Heat and Temperature?

Question

A very shallow pond contains 1.50 x 10^5 kg of water at 23°C. At the end of a windy day, 1.00 x 10^3 kg of water was lost by evaporation. It takes 2.26 x 10^6 for 1 kg of water to evaporate.

1. How much energy was removed from the pond by heat of evaporation? (Do you use the Q = mcΔT formula?)
2. How much water was left in the pond? (I just don't know how to this one)
3. By how much did the temperature of the water drop in the pond?(Do you use the Q = mcΔT formula, where c = 4.19 x 10^3 J/kg C°)
4. Assuming there were no other changes in energy, what was the temperature of the water at the end of the day? (Which formula do I use?)

Thank you!!

In the first paragraph, I wrote "It takes 2.26 x 10^6 for 1 kg of water to evaporate." I forgot to add the units for 2.26 x 10^6. It should be Joules.

Answer 1

1. No. The Q = mcΔT is for heating not for evaporation.
Use Q=mCv where Cv is the latent heat of evaporation.
2. Water left=water at start minus water evaporated
m(left) = m2-m1= 1.50 x 10^5 kg - 1.00 x 10^3kg = 149000
3. Yes . You compute Q=m1Cv = m2cΔT ΔT=m1Cv/(m2 c)
4.T=23+ΔT
You are welcome