12. Feb 14, 2019 at 8:05 AM

### stockzahn

Of course. Obvious, isn't it?

13. Feb 14, 2019 at 8:07 AM

### IshaanM8

Dang, I think I understand how to do this a bit. Thanks for giving this thread so much of ur time!

14. Feb 14, 2019 at 8:15 AM

### stockzahn

The specific evaporation enthalpy $r$ tells you how much heat has to be put into a fluid to evaporate 1 kg of it. The more heat $Q$ you have to cool away, the more mass $m$ has to be evaporated.

Better to understand it entirely. If you have more questions, just ask.