Bicycling on ?a ?Warm Day?. If the air temperature is the same as the temperature of your skin (about 30°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-kg person’s body produces energy at a rate of about 500 W due to metabolism, 80% of which is converted to heat. (a) 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 × 106 J/kg. (b) 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 1.0 kg.)

Solution to 42E Step 1 Rate of production of energy =500W Percentage of energy converted to heat =80% Amount of energy converted to heat = 500x0.8=400W Step 2 (a) 6 Heat of vaporization of water =2.42x10 J/kg Thus 2.42x10 Joules of heat is required for vaporizing 1kg of water. 6 Energy to be removed in 1h=400/3600 =1.44X10 J Therefore, Mass of water to be vaporized,M =1.44X10 J/2.42x10 J/kg 6 M=0.59kg