Final answer:
To cool an 80-kg person by 1°C through the evaporation of sweat, 80 grams of water would need to evaporate, not 539, 640, 800, or 400 grams as listed in the options. There is likely a typo in the provided choices.
Step-by-step explanation:
To determine the amount of water, in grams, needed to cool a person by 1 Celsius degree, we utilize the principle of heat transfer via the evaporation of sweat. The specific heat of water is 4.184 J/g°C. This means that to heat 1 gram of water by 1°C, 4.184 joules of energy are required. Conversely, the evaporation of 1 gram of water from the body would absorb the same amount of energy from the body, thereby cooling the body.
Using the given information, if an 80-kg person can be cooled by 1°C by evaporating 1 gram of water, then to cool the same person by 1°C, we'd need the equivalent amount of water in grams to their weight in kilograms, assuming a direct 1:1 gram-to-kilogram ratio. Therefore, to cool an 80-kg person by 1°C requires the evaporation of 80 grams of water.
However, since the options provided do not include 80 grams, it appears there might be a typographical error in the question. The correct answer, following the logic and proportional relationship stated, would be 80 grams, but based solely on the provided options, none of them are correct.