Final answer:
To cool an 84 kg body by 1.30°C, approximately 0.1574 kg of water must evaporate, based on the specific heat capacity and the heat required for evaporation at body temperature. This calculation yields a result significantly lower than the provided options, suggesting a possible error in the question or answer choices.
Step-by-step explanation:
To estimate the mass of water that must evaporate from the skin to cool the body by 1.30 °C for a person with an 84.0 kg body mass, first one must calculate the amount of heat that needs to be removed from the body using the specific heat capacity.
The formula to calculate the heat lost or gained is Q = mcΔT, where Q is the heat energy, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. Given this person's mass (84000 g) and the specific heat capacity (3.5 J/g°C), the body will lose:
Q = (84000 g)(3.5 J/g°C)(1.30 °C) = 382,200 J
Now, knowing that the heat required to evaporate 1 kg of water at body temperature is approximately 2428 kJ/kg (or 2,428,000 J/kg), we can calculate the mass of water that needs to evaporate:
m = Q / Lv = 382,200 J / 2,428,000 J/kg = 0.1574 kg
The closest answer choice to our calculation is option A which is 2.94 kg. However, our calculated value is significantly lower than all the provided options, so an error might have occurred either in the formulation of the question or the provided answer choices.
Therefore, the most accurate estimate based on this calculation is that approximately 0.1574 kg of water must evaporate to cool the body by 1.30 °C.