Question

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.

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 2.42×10⁶J/kg

Answer 1

To dissipate the excess heat produced during an hour of bicycling, the body must evaporate approximately 0.5982 kg of water, calculated using the given metabolic energy production rate and the heat of vaporization of water.

Step-by-step explanation:

The student's question involves calculating the amount of water that needs to evaporate from a person's body during an hour of bicycling to dissipate heat, applying concepts from thermodynamics, specifically the energy transfer through the process of evaporation. Since we're given the rate at which the body produces energy (500 W, with 80% of that converting to heat) and the heat of vaporization of water at body temperature (2.42×106 J/kg), we can find the mass of water that needs to evaporate to remove the excess heat.

The heat produced by the body in an hour is 0.80 × 500 W × 3600 s (since 1 hour = 3600 seconds). To find the mass of water that needs to evaporate, we use Q = m×Lv, where Q is the energy produced, m is the mass of water to evaporate, and Lv is the heat of vaporization.

The calculation goes as follows:

• Q = m×Lv
• (0.80 × 500 W × 3600 s) = m × (2.42×106J/kg)
• m = (0.80 × 500 W × 3600 s) / (2.42×106J/kg)
• m ≈ 0.5982 kg

The person's body must evaporate approximately 0.5982 kg of water in an hour to get rid of the excess heat produced by metabolism during bicycling.