Final answer:
Approximately 13 kg of water per hour are needed to be circulated in the LCVG to remove 2.0 MJ of heat generated by an astronaut, as calculated using the formula for thermal energy transfer with a specific heat capacity for water and a temperature change from body to freezing temperature.
Step-by-step explanation:
To determine how much water must be circulated in the liquid cooling and ventilating garment (LCVG) to remove 2.0 MJ of heat per hour, we need to calculate the mass of the water that can carry this amount of heat by using the specific heat capacity of water and the temperature change. The specific heat capacity of water is 4.186 J/g°C, and the temperature change is 37°C (from body temperature down to 0°C).
The heat quantity (Q) in joules is equal to the mass of the water (m) in grams, multiplied by the specific heat capacity (c) of water, and the temperature change (ΔT): Q = mcΔT.
Rearranging the formula to solve for the mass, we get m = Q / (cΔT).
Substituting the given values:
- Q = 2.0 MJ = 2,000,000 J
- c = 4.186 J/g°C
- ΔT = 37°C
m = 2,000,000 J / (4.186 J/g°C × 37°C) = 2,000,000 J / (154.882 J/g) ≈ 12,916 g or 12.916 kg per hour.
Therefore, the correct answer is approximately 13 kg/hr (Option A).