# Lesson Video: Specific Heat Capacity Physics • 9th Grade

In this lesson, we will learn how to use the formula 𝐸 = 𝑚𝑐Δ𝜃 to calculate the amount of energy needed to increase the temperature of a material or object by a given amount.

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### Video Transcript

In this video, we’re talking about specific heat capacity. As we’ll see, specific heat capacity helps us understand how different materials respond to changes in their internal energy level.

As we get started, say we have a piece of a given material. For our purposes, the material could be anything. It could be wood or stone or plastic. Using this material, we want to run an experiment. We wanna see what happens when we put energy into the material in terms of its temperature. Since one of the easiest ways to put energy into something is to heat it, let’s say we put a Bunsen burner underneath this block. Over time, the effect is that heat energy is transferred into the block.

Let’s say we let this process go on for one minute. Then over this time, an amount of energy — we’ll call 𝐸 — is input to the block. Over this one-minute interval, we’re also measuring the temperature of the block. And what we find is that it goes through a temperature change. We’ll call it Δ𝑇 that change in temperature is equal to the final temperature of the block — that is, its temperature after one minute of heating — minus its initial temperature, its temperature before the heating started.

So what we’ve seen so far is that if we put an energy 𝐸 into our block, then what we get out is a change in temperature Δ𝑇. But let’s imagine that we let our experiment continue on for another minute. At the end of the second minute, two times 𝐸 joules of energy have been input into the block. And as we continue to keep track of the block’s temperature, we find that after two 𝐸 joules of energy is input, the temperature change is now two times Δ𝑇. That is, it’s twice the temperature change that we found after one minute had elapsed.

What our measurements of this heated block are showing us is that the energy that we input to the block is proportional to the change in temperature that the block undergoes. That is, if we double the energy input, then the temperature change of the block doubles. If we quadruple the energy input, the temperature change quadruples, and so on. We can write out a relationship that summarises this relationship or discovering. We’re finding that the energy input to the block — we’ve called it 𝐸 — is directly proportional to the change in temperature of the block.

Now an important caveat for this relationship is that it’s only true if the state of matter of our block isn’t changing. In other words, it started out solid and it needs to remain as a solid in order for this relationship to hold. Assuming the block stays the way it is though, it doesn’t melt and it doesn’t turn into a gas, then this relationship is true. Looking at this expression, there is another way we can write it, a mathematically equivalent way to say this.

Whenever we have an expression that says that one value is directly proportional to a second value, then we can write that that first value is equal to some constant value — we’ll call that constant capital 𝐶 — multiplied by the second value. Even though these two expressions look different, mathematically they say the same thing. 𝐸 being directly proportional to Δ𝑇 means that 𝐸 is equal to some constant times Δ𝑇. Looking at this second form of the equation, the question naturally comes up “what is this constant capital 𝐶?” It turns out that the name of this constant is the heat capacity.

Here’s the definition of heat capacity. It’s the amount of energy that’s needed to raise a material’s temperature by one degree Celsius. So this makes sense. If we go back to our block, then we would see the heat capacity of this block is how much energy we need to put into it so that its temperature goes up by one degree Celsius. And an important thing to realise about heat capacity is that it varies from material to material. The heat capacity of stone is different from the heat capacity of brick, is different from the heat capacity of wood, is different heat capacity of water, and so on.

So the energy that’s input into a substance is equal to the heat capacity of that substance multiplied by its change in temperature. Knowing this, let’s imagine we want to expand our experiment. We want to make a comparison between different materials as they’re heated. To that end, we are adding these other materials of different shapes from our original block. And then, we start to heat each one, adding energy to the material. Just like before, as we heat these various materials we track their temperature.

But soon enough, we discover that making this comparison between various pairs of these materials isn’t as easy as we thought. That’s because in our experimental setup we have more than one independent variable. One independent variable is the material that each of our objects is made of. But another variable is the fact that all of these objects have different masses. The mass of our blue block is not the same as our orange cube, is not the same as our pink pyramid, is not the same as our green cylinder. This means there’s another variable that we want to account for if we want to establish a relationship between energy input and temperature change.

