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

Answer:

0.929%

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

$m L = Mc \Delta T \\m (2260000) = M (4200) (5)\\m (2260000) = M (21000)\\m = 0.00929 M \\\frac{m}{M} = 0.00929\\\frac{m(100)}{M} = (0.00929) (100)\\\\\frac{m(100)}{M} = 0.929$