In exercising, a weight lifter loses 0.185 kg of water through evaporation, the
ID: 2071748 • Letter: I
Question
In exercising, a weight lifter loses 0.185 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 1.64 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
Latent heat change for perspiration: 2.42 x 106 x 0.185 = 47.45 J = q
Let work done = w
Let change in internal energy = u
Now, using conservation of energy,
u = w - q
Hence, u = 172.2 - 47.45
u = 124.75 J (lost)
Number of nutritional calories reqd.: u/4186 = 0.03
(Assuming the 105 and 106 are not 10^5 and 10^6. In case they are, just redo the calculations with the exponents)