So here’s what we decide to do. We modify all of our shapes so that now they all have the same amount of mass. In each case, the material has a mass of exactly one kilogram. At this point, now that the rate of energy being put into our materials is equal across the materials and the mass of the materials is all the same, now as we track temperature increase of these materials over time, we can really compare one material to another. In order to make this comparison between materials, for the first time, we’ve been careful to notice and keep track of what the mass of each one is.

But this raises another question. How does the mass of an individual material, say our original blue block, affect the energy transfer into that block and how its temperature changes? In other words, if we once again put an amount of energy 𝐸 into this blue block and we measured a temperature change Δ𝑇, how does the mass of the block affect that relationship? To find out, we try another experiment. We tried doubling the mass of our block from one to two kilograms and then we put the same amount of energy 𝐸 into this block.

What we’ll find is that compared to the temperature change Δ𝑇 that corresponded to a mass of the block of one kilogram, the temperature change now is half of that: Δ𝑇 divided by two. Then trying something more, say that we double the size of this block once again. Now, it’s four kilograms. If we once again add the same amount of energy 𝐸 to the block, now the change in temperature isn’t Δ𝑇 over two. It’s Δ𝑇 over four. Here’s what we’re seeing. When we start to account for the mass of our material, we see that this mass and the change in temperature of the material are related. The more mass there is to a material, the smaller its temperature change given a fixed amount of energy input.

This makes some intuitive sense. Imagine heating a single ice cube using some fixed amount of energy. Then, imagine applying that same amount of energy to an entire iceberg. Which one — the ice cube or the iceberg — would increase more in temperature given the same amount of energy input? We know it will be the smaller mass of ice, the ice cube. And here, we’re finding the same thing that as the mass of a material increases, for a given fixed amount of energy input, the change in temperature that that material experiences decreases.

This most recent experiment we’ve performed means we can make a modification to our original proportionality expression here that energy is proportional to the change in temperature of a material. This expression is true. But notice that it has nothing to do with a mass of this material. It’s basically saying for some mass of material — we don’t know what that mass is, but we know that it’s fixed — the energy we put into it is proportional to its increase in temperature. But now, we have more information. We know that the mass of the material when we vary it affects what the change in temperature of that material is.

And even more than that, we can say that there is a relationship between the mass and the change in temperature as well as the energy input. That relationship is that the energy we input into our material is proportional to the product of the material’s mass and its change in temperature. We see that now we have this product, we start to account for our ice cube-iceberg example. In the case of the ice cube, the mass was very very small. And so that means for a given amount of energy input, the change in temperature will be relatively high, whereas for our iceberg, since the mass is very very high, for that same amount of energy input, the overall change in temperature of the iceberg would be low.

Now just like before, there is a second equivalent way we can express this mathematically. We can say that it means that the energy input into our material is equal to some constant — we’ll call this one lowercase 𝑐 — multiplied by 𝑚 times Δ𝑇. In the previous version of this equation, we saw that constant was the heat capacity of the material. But now that our equation has changed, now that it includes the mass of our material, this constant will have a bit of a different aspect. Rather than our constant being the heat capacity of the material, our constant now refers to something called the specific heat capacity. And this is the amount of energy needed to raise the temperature of one kilogram of a material by one degree Celsius.

Notice the only difference between heat capacity — what we had before — and specific heat capacity — what we have now — is that when we’re talking about specific heat capacity, we’ve specified that the material is one kilogram in mass. It’s this quantity the specific heat capacity of a material that we often use to compare one material to another one. One way to better understand specific heat capacity is to think of it in terms of its units. When we look at this expression 𝐸 is equal to 𝑐 times 𝑚 times Δ𝑇, we know some of the units involved here.

For example, we know that the base units of energy, the standard units, are joules. So we have some amount of joules is equal to the unit of specific heat capacity, whatever that is, multiplied by a mass. And we know the base unit of mass is kilograms. And then, that’s multiplied by a change in temperature. And a change in temperature will have units of degrees Celsius. And looking at this equation, we can now see what the units of specific heat capacity must be. Because this is an equation — because units of joules is equal to the units of specific heat capacity times kilograms times degrees Celsius — in order for this equation to be true, it must be the case that specific heat capacity has units of joules per kilogram degrees Celsius. And based on these units, we can see how this expression agrees with our description of specific heat capacity.

