A student measures the temperature of a 0.2-kilogram block of platinum and the change in its internal energy as it is heated. The results are shown in the graph. Use the values in the graph to calculate the specific heat capacity of platinum. Give your answer to three significant figures.
Taking a look at our graph, we see that it shows us the change in internal energy, measured in joules, against the temperature, measured in degrees Celsius, of this block of platinum. Taking a look at a couple of these data points, we see that when the change in internal energy of our block was zero joules, then at that point, the starting point, the temperature of our block was 20 degrees Celsius. Then, by the time we had added 50 joules of energy to the internal energy of the block, its temperature had gone up to 22 degrees Celsius. And then, once a 100 joules had been added to its internal energy, its temperature went up to 24 degrees Celsius, and so on.
Taking a look at our five plotted data points, we see that the line of best fit that connects them is a straight line. And not only that, each of the five points lies directly on it. This means that not only are these the data points we measured from this experiment. But they’re also exactly along our line of best fit, an unusual situation. And even more, notice that our line of best fit passes right through the origin. Now that we’re a bit familiar with what’s plotted in this graph, the question comes up. How will we use these data points to actually solve for the specific heat capacity of platinum? In order to see how we’ll do that, let’s clear a bit of space on screen.
Alright, so we wanna solve for the specific heat capacity of platinum. We can recall that specific heat capacity, in general, is the amount of energy we’d need to raise the temperature of a one-kilogram object by one degrees Celsius. We can also recall that specific heat capacity, symbolized using a lowercase 𝑐, can fit within a mathematical expression. This expression tells us that, given some object which goes through a temperature change Δ𝑇, if we multiply the mass of that object by its temperature change and then multiply that by the specific heat capacity of the material the object is made from, then that product altogether will equal the energy input into the object to affect this temperature increase. Or if it’s a temperature decrease, then energy will be the energy lost by the object.
Once we have this mathematical relationship that has the specific heat capacity in it, we can algebraically rearrange this expression to solve for that value, 𝑐. We can divide both sides of the equation by 𝑚 times Δ𝑇. When we do that, the 𝑚s and the Δ𝑇s on the right-hand side cancel out. And we find that the value we want to solve for, 𝑐, is equal to the energy change in our object divided by its mass times its change in temperature.
All this tells us that in order to calculate the specific heat capacity of platinum, what we’ll call 𝑐 sub 𝑝, we’ll need to know its change in internal energy, capital 𝐸, the mass of this platinum, 𝑚, as well as the temperature change it goes through, Δ𝑇. It’s important not to forget that we’ve actually already been told the mass of our object in the problem statement.
In this experiment, we’re working with a block of platinum with a mass of 0.2 kilograms. Once that’s substituted in, now all we need to do is solve for the internal energy and the change in temperature of our platinum. And to do that, we’ll rely on the data points plotted in our graph, even though there are five separate data points. Because they all lie exactly on the same line, we can choose any one of these five to give us our values for the change in internal energy and the change in temperature.
For example, let’s say we choose this data point right here, the one at a temperature of 26 degrees Celsius and a change in internal energy of 150 joules. Using this data point means that we’ll replace capital 𝐸 in our equation with 150 joules. After all, that’s the change in internal energy that our platinum block went through at this given data point. And then, what’s Δ𝑇? Well, we see that at this data point, the temperature is 26 degrees Celsius. But that’s not the change in temperature that our block went through. That’s because our block started not at a temperature of zero degrees Celsius, but at a temperature of 20 degrees Celsius. So, the change is the difference between 20 and 26. Based on that, we can see that the change in temperature is actually six degrees Celsius.
We’re now ready to calculate the specific heat capacity of platinum, the right-hand side of this expression. When we do, to three significant figures, our answer is 125 joules per kilogram degrees Celsius. And remember, we would’ve gotten the same result regardless of which of the five data points we chose to use in our calculation. Or indeed, we could’ve used any point along this line, since all the data points lay exactly along it. And if that claim seems dubious, then feel free to check it out.
In conclusion, we found that the specific heat capacity of platinum is 125 joules per kilogram degrees Celsius.