(a) How many kilograms of water must evaporate from a 60.0 kg woman to lower her body temperature by 0.750°C?
For human body c=3470 J/kg.
At 37.0ºC the heat of vaporization Lv for water is 2430 kJ/kg or 580 kcal/kg
m•c•Δt= m(cond) •L(v)
m(cond) = m•c•Δt/L(v) =
=60•3470•0.75/2430000 =6.43•10⁻² kg=64.3 g
posted by Elena