(adsbygoogle = window.adsbygoogle || []).push({}); 1. In exercising, a weight lifter loses 0.130 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 1.30 multiplied by 10^5 J.

(a) Assuming that the latent heat of vaporization of perspiration is 2.42 multiplied by 10^6 J/kg, find the change in the internal energy of the weight lifter.

_________J

(b) Determine the minimum number of nutritional calories of food (1 nutritional calorie = 4186 J) that must be consumed to replace the loss of internal energy.

_______nutritional calories

2. Q = m x L

n x 4186 = ∆E

3. (a) By Q = m x L

=>Q = 0.13 x 2.42 x 10^6 = 3.15 x 10^5 J

Thus the change in the internal energy of the weight lifter[∆E] = W + Q = 1.30 x 10^5 + 3.15 x 10^5

=>∆E = 4.45 x 10^5 J

(b) Let the the minimum number of nutritional calories of food that must be consumed to replace the loss of internal energy is = n

=>n x 4186 = ∆E

=>n = [4.45 x 10^5]/4186

=>n = 106.31

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: N exercising, a weight lifter loses 0.130 kg of water through evaporation

Can you offer guidance or do you also need help?

Draft saved
Draft deleted

**Physics Forums | Science Articles, Homework Help, Discussion**