Question

How many kilograms of water must evaporate from a 60.0-kg woman to lower her body temperature by 0.750ºC?

a) 0.25 kg
b) 0.50 kg
c) 0.75 kg
d) 1.00 kg

Answer 1

To lower the body temperature of a 60.0-kg woman by 0.750ºC, approximately 0.750 kg of water must evaporate.

Step-by-step explanation:

To determine the amount of water that must evaporate from a 60.0-kg woman to lower her body temperature by 0.750°C, we can use the specific heat capacity equation:

q = mcΔT

Where:
q = heat transfer
m = mass of the water
c = specific heat capacity of water
ΔT = change in temperature

Since we want to calculate the mass, rearranging the equation:

m = q / (cΔT)

Substituting the given values:
m = (60.0 kg)(4186 J/kg·°C)(0.750°C) / (4186 J/kg·°C)

Simplifying the equation gives the answer as 0.750 kg of water must evaporate from the woman.