Question

Evaporation of sweat requires energy and thus take excess heat away from the body. Some of the water that you drink may eventually be converted into sweat and evaporate. If you drink a 20-ounce bottle of water that had been in the refrigerator at 3.8 °C, how much heat is needed to convert all of that water into sweat and then to vapor?? (Note: Your body temperature is 36.6 °C. For the purpose of solving this problem, assume that the thermal properties of sweat are the sameas for water). Look up any constants you may need, Show all steps

Answer 1

?H_vap = 43.46 kJ/mol at 37 °C

water molar mass = 18.0 g/mol

1 ounce = 0.03L = 30 mL; 20 ounce = 600 mL = 0.6 L

1) First let water come to body temperature, for this purpose we have to calculate how much energy it will absorb by using formula q = m x Cs x ?T       

q--> heat absorbed by 0.6 L water,

m-->mass of water in grams = 0.6 L = 0.6 Kg = 600g (since water density is 1g/mL)

?T = temperature differece = 36.6^oC-3.8^oC = 32.8 ^oC; Cs = 4.184 J/g^oC

q = (600g) x (4.184 J/g^oC) x 32.8 ^oC = 82341.12 J = 82.34 kJ

2) now we will calculate how much energy is required to evaporate water as sweat from 36.6 ^oC

adding 1) and 2) gives = 82.34 kJ + 1450 kJ = 1532.34 kJ

1532.34 kJ of heat is needed to convert all of that water into sweat and then to vapor.

Thank You!

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