Answer:
m=0.612kg
0.8164 bottles
Explanation: Â 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