In exercising, a weight lifter loses 0.100 kg of water through evaporation, the
ID: 1985370 • Letter: I
Question
In exercising, a weight lifter loses 0.100 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 1.75 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).Explanation / Answer
Energy added from the work is 1.79 x 10^5
Removed from sweat is 0.168 x 2.42 x 10^6
change in internal energy is 1.79 x 10^5 - 0.168 x 2.42 x 10^6
= -227560 J
we know 1 kcal=4186
= 227560 /4186
= 54.36 K cal