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 71.0 kg person’s body produces energy at a rate of about 502 w due to metabolism, 82.0 % of which is converted to heat. [recall that the normal internal body temperature is 98.6 ∘f and the specific heat capacity of the body is 3480 j/(kg⋅∘c) .] 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 \times 10^6\; {\rm j/kg}.the evaporated water must, of course, be replenished, or the person will dehydrate. how many 750 {\rm 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 {\rm kg}.)

m=0.612kg

0.8164 bottles

Explanation:  the last part of the question says

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 \times 10^6\;{\rm J/kg}.The evaporated water must, of course, be replenished, or the person will dehydrate. How many 750 {\rm 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 {\rm kg}.)

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.

502 J/sec x (60 sec/min) x (60 min/hour)

1807200J

Rcallit is 82% efficient

1481904J

q=mlv

q=heat produced by the body in Joules

m=mass of water loss

lv=latent eat of vaporization in

1481904J=m*2.42*10^6j/kg

m=0.612kg

the same as 0.612L

0.612l/0.750L/bottle

0.8164 bottles