# Video: Determining the Thermal Expansion of a Liquid and Its Solid Container

Most cars have a coolant reservoir to catch radiator fluid that may overflow when the engine is hot. A radiator is made of copper and is filled to its 16.0-L capacity when at 10.0°C. What volume of radiator fluid will overflow when the radiator and fluid reach a temperature of 95.0°C given that the fluid’s volume coefficient of expansion is 𝛽 = 400 × 10⁻⁶°C⁻¹ and copper’s volume coefficient of expansion is 51.0 × 10⁻⁶°C⁻¹?

10:53

### Video Transcript

Most cars have a coolant reservoir to catch radiator fluid that may overflow when the engine is hot. A radiator is made of copper and is filled to its 16.0-liter capacity when at 10.0 degrees Celsius. What volume of radiator fluid will overflow when the radiator and fluid reach a temperature of 95.0 degrees Celsius given that the fluid’s volume coefficient of expansion is 𝛽 is equal to 400 times 10 to the power of negative six per degrees Celsius and copper’s volume coefficient of expansion is 51.0 times 10 to the negative six per degrees Celsius?

Okay, so lots of information in this one, so let’s start by underlining all the important bits. The first sentence tells us that there’s usually a coolant reservoir to catch radiator fluid that overflows. The radiator itself is made of copper and it’s filled to it 16.0-liter capacity. This happens at 10.0 degrees Celsius. We need to find the volume of radiator fluid that overflows when both the radiator and the fluid reach a temperature of 95.0 degrees Celsius.

We’ve also been given the fluid’s volume coefficient of expansion, which is 𝛽 is equal to 400 times 10 to the power of negative six per degrees Celsius. And we’ve also been given copper’s volume coefficient of expansion, which is 51.0 times 10 to the negative six per degrees Celsius. So this is indeed a lot of information, which is why a diagram will probably help us visualize the problem a bit better.

So let’s say that this is the radiator which is made of copper. It’s essentially a tank that can hold 16.0 liters of fluid. We know this because we’ve been told that the capacity of the tank is 16.0 liters. In other words, the volume of the tank on the inside, we’ll call this 𝑉 sub 𝑐 for volume sub copper, is 16.0 liters. And because the tank can hold 16.0 liters of fluid, we’ve put 16.0 liters of fluid in the tank. We’ll call this 𝑉 sub 𝑓 for volume sub fluid.

So we’ve got a copper tank that holds 16.0 liters, and we’ve got a fluid that’s 16.0 liters inside the copper tank. Now this is what we have at the beginning, when the temperature of both the tank and the fluid is 10.0 degrees Celsius. We’ll call this 𝑇 sub 𝑏 for temperatures sub before. Let’s also label the volume coefficients of expansion that we’ve been given.

We’ve been told that the volume coefficient of expansion of copper is 51.0 times 10 to the power of negative six per degrees Celsius. And we’ll call this 𝛽 sub 𝑐 because 𝛽 stands for the volume coefficient of expansion and the subscript tells us that it’s for the copper tank. Likewise, we’ve been told that the fluid’s volume coefficient of expansion is 400 times 10 to the power of negative six per degrees Celsius. We’ll call this 𝛽 sub 𝑓 for the fluid.

Now what the question tells us happens to these two things, the fluid and the tank, is that the temperature increases to 95.0 degrees Celsius. We’ll call this temperature 𝑇 sub 𝑎 for temperature sub after. When this temperature is increased, thermal expansion is going to occur for both the copper and the fluid. In other words, the capacity of the copper tank will increase as we increase the temperature, and so will the volume of the fluid.

What we need to work out is how much fluid overflows, taking into account the fact that the copper tank will expand as well. In other words, when we’ve increased the temperature, due to thermal expansion, the volume of the copper tank will no longer be 16.0 liters. It will most likely be a little bit more than this. Similarly, the volume of the fluid will no longer be 16.0 liters either because the fluid expands too.

However, if we compare the volume coefficients of thermal expansion, then just by a qualitative argument, we can see that the expansion of the fluid is gonna be much greater than the expansion of the copper tank. This is because the coefficient of expansion of the fluid is 400 times 10 to the power of negative six. However, the coefficient of expansion of copper is only 51.0 times 10 to the power of negative six.

In other words, for every degree Celsius increase in temperature, the fluid will expand a lot more than the copper will. So that’s just a qualitative argument. And we found out that the copper tank will expand a little bit, but not enough to contain all the fluid because the fluid is also expanding. And in fact, the fluid is expanding even more than the copper is. And so we need to find out how much of that fluid overflows. So let’s go about doing that!

First of all, we can recall that the volume coefficient of expansion of any substance is given by the following equation: 𝛽, the coefficient of expansion, is equal to one over 𝑉, the volume, times d𝑉 by d𝑇, where d𝑉 by d𝑇 is the rate of change of volume with respect to the temperature. We can apply this equation to the two substances we’ve been given. We can say that the coefficient of expansion of the fluid is equal to one over 𝑉 sub 𝑓, which is the volume of the fluid, times d𝑉 sub 𝑓 by d𝑇.

And similarly for the copper, we said that 𝛽 sub 𝑐 is equal to one over 𝑉 sub 𝑐 multiplied by d𝑉 sub 𝑐 by d𝑇. Now the reason we haven’t put subscripts next to the 𝑇s, the temperatures, is because both of these substances, the copper and the fluid, go through the same temperature change. They both start at 10.0 degrees Celsius and finish at 95.0 degrees Celsius. So the change in temperature for both of these substances is the same. Now let’s notice something about the values we’ve been given for the volume coefficients of expansion. They are both constant.

