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A lady perspires. How much liquid water (as a percentage of the mass of the lady) must evaporate to lower the temperature of said lady by 5°C? [Assume that the specific heat of the human body is approximately that of water, 4200 J K-1kg-1.]

1 Answer

3 votes

Answer:

0.929%

Step-by-step explanation:


m = mass of water evaporated


M = mass of the lady


c = specific heat of human body = 4200 JK⁻¹kg⁻¹


L = Latent heat of evaporation of water = 2260000 Jkg⁻¹


\Delta T = Drop in temperature = 5 C

Using conservation of heat


m L = Mc \Delta T \\m (2260000) = M (4200) (5)\\m (2260000) = M (21000)\\m = 0.00929 M \\(m)/(M) = 0.00929\\(m(100))/(M) = (0.00929) (100)\\\\(m(100))/(M) = 0.929

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