Question

One of the challenges of spaceflight is keeping astronauts cool. The suits worn by the Apollo astronauts when they walked on the moon were heavily insulated to protect them from the lunar environment. But, that meant the suits needed some sort of cooling system to radiate away the astronauts' body heat. NASA solved this problem using a system that circulated water through pipes that lined the suit. The water was cooled by a system that allowed the water to evaporate away into the vacuum of the lunar surface. If we assume the astronaut was generating heat at a rate of 120 watts (in other words, 120 joules/second) what is the minimum amount of water, in grams, that the cooling system would have to carry in order to last for 3.0 hours of activity on the lunar surface? (Note: the water did not have to reach the boiling point in order to evaporate. It just evaporated because it was exposed to the vacuum of space. So, don't worry about changes in temperature: Just focus on the heat absorbed when water vaporizes.)

Answer 1

(We know that specific heat capacity) a of water s 4.2 J/kg per gram generates heat here, body at rate of 120 J/ seconds Total heat in 3 hours = 120x60x60x3 > 12960000 grams of water needed 7 1296000, 308571.42 grami to absorb the heat s 4.2