Can someone help me with vaporization?

Question

For many years, drinking water has been cooled in hot climates by evaporating it from the surfaces of canvas bags or porous clay pots. How many grams of water can be cooled from 35°C to 25°C by the evaporation of 38 g of water? (The heat of vaporization of water in this temperature range is 2.4 kJ/g. The specific heat of water is 4.18 J/g·K.) Answer should be in KJ.

For many years, drinking water has been cooled in hot climates by evaporating it from the surfaces of canvas bags or porous clay pots. How many grams of water can be cooled from 35°C to 25°C by the evaporation of 38 g of water? (The heat of vaporization of water in this temperature range is 2.4 kJ/g. The specific heat of water is 4.18 J/g·K.)

Answer 1

In this problem, you are assuming that the energy used to evaporate the water outide is the same amount as the energy lost in the cooling of the water inside the bag.

You will need to remember two equations for this:

q = m(inside) x C x detla(T)

q = m(outside) x delta(H)vap

C = 4.18 J/gK
m(inside) = ? g
m(outside) = 38 g
delta(H)vap = 2400 J/g
delta(T) = 35 - 25 = 10

Combining the two equations, we get:

m(inside) x C x delta(T) = m(outside) x delta(H)vap
m(inside) x 4.18 x 10 = 38 x 2400
m(inside) x 41.8 = 91200
m(inside) = 2200 g or 2.2 kg

Answer 2

Given the HOV of 2.4 kJ/g multiplied by 38g (notice the gram in the denominator of your HOV will cancel the gram in the mass of water) 91.2 kJ which equals 91200J
This number is a measure of how much energy (in J) is being lost with the evaporation.
Now that you know how much energy you're losing, AND how much it's going to affect the temperature (in this case, 10ºC from 35 to 25ºC), you can simply divide the amounts.
(91200J)(10 K)
--------------------------=
4.18 J/g*K

(912000J*K)
----------------------=
4.18 J*K*1/g

will give you your answer in grams of water. You'll be surprised at how much water that'll cool!