### Video Transcript

Water in a pan reaches 100 degrees Celsius. But the pan is still left on the heat. So eventually, all of the water turns to water vapor. Calculate the energy needed to evaporate the 1.2 kilograms of water contained by the pan. Use a value of 2258 kilojoules per kilogram for the specific latent heat of vaporization of water. Give your answer to two significant figures.

Alright. So this is a long question. So we should start by underlining all the important bits, so we don’t miss anything out. Okay, so firstly, we’ve been told that we’ve got water in a pan. And it reaches 100 degrees Celsius which is its boiling point. But the pan is still left on the heat. So eventually, all of the water turns to water vapor. Now, it’s important that we’ve been told that all of the water is turning to water vapor. We’ll see why this is the case in a second. Moving on though, we’ve been asked to calculate the energy needed to evaporate the 1.2 kilograms of water contained by the pan. We’ve also been told to use a value of 2258 kilojoules per kilogram for the specific latent heat of vaporization of water. And the final important piece of information we need to know is that we need to give our answer to two significant figures.

Okay, so let’s start by recalling the definition of specific latent heat of vaporization. The specific latent heat of vaporization then, which we’ll call 𝐿, is the energy needed to convert a substance entirely from liquid to vapor per unit mass. In other words, the specific latent heat of vaporization is how much energy you need to convert one kilogram, or one unit mass, of any substance from liquid to vapor. Now, in this case, we’ve got water. And we’ve been told that the specific latent heat of vaporization of water is 2258 kilojoules per kilogram. We’ve also been told that the mass of the water in the pan is 1.2 kilograms. And we’re trying to find the energy needed.

Now, we’re trying to find the energy needed to evaporate the entire 1.2 kilograms of water. And as we said earlier, this is important because the specific latent heat of vaporization is defined as the energy needed per unit mass to entirely convert a substance from liquid to vapor. Now, what this means is that we’ve been given 𝐿 and 𝑚. And we’re trying to work out 𝐸. So we need to rearrange the equation. We do this by multiplying both sides by the mass 𝑚. When we do this, we find that the mass multiplied by the specific latent heat of vaporization is equal to the energy needed. And which one we can substitute in our values. We can say that the energy needed is equal to the mass, which is 1.2 kilograms, multiplied by the specific latent heat of vaporization, 2258 kilojoules per kilogram.

Now, in situations like this, it’s actually really useful to include the units in our calculation. This is because we’re not working in standard units in this case. The specific latent heat of vaporization has been given to us in kilojoules per kilogram. Now, the standard unit of energy is joules. And the standard unit of mass is kilograms. So the standard unit of specific latent heat of vaporization should be joules per kilogram not kilojoules per kilogram. However, we’ve been given kilojoules per kilogram. So we can work with that.

If we include the units in our calculation, we can see that the kilograms from the mass cancels with the per kilograms from the specific latent heat of vaporization. And what we’re left with is kilojoules. Now, this makes sense because kilojoules is a unit of energy. It’s just not the standard unit. And it’s just something that we need to be careful of. So when we multiply 1.2 by 2258, whatever the answer to that is, will be our energy in kilojoules not in joules.

Anyway, so evaluating the right-hand side of this equation, we find that the energy needed is 2709.6 kilojoules. However, this is not our final answer because, remember, we’ve been told to give our answer to two significant figures. So here’s significant figure number one. Here’s significant figure number two. We’ll look at the next one, this zero, to tell us what happens to the seven. Now, obviously, zero is less than five. So this seven will stay the same. It’s not going to round up. And once the rounding stage is complete, we have our final answer.

The energy needed to evaporate the entire 1.2 kilograms of water contained by the pan is 2700 kilojoules, to two significant figures.