Question

How much sweat (in ml) would you have to evaporate per hour to remove the same amount of heat a 90.0 w light bulb produces? (1w=1j/s.)?

Answer 1

1 W = 1 J/s
Therefore;
Total energy = 90*1*60*60 J = 324000 J = 324 kJ

Such an evaporation must take place at a temperature near the body temperature. Assuming a body temperature of 37°C at which heat of vaporization is approximately 2413.1 kJ/kg (again assuming sweet behaves like water).

Then,
mC = 324 kJ, where m = mass of sweet and C = heat of vaporization
Therefore,
m = 324/C = 324/2413.1 = 0.13427 kg

Density = m/v => volume = m/Density = 0.13427/1000 = 1.3427*10^-4 m^3
1 m^3 = 1000 liters
Then,
1.3427*10^-4 m^3 = 0.13427 liters of sweet = 134.27 ml