Question

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.]

Answer 1

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`