Question Video: Finding the Specific Latent Heat of fusion of Water given Its Mass, the Rate of Energy Transfer, and the Time Taken to Change State

A 10 g cube of ice at 0°C is placed outside in direct sunlight on a hot day. The sunlight and the hot air around the cube heat it at a rate of 10 W. It takes 5.56 minutes for the ice to completely melt. Calculate the specific latent heat of fusion for water. Give your answer to 3 significant figures.

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Video Transcript

A 10-gram cube of ice at zero degrees Celsius is placed outside in direct sunlight on a hot day. The sunlight and the hot air around the cube heat it at a rate of 10 watts. It takes 5.56 minutes for the ice to completely melt. Calculate the specific latent heat of fusion for water. Give your answer to three significant figures.

Okay, so, what we have here is a cube of ice which we’ve been told has a mass, which we’ll call 𝑚, of 10 grams. As well as this, we know that the temperature of the ice, which we’ll call capital 𝑇, is zero degrees Celsius. Now we know that this cube of ice has been placed in direct sunlight on a hot day, and that this sunlight and the hot around the ice cube heat the cube at a rate of 10 watts. In other words, they provide a power to the cube, which we’ll call 𝑝, of 10 watts because watts is a unit of power.

Now in order for the cube to go from being a cube to completely melting, we’ve been told it takes a time, which we’ll call 𝑡, of 5.56 minutes. And what we need to do is to calculate the specific latent heat of fusion for water. So, let’s call the specific latent heat of fusion capital 𝐿. And this is what we’re trying to calculate.

Now the specific latent heat of fusion of any material is simply a measure of how much energy it takes the change one unit mass of that substance from a solid to a liquid. In other words, the specific latent heat of fusion of a material is defined as the energy required per unit mass to change that material from solid to liquid. Now in this case, we’re trying to calculate the specific latent heat of fusion of water, which is perfectly fine. Because what we have is solid water, that’s ice, and liquid water at the end.

Also, we already know the mass of the water that we’re dealing with. We’ve been told in the question that the mass of the ice cube is 10 grams. So, to calculate this latent heat of fusion, all we need to know now is the amount of energy transferred to this ice cube.

Now it’s also important that we know, by the way, that the ice cube is originally at a temperature of zero degrees Celsius. Because the way that melting a substance works is that if that substance is at a temperature lower than its melting point, then the energy provided to that substance will first go towards increasing the temperature of the substance. This happens until the substance reaches its melting point. And once the substance is at its melting point, all of the energy provided goes towards melting the object. Then once the object is completely melted, then the temperature begins to increase again after more energy is provided.

However, in this case, we know that the temperature of the ice block is zero degrees Celsius. This means that it’s already at its melting point, which happens to be zero degrees Celsius. Therefore, all of the energy provided initially to this ice block goes towards melting the ice. And none of it is wasted in increasing its temperature. So, all we need to do here is to work out how much energy is provided to the ice block over a time period of 5.56 minutes in order to convert all 10 grams of the ice block into water.

Now it’s at this point that we should realise that we’ve been told the amount of power transferred to the ice cube. This is because power is defined as the energy transferred, in this case to the ice cube, per unit time. And this energy is transferred by the sunlight and the heat in the air. So, this is the energy that goes towards melting the ice cube. And hence, we can rearrange this equation since we already know the power transmitted and the time for which this power is transmitted.

We can use this equation to work out the energy transferred to the ice block. And this, of course, is the total energy transferred to the ice block over the 5.56 minutes that causes the ice block to melt. So, we can rearrange the equation by multiplying both sides by 𝑡, so that 𝑡 on the right-hand side cancels. And what that gives us is that the amount of time taken for the block to melt multiplied by the power transferred to the block is equal to the total energy transferred to the block causing it to melt.

But then, this energy 𝐸 is the same as the energy 𝐸 in the latent heat of fusion equation. And so, what we can say is that the latent heat of fusion of water is equal to the total energy transferred to the ice block divided by the mass of the ice block. Because this way what we’re finding is the amount of energy required per unit mass to melt ice, and of course turn that ice into water therefore.

So, at this point, we’ve got an equation now where we know all of the quantities on the right-hand side. We know the amount of time taken for the ice block to melt. We know the power transferred to it. And we know the mass of the ice block. So, we can work out the latent heat of fusion. However, before we plug in any value, let’s consider units quickly.

