In exercising, a weight lifter loses 0.175 kg of water through evaporation, the
ID: 1386558 • Letter: I
Question
In exercising, a weight lifter loses 0.175 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 1.58 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
U = Q + W (1st law of thermo)
U = (change) in internal energy
Q = heat flow in J
W = work in J
Q = m*L = 0.175kg * 2.42 e6 J/kg = -4.235 e5 J ( Negative because heat flowed out of the system )
Mechanical work done = -1.58 e5J. Negative because the weight lifter did work
U = Q + W = -4.235 e5 J -1.58 e5J = -5.815 e5J
PART B )
to compensate for the loss, the needed energy is minus the energy lost, which becomes +5.815 *10? J
number of calories=Total energy/energy in one Calorie
=+5.815 *10?/4186
=138.91