In exercising, a weight lifter loses 0.215 kg of water through evaporation, the
ID: 1471056 • Letter: I
Question
In exercising, a weight lifter loses 0.215 kg of water through evaporation, the heat required to evaporate the water coming from the weight lifter's body. The work done in lifting weights is 1.19 x 105 J. (a) Assuming that the latent heat of vaporization of perspiration is 2.42 x 106 J/kg, find the change in the internal energy of the weight lifter. (b) Determine the minimum number of nutritional Calories of food that must be consumed to replace the loss of internal energy. (1 nutritional Calorie = 4186 J).
I've tried working out this problem from others that were similarly posted and have not been able to get the correct answer. Please use the figures above in the problem. This is the complete question. If you need additional information for the question please leave it for someone else to answer. Thanks!
Explanation / Answer
(a) By Q = m x L
=>Q = 0.215 x 2.42 x 10^6 = 520300 J
Thus the change in the internal energy of the weight lifter[E] = W + Q = 1.19 x 10^5 + 520300
=>E = 639300 J
(b) Let the the minimum number of nutritional calories of food that must be consumed to replace the loss of
internal energy is = n
=>n x 4186 = E
=>n = [639300 ]/4186
=>n =152.723