Now normally latent heat of fusion is given in its base unit. And we can work out what this base unit is if we look at this equation. Latent heat of fusion is defined as the energy per unit mass. And energy’s base unit is joules. And mass’ base unit is kilograms. So, we can give our answer in joules per kilogram for the latent heat of fusion. However, most commonly, latent heat of fusion is actually given in kilojoules per kilogram. And so, before we plug values into the right-hand side of this equation, we need to convert so that our final answer will either be in joules per kilogram or kilojoules per kilogram.

So, let’s say we choose to convert everything so that our final answer ends up being in joules per kilogram, the base unit of latent heat of fusion. This is the easiest thing to do because if our answer for 𝐿 ends up being in base units, then everything else needs to be in base units as well. In other words, the amount of time for melting needs to be in seconds. The power needs to be in watts. And the mass needs to be in kilograms.

So, let’s go back converting all the quantities we’ve been given. Now the power transferred to the ice block is already in watts. So, we don’t need to convert this. However, the amount of time taken for the ice block to melt is in minutes not seconds. So, we can make recall the conversion that one minute is equivalent to 60 seconds. And so, when we say that the time 𝑡 is equal to 5.56 minutes, what we can choose to do is to multiply this entire thing by the fraction 60 seconds divided by one minute.

Now the reason that we can do this is because this fraction is equivalent to one. So, essentially what we’re doing is multiplying 5.56 minutes by just one. And the reason that this fraction is equivalent to one is because 60 seconds is the same thing as one minute. Or another way to think about it is that there is 60 seconds per minute. And so, what we’re really multiplying by is one minute divided by one minute, which is just one.

Then, we’ve converted the numerator one minute into 60 seconds. And what this allows us to do is to cancel the unit of minutes in the numerator with the unit of minutes in the denominator. Then we’ll have our time in seconds, which is equivalent to 5.56 times 60 seconds. And when we evaluate, it ends up being 333.6 seconds. So, now we’ve converted our time into seconds. Let’s think about changing our mass into kilograms.

We can recall that one gram is equivalent to one thousandth of a kilogram. And therefore, if we take this entire equation and multiply it by 10 on both sides of the equation — and remember, we can do this because we’re multiplying both sides by the same thing so whatever is done to one side is done to another. Unless we want to multiply one of the sides by one, which is totally fine. But anyway, multiplying both sides of the equation by 10 tells us that 10 grams is equivalent to ten thousandths of a kilogram. Or in other words, this is 0.01 kilograms. Therefore, we can say that the mass of the ice block is 0.01 kilograms. And at this point, we’ve converted everything into base units.

So, now let’s plug in all the values on the right-hand side of this equation. And since 𝐿 is equal to 𝑡𝑝 divided by 𝑚, what we have here is 𝑡 in seconds, 𝑝 in watts, and 𝑚 in kilograms. Then confirming the units, we can see that in the numerator we’ve got seconds multiplied by watts, which is equivalent to a joule. Simply because of the equation for power, once again, the power in watts is equal to a joule per second. And so, a second multiplied by a watt is equal to a joule. And in the denominator of our fraction, we’ve got kilograms.

So overall, what we’ve got is joules per kilogram, which is exactly the base unit for latent heat of fusion. So, evaluating everything on the right-hand side, we find that the latent heat of fusion for water is 333600 joules per kilogram. However, this is not our final answer because we need to give our answer to three significant figures.

So, here’s significant figure number one, number two, number three. Now we need to look at the next one, significant figure number four, to see what happens to number three. Well, the fourth significant figure is actually a six. Six is larger than five. And so, our third significant figure, this three here, is going to round up to a four.

In other words then, to three significant figures, the latent of fusion of water is 334000 joules per kilogram. And we can choose to give this as our final answer. Or we can convert it to kilojoules per kilogram, which is the most commonly used unit for latent heat of fusion.

To do this, we can recall that one joule is equivalent to one thousandth of a kilojoule. And so, 334000 joules per kilogram are going to be equivalent to 334 kilojoules per kilogram because, of course, the prefix kilo represents a 1000. And so, what we have here is 334000 joules per kilogram. Now we use kilojoules per kilograms just to make the answer a bit more concise. And that we don’t have to write three more zeros. But essentially our final answer is either 334 kilojoules per kilogram or 334000 joules per kilogram.

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