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Water in a pan reaches 100 degrees Celsius.
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But the pan is still left on the heat.
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So eventually, all of the water turns to water vapor.
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Calculate the energy needed to evaporate the 1.2 kilograms of water contained by the pan.
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Use a value of 2258 kilojoules per kilogram for the specific latent heat of vaporization of water.
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Give your answer to two significant figures.
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Alright.
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So this is a long question.
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So we should start by underlining all the important bits, so we don’t miss anything out.
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Okay, so firstly, we’ve been told that we’ve got water in a pan.
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And it reaches 100 degrees Celsius which is its boiling point.
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But the pan is still left on the heat.
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So eventually, all of the water turns to water vapor.
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Now, it’s important that we’ve been told that all of the water is turning to water vapor.
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We’ll see why this is the case in a second.
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Moving on though, we’ve been asked to calculate the energy needed to evaporate the 1.2 kilograms of water contained by the pan.
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We’ve also been told to use a value of 2258 kilojoules per kilogram for the specific latent heat of vaporization of water.
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And the final important piece of information we need to know is that we need to give our answer to two significant figures.
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Okay, so let’s start by recalling the definition of specific latent heat of vaporization.
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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.
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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.
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Now, in this case, we’ve got water.
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And we’ve been told that the specific latent heat of vaporization of water is 2258 kilojoules per kilogram.
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We’ve also been told that the mass of the water in the pan is 1.2 kilograms.
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And we’re trying to find the energy needed.
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Now, we’re trying to find the energy needed to evaporate the entire 1.2 kilograms of water.
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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.
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Now, what this means is that we’ve been given 𝐿 and 𝑚.
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And we’re trying to work out 𝐸.
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So we need to rearrange the equation.
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We do this by multiplying both sides by the mass 𝑚.
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When we do this, we find that the mass multiplied by the specific latent heat of vaporization is equal to the energy needed.
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And which one we can substitute in our values.
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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.
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Now, in situations like this, it’s actually really useful to include the units in our calculation.
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This is because we’re not working in standard units in this case.
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The specific latent heat of vaporization has been given to us in kilojoules per kilogram.
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Now, the standard unit of energy is joules.
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And the standard unit of mass is kilograms.
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So the standard unit of specific latent heat of vaporization should be joules per kilogram not kilojoules per kilogram.
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However, we’ve been given kilojoules per kilogram.
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So we can work with that.
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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.
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And what we’re left with is kilojoules.
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Now, this makes sense because kilojoules is a unit of energy.
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It’s just not the standard unit.
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And it’s just something that we need to be careful of.
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So when we multiply 1.2 by 2258, whatever the answer to that is, will be our energy in kilojoules not in joules.
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Anyway, so evaluating the right-hand side of this equation, we find that the energy needed is 2709.6 kilojoules.
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However, this is not our final answer because, remember, we’ve been told to give our answer to two significant figures.
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So here’s significant figure number one.
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Here’s significant figure number two.
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We’ll look at the next one, this zero, to tell us what happens to the seven.
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Now, obviously, zero is less than five.
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So this seven will stay the same.
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It’s not going to round up.
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And once the rounding stage is complete, we have our final answer.
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The energy needed to evaporate the entire 1.2 kilograms of water contained by the pan is 2700 kilojoules, to two significant figures.