On a summer day sunlight may put out 100 million calories into an 80,000 liter swimming pool.?

Question

On a summer day sunlight may put out 100 million calories into an 80,000 liter swimming pool.
A.) What temperature rise occurs if no other heat enters or leaves the pool?
B.) How many liters of water would have to evaporate from the pool to keep its temperature constant?

Answer 1

to heat one gram of liquid up by one degree, you need one Calorie

Heat, in calories = (mass) (temperature change) (specific heat)

The amount of heat needed to change one gram of a substance from the liquid phase to the gas phase - Hvap of water is about 540 calories / gram.

if 1 cal = heat to increase 1 gm of water 1deg C.
then convert 80,000 liters into Grams
=80,000 kg= 80 x 10^6 g.

So the heat increase is 100x10^6 calories/80x10^6 g = 1.25 C

Rates of evaporation of water depend on things like the temperature, humidity, and wind. You do not what the temprature of the pool was originally, so I assume it is we "use" heat to only evaporate water.

100x10^6/540=
1.83x10^5 grams= 183 kg= 183

As far as evaporation go, in the real world, only two things matter: the temperature, and the amount of surface
area available for the water to evaporate from. Some additives
to water may have special effects at the surface (for example adding oil leaves a film of oil on top of the surface, making it impossible for water to evaporate) - so go careful on that sun tan lotion. And coloured additives may change the rate at which the water absorbs or releases heat, thus modifying the temperature.

Answer 2

This sounds imposing, but it really isn't .
1 cal = heat to increase 1 gm of water 1deg C.
80,000 L=80,000 kg= 80 x 10^6 g. So
heat increase is 100x10^6 calories/80x10^6 g = 1.25 C

In part 2, we "use" heat to only evaporate water. If I remember right, heat of vaporization/ gm water is about 550 calories. So grams to evaporate= 100x10^6/550=
1.8x10^5 grams= 180 kg= 180 L