Answer: Evaporative Cooling: Body Temp Drops 0.05°C w/ 10g Water

[Qube]

Homework Statement

By how many degrees will your body cool if you evaporate away 10g of water? Specific heats of water and water vapor are, respectively, 4.18 J/gC and 2.00 J/gC. Heat of vaporization of water is 40.7 kJ/mol. Heat of fusion of water is 6.03kJ/mol. Assume a 100kg person who is mostly water.

Homework Equations

The heat of fusion and specific heat of water vapor are irrelevant. We don't have a person full of hot air in this hypothetical.

The Attempt at a Solution

First we calculate the heat of vaporization of 10g of sweat. The units are kJ/mol so we convert 10g to moles of water (10/18) and multiply by 40.7kJ to get the number of kJ, which is about 22 kJ.

Then we find how much the body is cooled. Convert kg of the body to grams since specific heat is given in terms of Joules per gram*Celsius. Convert kJ to Joules.

q = mcΔt = 100(1000) * 4.184 * Δt = 22,000 J

The change in temperature is about 0.05 degrees Celsius (I solved for delta T).

Question:

1) Am I correct in ignoring the specific heat of water vapor and the heat of fusion in the problem? We are, after all, talking about a body which is mostly water. Are these other constants just distractions?

 

[Borek]

Am I correct in ignoring the specific heat of water vapor and the heat of fusion in the problem?

Yes.