Question

In exercising, a weight lifter loses 0.110 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 1.49 x 10^5 J. (a) Assuming that the latent heat of vaporization of perspiration is 2.42 x 10^6 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).

Answer 1

If the latent heat of vaporization of perspiration is 2.42 x 10^6 J/kg, then after loosing 0.110 kg of water through evaporation the weight lifter loses the following amount of energy:

$$Q_{evap} = 2.42\times10^6J/kg\cdot 0.11kg = 2.662\times10^5J$$

Total change in the internal energy will be then:

$$Q = Q_{evap} + 1.49\times 10^5J = 4.152\times 10^5J$$

As far as 1 nutritional Calorie = 4186 J, it is required the following amount of Calories:

$$4.152\times 10^5J/4186 \approx 99.16\space Calories$$

Answer. 99.16 Calories.