[viet]

Evaporating sweat cools the body because evaporation is an endothermic process: H₂O (l) => H₂O (g)  /\H_rxn = +44.01 kJ

Estimate the mass of water that must evaporate from the skin to cool the body by 0.50C. Assume a body mass of 85 kg and assume that the specific heat capacity of the body is 4.0 J/g*C.

m = ____ g

heres what i have:
q = m * C_s * /\T
q = 85000 * 4.0 J/gC * 0.50C
q = 170,000 J

im not sure what to do next, is there a way to solve this problem without the heat of vaporization formula? havent learn that yet.

[Astrokel]

q = 170,000 J

If you required this amount of energy to evaporate and that per mole of evaporation reaction is +44.01KJ, how many moles of water required, thus its mass?

[viet]

so i divide 170,000 J by 44.01 kJ:  170000/44.01 = 3862.76 K. how do i get it in to grams?

[Astrokel]

170000/44.01 = 3862.76

check your unit

how do i get it in to grams?

mole ---> mass

[viet]

is this right?

170000/ 44010 = 3.9
3.9 (18.01g H₂O/1mol) = 69.6g

[Astrokel]

Yes, correct, maybe you want to use more sig fig(3-4) in your intermediate working.