# Video: Comparing the Energy of Phase Change to the Energy of Temperature Change

On a certain dry sunny day, a swimming pool’s temperature would rise by 1.50°C 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 J/kg ⋅ °C for the specific heat capacity of water and use a value of 2430 kJ/kg for the latent heat of vaporization of water.

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### 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.