Final answer:
Approximately 226 grams of water need to evaporate from the skin to cool a body of 86.0 kg by 1.60°C, assuming a specific heat capacity of the body is 4.00 J/g°C.
Step-by-step explanation:
To estimate the mass of water that must evaporate from the skin to cool the body by 1.60°C, we first need to calculate the amount of heat that needs to be removed from the body. We use the formula Q = mcΔT, where Q is the heat, m is the mass of the body, c is the specific heat capacity, and ΔT is the change in temperature. Given that the body mass m is 86.0 kg and the specific heat capacity c is 4.00 J/g°C, and the temperature change ΔT is 1.60°C, we first convert the body mass to grams (86000 grams), then calculate the heat as follows:
Q = (86000 g)(4.00 J/g°C)(1.60°C)
Q = 5.488×105 J
Next, we need to know how much heat is required to evaporate 1 kg of water at body temperature. From the example given, the heat input required for evaporation of sweat at body temperature is 2428 kJ/kg. To find the mass of water that must evaporate, we divide the total heat by the heat input per kilogram:
mass of water = Q / (heat input per kilogram)
mass of water = (5.488×105 J) / (2428×103 J/kg)
mass of water = 0.226 kg or 226 grams of water
Therefore, approximately 226 grams of water need to evaporate to cool the body by 1.60°C.