How much sweat (in mL) would you have to evaporate per hour to remove the same amount of heat a 120 W light bulb produces? (1W=1J/s.)
To find out how much sweat would need to evaporate per hour to remove the same amount of heat as a 120 W light bulb, we need to know the amount of heat energy produced by the light bulb per hour.
The power output of the light bulb is given as 120 W, which means it produces 120 Joules of energy every second.
To calculate the amount of heat produced per hour, we multiply the power output by the number of seconds in an hour:
120 W x 3600 s = 432,000 J/h
Since 1 mL of water absorbs approximately 2.4 J of heat when it evaporates, we can calculate the amount of sweat (in mL) needed to absorb the same amount of heat:
432,000 J/h ÷ 2.4 J/mL = 180,000 mL/h
Therefore, you would need to evaporate approximately 180,000 mL (or 180 liters) of sweat per hour to remove the same amount of heat that a 120 W light bulb produces.
120 W = 120 J/s. For 1 hour that is
120 J/s x 60 s/hr = 7200 J.
7200 = mass H2O x delta H vaporization.
Substitute and solve for mass H2O (sweat)