Specific heat capacity is how much energy measured in joules is needed to raise the temperature of a one-kilogram object by one degree Celsius. So the great thing about knowing the specific heat capacity for a given material, we often either look it up in a table or are given that information in our problem statement, is that once we know it based on other information that might be given we could solve for say the mass of a sample or the change in temperature that undergoes or how much energy is input into a material.

Before getting some practice doing this, it’s helpful to know just as a side note that this equation for specific heat capacity sometimes shows up in a different form. Occasionally, instead of representing a change in temperature as Δ𝑇, it’s represented as a capital 𝜃 symbol. Which form of the equation we encounter depends on which textbook we use. But just know that the two equations are entirely equivalent and they say the same thing. It’s just that one uses Δ𝑇 to represent change in temperature and the other uses capital 𝜃. Now that’s it. Let’s look at an example involving specific heat capacity.

Asphalt concrete is used to surface roads. When under direct sunlight for a long time, it can get very hot. If 2500 kilograms of asphalt increases in temperature from 18 degrees Celsius to 40 degrees Celsius, absorbing 50 megajoules of energy from sunlight, what is the specific heat capacity of asphalt concrete? Give your answer to two significant figures.

Okay, so say we have this huge slab of asphalt. We’re told it has a mass of 2500 kilograms and that this mass is heating up in the sun. Thanks to all the energy it absorbs from the sun, the asphalt goes from a temperature of 18 degrees Celsius to 40 degrees Celsius. That tells us that the change in temperature undergoes or we can call Δ𝑇 is equal to 40 degrees Celsius minus 18 degrees Celsius or 22 degrees Celsius. We’re told that the amount of energy the asphalt absorbs in order to effect this change in temperature is 50 megajoules. That’s 50000000 joules of energy.

Based on all this information, we want to solve for the specific heat capacity of asphalt concrete. At this point, we can recall that specific heat capacity is the amount of energy needed to increase the temperature of one kilogram of a given material by one degree Celsius. As a mathematical equation, we can say that this amount of energy — what we can call 𝐸 — is equal to the mass of our substance multiplied by the specific heat capacity multiplied by the change in temperature of the substance, Δ𝑇.

Like we said, it’s the specific heat capacity 𝑐 that we want to solve for in this case. So to do that, let’s rearrange this equation algebraically. Let’s divide both sides of the equation by the mass of our substance multiplied by its change in temperature. When we do that, we find the specific heat capacity is equal to the energy added into our material divided by the mass of that material multiplied by its change in temperature.

In our case, we know the mass of our substance. It’s given as 2500 kilograms. We also know its change in temperature. We calculated that to be 22 degrees Celsius. And moreover, we know the amount of energy input to our substance, 50000000 joules. So to calculate specific heat capacity, it’s just a matter of substituting in these values into this expression. When we do that, our mass of 2500 kilograms and our change in temperature of 22 degrees Celsius can go straight in as is.

But our input energy, 50 megajoules, needs to change a bit. That’s because mega is a prefix that means million. So to express our energy in base units of joules, we need to convert it from 50000000 joules and write it as 50 times 10 to the six joules. With this conversion done, we can see that the final units of our specific heat capacity will be joules per kilogram degrees Celsius, just what we want. When we calculate this value to two significant figures, we find a result of 910 joules per kilogram degrees Celsius.

This answer means that in order to heat one kilogram of asphalt concrete by one degree Celsius, we would need to input 910 joules to the concrete. That’s the specific heat capacity of the asphalt.

Let’s take a moment now to summarise what we’ve learnt about specific heat capacity.

In this lesson, we first saw that a material’s heat capacity is how much energy it needs to go up in temperature by one degree Celsius. But then as we experimented with this idea, we saw there’s a way to extend it. By accounting for material mass, we learn that a material’s specific heat capacity is how much energy one kilogram of the material needs to go up in temperature by one degree Celsius.

Furthermore, we’ve learned that specific heat capacity, often symbolised using a lowercase 𝑐, fits into a mathematical equation. This equation tells us that if we multiply the specific heat capacity of a given material multiplied by a particular chunk of mass of that material times the change in temperature that mass has undergone, then that overall product is equal to the total amount of energy input into that particular mass of that particular material. And we also learned that another way to write this expression is 𝐸 is equal to 𝑚 times 𝑐 times 𝜃.

Specific heat capacity is useful for comparing materials and showing how they respond to changes in internal energy.