Question

A Sweaty Weight Lifter

Due this Friday, Dec 4 at 11:59 pm (EST)

In exercising, a weight lifter loses 0.142 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 1.38E+5 J. Assuming that the latent heat of vaporization of perspiration is 2.45E+6 J/kg, find the change in the internal energy of the weight lifter.

Answer 1

Calculating the heat lost by the body in the form of water = mass of water × latent heat of vaporization

                                                                                                = 0.142 Kg × 2.45E+6 J/kg

                                                                                                = 3.479E+5 J

Since heat is given out by the body it is taken negative according to sign conventions

Therefore, Q = -3.479E+5 J

As work is done by the body it is considered positive

W = + 1.38E+5 J

Change in internal energy = (delta) U = Q + W

                                                                = -3.479E+5 J + 1.38E+5 J

                                                                = -2.099E+5 J

As (delta) U is negative we can say that internal energy of the body decreases