physics!
posted by alex on .
while running a marathon, alex is producing about 900 calories per hour. 1 calorie is 4184 j. if the specific heat of the human body is 3500 j/kg x c, how many degrees celsius would alexs temperature increase if he did not sweat and ran the marathon in 3 h and 10 min? he has a mass of 75 kg.
thank god for sweat. if alex wants to release all of the caloric energy he produced so he doesnt heat up, how many litres of sweat would he need to produce and evaporate? the heat of vaporization for the water in sweat is 2.26 x 10^6 j/kg. the density of water is 1 g/ml

You are talking about kilocalories, not calories. (It is written as capitalized Calories when talking about nutrition and metabolic rates)
In a 3.17 hour race, the body heat generated will be
Q = 3.17h*900Cal/h*4184J/Cal = 1.19*10^7 Joules
Temperature would increase by
delta T = Q/M*C = 45.5 C
For the second question, divide Q by the heat of vaporization in J/kg. Then convert that mass to volume in liters, using the density.