Posted by **Mary** on Wednesday, April 18, 2007 at 11:16pm.

In exercising, a weight lifter loses 0.100 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 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 (1 nutritional calorie = 4186 J) that must be consumed to replace the loss of internal energy.

It would be better if you could post your answer first so that we can help. This is not a forum to do the homework for you.

If mass of water lost =m and latent heat =L

then energy used to evaporate water is mL

If work done in lifting weights is W then the total energy used is mL+W

Assuming that this is the energy that needs to be replaced then the number of calories is (mL+W)/4186

If you want to post your calculations we can check these.

## Answer This Question

## Related Questions

- Thermodynamics - In exercising, a weight lifter loses 0.100 kg of water through ...
- Physics - In exercising, a weight lifter loses 0.236 kg of water through ...
- Physics - In exercising, a weight lifter loses 0.236 kg of water through ...
- Physics please check repost - In exercising, a weight lifter loses 0.100 kg of ...
- Biomechanics - An 88.9 kg weight lifter is required to perform a snatch movement...
- Physics - A weight lifter is bench-pressing a barbell whose weight is 710 N. He ...
- Physics - The box of a well-known breakfast cereal states that one ounce of the ...
- physics - A weight lifter lifts a set of weights a vertical distance of 2.07 m. ...
- biomechanics - When a weight-lifter lifts a weight equal to his upper body ...
- biomechanics - When a weight-lifter lifts a weight equal to his upper body ...

More Related Questions