Final answer:
Approximately 0.071 kg of water must evaporate from a 55.6 kg woman to lower her body temperature by 0.745°C.
Step-by-step explanation:
To calculate the amount of water that must evaporate from a 55.6 kg woman to lower her body temperature by 0.745°C, we need to use the specific heat capacity of water and the heat of vaporization.
The formula to calculate the heat required to change the temperature of a substance is Q = mcΔT, where Q is the heat, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
In this case, we need to find the mass of water that will lose enough heat to lower the body temperature by 0.745°C.
We can use the formula Q = mL, where Q is the heat, m is the mass, and L is the heat of vaporization.
Since the heat required to evaporate water is given by the equation Q = mL, we can rearrange the equation to solve for the mass of water, m = Q/L. Plugging in the values, we have:
m = Q/L = mcΔT/L
where m is the mass of water, c is the specific heat capacity of water, ΔT is the change in temperature, and L is the heat of vaporization of water.
Plugging in the known values:
- m = (55.6 kg)(4.18 kJ/kg°C)(0.745°C)/(2260 kJ/kg) ≈ 0.071 kg
So, approximately 0.071 kg of water must evaporate from a 55.6 kg woman to lower her body temperature by 0.745°C.