Improved lab on Thursday

Based on what we got on last Saturday, we performed another experiment on Thursday. It was a very nice afternoon and it was sunny.
Aim:
We wish to study how much solar power is needed to evaporate water, and compare the results with the fresh water system in race rocks.
Procedure:
This lab is quite straightforward and easy.
Put the same amount of ocean water into 2 small beakers
Measure the weight of beakers and water with a electronic weight scale
Put the beakers at sunny place on the dock
Take down the data of humidity, outside temperature, and the solar power
Leave them for 3 hours
Take down the data of humidity, outside temperature, and the solar power after 3 hours
Measure the weight of beakers and water with a electronic weight scale again
Data:
The diameter of the beaker: 6.52cm
Before 3 hours
Time 12:35 Temperature 12⁰C Humidity 57% Solar energy: 856W/m^2.
Weight 1: 74.486g Weight 2: 75.034g
After 3 hours
Time 3:35 Temperature 11.6 ⁰C Humidity 54% Solar energy: 543W/m^2
Weight 1: 63.534g Weight 2: 65.869g
Data Analysis:
Let us assume that the change of solar energy is linear, so that the power can be calculated as below:
The surface area of the beaker = 3.14×〖(6.52×〖10〗^(-2))〗^2=0.0133 m^2
Solar power received=surface area×solar energy×time=0.0018×((856+542))/(2×〖10〗^(-3) )×3=1.26×〖10〗^(-3) kw∙h
For group 1, the fresh water that produce=74.486g-63.534g=10.592g
For group 2, the fresh water that produce=75.034g-65.869g= 9.165g
Average fresh water that produced=(10.592+9.165)/2=9.879g
Conclusion
To sum up, we managed to produce 9.879g fresh water with 1.26×〖10〗^(-3) kw∙h solar power.
Therefore, to produce 1kg fresh water in 3 hour, we will need:
0.128 kw∙h of energy
Surface area of 1.346m^2
(Note that this lab is done on the afternoon of a sunny day)


(The blogger is a little bit retarded at showing equations, you can access if you email: peirui422@hotmail.com rui)