Posted by **Jamie** on Monday, December 3, 2007 at 12:30pm.

If the air temperature is the same as the temperature of your skin (about 30.0 C), your body cannot get rid of heat by transferring it to the air. In that case, it gets rid of the heat by evaporating water (sweat). During bicycling, a typical 70.0 kg person's body produces energy at a rate of about 500 W due to metabolism, 80.0 % of which is converted to heat.

How many kilograms of water must the person's body evaporate in an hour to get rid of this heat? The heat of vaporization of water at body temperature is 2.42*10^6 J/kg.

m=??kg

The evaporated water must, of course, be replenished, or the person will dehydrate. How many 750 mL bottles of water must the bicyclist drink per hour to replenish the lost water? (Recall that the mass of a liter of water is 1.00 kg.)

=?? bottles per hour

Our teacher hasn't gone over a problem this complex, and the book isn't any help for something this hard either...I'm extremely lost!

- Physics- LOST! -
**DrBob222**, Monday, December 3, 2007 at 1:24pm
Convert 500 watts to joules/sec.

That is 500 J/sec. The problem states about 80% is converted to heat so 500 J/sec x 0.8 = ?? heat produced in 1 sec.

How many J is that per hour.

500 J/sec x (60 sec/min) x (60 min/hour) = ?? J/hour.

Then q = mass H2O in kg x heat of vap of water at 30^{o} C. You have q and you have heat of vap of water, solve for mass H2O in kg.

Then convert to # bottles of water to replenish.

Post your work if you get stuck.

- Physics- LOST! -
**drwls**, Monday, December 3, 2007 at 1:28pm
It isn't that hard if you think it through one step at a time.

80% of 500 W converted to heat is 400 W (Joules per sec) In one hour, that is 1 hr x 400 J/s x3600 s/hr= 1.44*10^6 J to be removed

Each kg of water evaporated removes

2.42*10^6 J . So.. divide 1.44*10^6 J by 2.42 J/kg to get the number of kg of sweat to be evaporated. Each kg is one liter. For the number of bottles of water, divide the # of kg by 0.75 kg/bottle.

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