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/

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

At maximum height, kinetic energy is converted in potential energy  formula: at max height, ke = ½ * 0.27kg * (16m/s) ² = 34.56 j  so the remaining pe = (41.70 – 34.56)= 7.14 j = 0.27 kg * 9.8m/s² * h  = (9.8/7.14 * 0.27)so max height h = 5.08 or 5 m