Specific heat capacity

Question

A mass of water must evaporate from the skin of a 66.0 kg man to cool his body by 1.30°C. The heat of vaporization to water at body temperature (37°C) is 2.42×10⁶J/Kg the specific heat capacity of a typical human body is 3480 J/Kg. How can I find the mass of water evaporated and the volume of water the man needs to drink to replenish the water in his system?

Answer 1

First we need to find the heat loss required to lower the man's temperature by 1.30 C.

This is done by using the equation Q = mass x specific heat x Δt.

So, Q = 66.0 kg x 3480 j/kg-C x 1.30 C = 2.98 x 10^5 joules.

This heat is used to evaporate m kg of water, so Q = m kg x 2.42 x 10^6 j/kg. substitute for Q from the calculation above and solve for mass of water. m = Q / 2.42 x 10^6 = 0.123 kg ( 123 g) of water evaporated.

Since the density of water is 1.00 g / mL, the volume of water would be m / d = 123 g / 1.00 g/mL = 123 mL of water to be consumed.