They’re just values they don’t change with respect to temperature. They’re just a number. So if the left-hand sides of these equations are constant, then so must the right-hand sides be. So if we look at the blue equation, 𝛽 sub 𝑓 is equal to one over 𝑉 sub 𝑓 multiplied by d𝑉 sub 𝑓 by d𝑇, we know that one over 𝑉 sub 𝑓 multiplied by d𝑉 sub 𝑓 by d𝑇 must also be a constant. In other words, this whole part should be a constant.

Now we already know that one over 𝑉 sub 𝑓 is a constant because, again, 𝑉 sub 𝑓 is 16.0 liters. We’ve already defined that, which means that d𝑉 sub 𝑓 by d𝑇 should also be a constant. Similarly, d𝑉 sub 𝑐 by d𝑇 should also be a constant. And this simplifies our calculations quite a lot, because a constant d𝑉 by d𝑇 means that a graph of 𝑉 against 𝑇 will be a straight line graph. And we can draw a little diagram for 𝑉 sub 𝑓 against 𝑇. And we can do the same thing for 𝑉 sub 𝑐 against 𝑇.

Now we don’t actually know what the gradients of the straight lines are. All that we know is that these gradients are constant because d𝑉 by d𝑇 in both cases is constant. But if these values of d𝑉 by d𝑇 are constant in both cases, then we can change this to Δ𝑉 by Δ𝑇, again in both cases. So what does Δ represent? Well Δ represents a change in quantities.

Let’s say we take two points on the line. The change in volume between the two points is Δ𝑉 sub 𝑓. Similarly, a change in temperature between two points is Δ𝑇, which means we’re no longer looking at a derivative of 𝑉 sub 𝑓 with respect to the temperature, we can instead just look at changes between two points. And we can only do this when the gradient is constant. This is because calculating Δ𝑉 by Δ𝑇 gives us the gradient of a straight line.

So how does this make life easier for us? Well we can calculate what Δ𝑇 is for both cases. Δ𝑇 is simply the temperature after minus the temperature before. In other words, it’s how much the temperature has changed by. Substituting in the values of 𝑇 sub 𝑏 and 𝑇 sub 𝑎, we find that Δ𝑇 is equal to 85.0 degrees Celsius in both cases. Now we’ve already said that we can replace in the equations d𝑉 by d𝑇 with Δ𝑉 by Δ𝑇. So we’ve got this part with the Δs in it.

Now at this point, let’s take a moment to remember what we’re actually trying to do. We’re trying to find out how much each substance expands by. In other words, we’re trying to find out the value of Δ𝑉 sub 𝑓 and Δ𝑉 sub 𝑐. So we can rearrange the equations. Let’s start with the one in blue. All we need to do is multiplying both sides of the equation by 𝑉 sub 𝑓 multiplied by Δ𝑇 and similarly, for the orange equation, multiplied by 𝑉 sub 𝑐 multiplied by Δ𝑇.

This way we’ve isolated the quantities that tell us how much the volume changes by for both the fluid and the copper tank. So let’s plug in our values. Since we already know the values of the 𝑉 sub 𝑓, Δ𝑇, and 𝛽 sub 𝑓, we can plug those in to find our answer of Δ𝑉 sub 𝑓. So we’ve plugged those values in now. And we can see that the degrees Celsius here cancels with the per degrees Celsius here. That leaves us with an overall unit of liters.

That’s exactly what we expect because we’re trying to calculate Δ𝑉 sub 𝑓; that’s the change in volume of the fluid. So putting these numbers into our calculator gives us Δ𝑉 sub 𝑓 is equal to 0.544 liters. In other words, when we heat up the fluid from 10 degrees Celsius to 95 degrees Celsius, the volume of the fluid changes by 0.544 liters. We can do the same thing by plugging in values to find out Δ𝑉 sub 𝑐. 𝑉 sub 𝑐 initially with 16.0 liters, the capacity of the tank, Δ𝑇 is 85.0 degrees Celsius, and 𝛽 sub 𝑐 is 51.0 times 10 to the negative six per degrees Celsius.

And so we find that Δ𝑉 sub 𝑐 is 0.06936 liters. So we’re heating these two substances from 10 degrees Celsius to 95 degrees Celsius. And the capacity of the copper tank increases by 0.06936 liters. However, the volume of the fluid increases by 0.544 liters. Now out of these 0.544 liters extra, the tank will be able to contain 0.06936 liters, because if the tank is expanded so there’s a bit more space in it, it will be able to contain those 0.06936 liters.

However, whatever fluid is left over will overflow. In other words, we can call the volume of fluid that overflows 𝑉 sub 𝑜. And this is equivalent to 0.544 minus 0.06936 liters. And this ends up being 0.47464 liters. However, this is not our final answer. If we look carefully, all of the values in the question have been given to three significant figures, which means that we should give our answer to three significant figures as well.

So here are the first three significant figures. It’s the fourth one that will tell us what happens to the third significant figure, whether it rounds up or stays the same. Now this fourth significant figure is a six. That number is greater than five. Therefore, this significant figure will round up to 0.475. And so our final answer is that the volume overflow is 0.475 liters to three significant figures.