If the air temperature is the same as the temperature of your skin (about 30.0 oC), 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 70.0kg person's body produces energy at a rate of about 67.3 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 3.7 ¡Ñ106 J/kg.

Pt (.80) = mL

You have to convert hours and find convert mass of H2O into liters.

To solve this problem, we need to calculate the amount of heat that needs to be dissipated and then use the heat of vaporization of water to find the mass of water that needs to be evaporated.

1. Calculate the amount of heat that needs to be dissipated:
- Given: Energy produced by metabolism = 67.3 W
- Conversion efficiency to heat = 80%
- Heat produced = Energy produced x Conversion efficiency
- Heat produced = 67.3 W x 0.8 = 53.84 W (watts)

2. Convert the heat produced from watts to joules per hour:
- 1 watt (W) = 1 joule per second (J/s)
- 1 hour = 3600 seconds
- Heat produced in joules per hour = Heat produced in watts x Time in seconds
- Heat produced in joules per hour = 53.84 W x 3600 s = 193,824 J/h

3. Use the heat of vaporization to find the mass of water:
- Heat of vaporization of water at body temperature = 3.7 x 10^6 J/kg
- Mass of water evaporated = Heat produced / Heat of vaporization
- Mass of water evaporated = 193,824 J/h / (3.7 x 10^6 J/kg)
- Mass of water evaporated ≈ 0.052 kg

Therefore, the person's body needs to evaporate approximately 0.052 kilograms (or 52 grams) of water in an hour to get rid of the heat generated.