Final answer:
The total heat required to heat a 20-ounce bottle of water from 3.8 °C to body temperature and then evaporate it at body temperature is approximately 1,516,000 Joules. This calculation involves using the specific heat capacity of water and the latent heat of vaporization at body temperature.
Step-by-step explanation:
To calculate the heat needed to convert all of the 20-ounce bottle of water into sweat and then to vapor, we must consider two stages: heating the water from 3.8 °C to body temperature (36.6 °C) and then evaporating it at body temperature.
Using the specific heat capacity for water (4.18 J/g°C), the mass of the water (20 ounces is approximately 591.47 g), and the latent heat of vaporization at body temperature (2428 kJ/kg), we can calculate the total energy required.
First, we calculate the energy required to heat the water from 3.8 °C to 36.6 °C:
Q = m • c • ΔT
Q = 591.47 g • 4.18 J/g°C • (36.6 °C - 3.8 °C)
Q = 591.47 g • 4.18 J/g°C • 32.8 °C
Q = 79861.24 J
Then, we calculate the energy needed to vaporize the water at body temperature:
Qvap = m • Lv
Qvap = 591.47 g (0.59147 kg) • 2428 kJ/kg
Qvap = 0.59147 kg • 2428 kJ/kg • 1000 J/kJ
Qvap = 1436137.56 J
Finally, we sum both energies to find the total heat required:
Total Heat = Q + Qvap
Total Heat = 79861.24 J + 1436137.56 J
Total Heat = 1515998.80 J
Therefore, the closest answer to the energy required to heat and vaporize the entire 20-ounce bottle of water is 1,516,000 Joules, which is not explicitly listed in the provided options.