Final answer:
Using the specific heat capacity formula and the stated specific heat, we calculated that 95000 g multiplied by 4 J/g°C multiplied by 0.50°C equals 190000 J of energy to be removed. Then, using the enthalpy of vaporization at approximately 2400 J/g, we found that 79.16 g of water must evaporate, but this does not match any of the provided answer options.
Step-by-step explanation:
To estimate the mass of water that must evaporate from the skin to cool the body by 0.50°C, we need to calculate the amount of heat that needs to be removed from the body and then connect this to the enthalpy of vaporization for water. First, we calculate the heat removed using the specific heat capacity formula, Q = mcΔT, where 'm' is mass, 'c' is specific heat capacity, and ΔT is the change in temperature.
For a 95 kg person and a body temperature change of 0.50°C:
Q = m × c × ΔT
Q = 95000 g × 4 J/g°C × 0.50°C
Q = 190000 J
To find the mass of water that needs to evaporate, we use the heat of vaporization. The enthalpy of vaporization is
Q = mass × Hvap
190000 J = mass × 2400 J/g
mass = 190000 J / 2400 J/g
mass ≈ 79.16 g
However, none of the provided options (a) 38,000 grams, (b) 11,250 grams, (c) 47,500 grams, and (d) 19,000 grams are close to the calculated mass. Therefore, there may be an error in the options given, as the calculation based on the specific heat capacity and enthalpy of vaporization indicates a different result.