Evaporating sweat cools the body because evaporation is an endothermic process:  

 

 

 

Calculate the number of moles of water that must evaporate from the skin to cool the body by 0.50 °C. Assume a body mass of 110 kg and assume that the specific heat capacity of the body is 4.0  

       

Calculation of Moles of Water Evaporated for Cooling

An endothermic process: evaporation of water

Given Data:

- Enthalpy of vaporization ΔH = 44 kJ/mol

- Temperature drop ΔT = 0.50 °C

- Body mass = 110 kg (convert to grams = 110,000 g)

- Specific heat capacity of body = 4.0 J g^-1 °C^-1

Step 1: Calculate heat lost by body

Q = mass × specific heat × temperature change

Q = 110,000 g × 4.0 J g^-1 °C^-1 × 0.50 °C = 220,000 J = 220 kJ

Step 2: Calculate moles of water evaporated

Using Q = n ΔH, where n = number of moles

n = Q / ΔH = 220 kJ / 44 kJ/mol = 5 mol

Answer: 5 moles of water must evaporate to cool the body by 0.50 °C.