Homework Help: Physics please check repost

Posted by Mary on Thursday, April 19, 2007 at 7:40am.

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.

This is what i've done so far but I know it is wrong because I don't know where the weight of the water lost comes in.

a)
delta U = Q -W
delta U = (2.42 x 10^6) - (1.30 x 10^5)
delta U = 2290000 J

b)

if
1 cal = 4186 J

then
547.062 cal = 2290000 J

The heat loss Q is too high. You multiply 0.1 kg by 2.42 x 10^6 J/kg. That will give you a different delta U.

You did part b correctly but the question is misleading. UYse the correct delta U next time. Additional Calories are needed to maintain body heat and blood flow and other metabolic processes. The number you get will be low, so perhaps that is why they call it the minimum.

Physics please check repost - jim, Friday, November 21, 2008 at 2:08pm
you're on the right track.. try this
Q=(.1kg)*(2.42x10^6)
delta U = -Q - W
delta U should be negative.

for part b. it's just Q/4186

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