### Video Transcript

On a certain dry sunny day, a
swimming pool’s temperature would rise by 1.50 degrees Celsius if the water in the
pool did not lose some internal energy by evaporation. Determine the percent of the water
in the pool that must evaporate to remove precisely enough internal energy to keep
the water temperature constant. Use a value of 4182 joules per
kilogram degrees Celsius for the specific heat capacity of water. And use a value of 2430 kilojoules
per kilogram for the latent heat of vaporization of water.

In this example, we have a pool of
water with sunlight shining on the pool and adding energy to it. When that energy from the Sun
reaches the water, it can do one of two things. First, it could heat the water up
according to the heat capacity of the water in the pool. The second thing that Sun’s energy
could do is vaporize the water, make some evaporate. This happens according to the heat
of vaporization of water.

In a typical scenario, both of
these things would be going on, the water would be heating up and some would
evaporate. But in our case, we want to imagine
that one or the other happens. In particular, we want to calculate
the percent of the pool water that would have to evaporate, thereby taking energy
away from the pool, in order to perfectly counteract the amount of energy that would
be needed to heat up the pool 1.50 degrees Celsius.

As a first step, let’s figure out
just how much energy it would take to heat up the pool 1.5 degrees Celsius. To figure that out, we’ll take our
heat capacity of the water in the pool, multiply it by the change in the pool’s
temperature that would happen, and then multiply that by the amount of water in the
pool. We don’t know how much water is in
the pool. So let’s leave it undefined. We’ll just call it 𝑋 kilograms of
water are in this pool.

Looking at this expression then, we
see that the units of degrees Celsius as well as the units of kilograms cancel
out. And we’re left with an overall
energy. In order to heat 𝑋 kilograms of
water 1.5 degrees Celsius, it would take 6273 times 𝑋 joules of energy. But what if we took all this energy
and instead of using it to heat the water, we used it to evaporate the water. In that case, how much water would
evaporate?

Our heat of vaporization value
tells us that in order to evaporate one kilogram of water, move it from being a
liquid to a gas, it would take 2430 times 10 to the third joules of energy. If we multiply this heat of
vaporization by the amount of energy that could be dumped into the pool to increase
its temperature 1.5 degrees Celsius, then we see that the units of joules cancel
out. And we’re left with a mass. That mass is equal to 2.58 times 10
to the negative third times 𝑋 kilograms.

This result then is the mass of
water that would need to evaporate in order to offset a 1.5 degrees Celsius increase
in a pool of original mass 𝑋 kilograms. The percent then of the pool that
would need to evaporate would be this result divided by the original volume of the
pool times 100 percent. In this division, we lose our units
of kilograms as well as our amount 𝑋, whatever that number is. Multiplying these values together,
we find a result of 0.258 percent. This is the percent of the pool
water that would have to evaporate to offset this temperature change.