Question

If the clothes you are wearing absorb 1.00 kg of water and then dry in a cold wind on Mount Washington, how much heat would your body lose during this process? ​cp,H2O = 75.3 J/(mol*C) and delta H vap, H2O = 40.67 kJ/mol. If the heat lost  by your body is not replaced, what will the final temperature of your body be after the 1.00 kh. of water has evaporated.The hiker is 160 lbs and the specific heat of their body is 4.25 J/g.

Answer 1

mass of water absorbed by clothes = 1 kg = 1 Kg x (1000 g / 1 Kg) = 1000 g

moles  of water absorbed by clothes = mass / molar mass = 1000 g / 18 g/mol = 55.56 mol

Hence amount of heat lost from our body to evaporate 55.55 mol water

= 40.67 kJ/mol x 55.56mol = 2260 KJ (answer)

Now 2260 KJ of heat will be lost from our body.

inital temperature of body, Ti = 37 DegC

Let the final temperature be T DegC

mass of body, m = 160 lbs = 160 lbs x (453.6 g / 1 lb) = 72576 g

Hence heat lost by body = mS(body)xT = - (heat gained by water)

=> 72576 g x 4.25 J/g x (T - 37) = - 2260 KJ = 2260000 J

=> (T - 37) = - 7.33 DegC

=> T = 29.67 DegC (answer)