An iron cylinder has a mass of 0.80 kilograms and a temperature of 1.00 times 10 to the power of three degrees Celsius. The cylinder is dropped into an insulated chest containing 1.0 kilograms of ice at its melting point. What is the equilibrium temperature of the chest’s contents? Use the value of 4180 joules per kilogram degrees Celsius for the specific heat capacity of water, a value of 334 kilojoules per kilogram for the latent heat of fusion of water, and a value of 452 joules per kilogram degrees Celsius for the specific heat capacity of iron.
Okay so in this question, there are three objects that we need to think about: firstly, the iron cylinder; secondly, the ice; and thirdly, the chest which contains the ice. Let’s start by thinking about the iron cylinder. So here is our iron cylinder. Now let’s write down all the information that we have about this cylinder from the question.
Firstly, we can say that it has a mass, which we’ll call 𝑚 sub 𝑖 for mass sub iron, of 0.80 kilograms. Secondly, it has a temperature 𝑇 sub 𝑖 of 1.00 times 10 to the power of three degrees Celsius. In other words, this is 1000 degrees Celsius. And finally, we know that it has a specific heat capacity which, we’ll call 𝑐 sub 𝑖 of 452 joules per kilogram degrees Celsius. All right! So that’s the iron cylinder.
Let’s now think about the ice that’s in the chest okay. So here are our cubes of ice. Now we know that the ice that’s in the chest has firstly a mass of 1.0 kilograms. And by the way we’re calling the mass 𝑚 sub 𝑤 because we’ve already used the subscript 𝑖 to represent iron. So we use the subscript 𝑤 to represent water rather than ice. Anyway, so we’ve also been told in the question that the ice is at its melting point. So the temperature of the ice, initially, 𝑇 sub 𝑤 is zero degrees Celsius, because that’s the melting point of ice.
Thirdly, we’ve been told that the specific heat capacity of water is 4180 joules per kilogram degrees Celsius. And finally, we’ve been told that the latent heat of fusion of water, which we’ll call 𝐿 sub 𝑤, is 334 kilojoules per kilogram. So that’s all the information we’ve got about the ice from the question. Now let’s finally think about the insulated chest. Now we don’t have a lot of information about this chest, but we do know that it’s insulated. This means that once we drop the iron cylinder into the chest, the iron cylinder can only interact with the ice in the chest.
Because the chest is insulated, the only heat transfer is between the contents of the chest, in other words, the iron cylinder and the ice. The two can interact with each other, but not anything outside of the chest. And so there is no heat transfer either into the chest or out of the chest. And with all of this in mind, we need to try and find the equilibrium temperature once the iron has been dropped into the chest of ice. Okay, so at this point we’ve got all the information we can from the question. So let’s clear it off the screen.
So now let’s take our iron cylinder and drop it into our chest, and then let’s close our chest thus making it fully insulated. Now the thing that we need to remember is that objects in thermal equilibrium are at the same temperature. In other words, even though the iron starts out at a temperature of 1000 degrees Celsius and the ice starts out at a temperature of zero degrees Celsius, they’re both gonna end up at the same temperature once they’re in thermal equilibrium. And so we’ll say that the temperature of the iron at the end of the process, which we’ll call 𝑇 sub 𝑖 superscript fin, is going to be equal to the temperature of the water at the end of the process as well.
So that’s one important piece of information that we’ll need in order to solve our problem. Now we can relate these final temperatures of the iron and of the water to the initial temperatures of the iron and water. Let’s say that the final temperature of the iron is equal to the initial temperature of the iron, which we know to be 1000 degrees Celsius, plus the change in temperature that the iron undergoes in this process, which we’ll call Δ𝑇 sub 𝑖. Similarly, we’ll say that the final temperature of the water is equal to the initial temperature of the water, which happens to be zero degrees Celsius, plus the change in temperature that the water undergoes.
But then this very first equation is telling us that the final temperature of the iron is the same as the final temperature of the water. Therefore, this is equal to this, and so we can equate the two right-hand sides of both of these equations. So instead of talking in terms of 𝑇 sub 𝑖 superscript fin and 𝑇 sub 𝑤 superscript fin, let’s just say that 𝑇 sub 𝑖 plus Δ𝑇 sub 𝑖 is equal to 𝑇 sub 𝑤 plus Δ𝑇 sub 𝑤. Now we can intuitively realize that one of these Δ𝑇 sub 𝑖 or Δ𝑇 sub 𝑤 is going to be negative, because what’s going to happen is that heat is going to be transferred between the iron and the ice such that one of them increases in temperature and the other one decreases in temperature.
But we’ll come to that when we actually do the maths. For now, let’s consider something else. Let’s realize that because the chest is insulated, there is going to be no heat transfer between the contents of the chest and the external environment of the chest. In other words then, we can say that any change in the heat of the contents of the chest is equal to zero. But then, we can split up the heat transfer of the components of the chest into two parts: one is the heat transferred by the iron and the second is the heat transferred by the water. And when we add them together this is equal to zero.
Now at this point, let’s look individually at the heat transfer of the iron and the heat transfer of the water. But before we do, let’s very quickly recall that the specific heat capacity of a substance is defined as the amount of heat change required per unit mass to increase the substance’s temperature by one unit — in this case, one degree Celsius — and the latent heat of fusion of a substance is defined as the heat required to convert one unit mass from solid to liquid. Okay armed with that information then, let’s look at the heat transfer of the iron.
Now in this situation, iron is not anywhere close to its melting point. So we don’t need to consider the melting of iron. We only need to consider the change in temperature of the iron. And having been given the specific heat capacity of the iron, we can then rearrange this equation for the iron to say that the heat transferred by the iron is equal to its mass multiplied by its specific heat capacity multiplied by its change in temperature. And so we can say that Δ𝑄𝑖 becomes 𝑚𝑖𝐶𝑖 Δ𝑇𝑖. And then to this we need to add the heat transfer of the water, but this is a little bit more complicated, because remember we’ve been told that the water is at its melting point. That’s zero degrees Celsius.
Therefore, any heat transferred to the water, or ice in this case, will go towards melting the ice first. And then once all of the ice has completely melted, then any remaining heat will go towards increasing the temperature of the water. So we can take Δ𝑄 sub 𝑤 and split it into two parts. We can split it into the heat transferred into the ice that causes the melting of the ice plus the heat transferred to the water once all of the ice has melted that goes towards increasing the temperature of the water. Now this second bit, Δ𝑄 sub 𝑤 superscript temp, the heat transferred in order to increase the temperature of the water is going to have a similar expression as Δ𝑄 sub 𝑖, because when we’re raising the temperature, the heat transferred is equal to the mass of the water multiplied by the specific heat capacity of the water multiplied by the temperature change of the water.
However, to talk about the heat transferred to the ice that causes it to melt, we need to look at the latent heat of fusion. And once we rearrange this equation, we find that the heat transferred is equal to the latent heat of fusion of water multiplied by the mass of the water. And so we have an expression for Δ𝑄 sub 𝑤 superscript melt as well. So, let’s now sub in our expressions for Δ𝑄 sub melt and Δ𝑄 sub temp. Now at this point, we can see that in this equation the only things that we don’t know the values of are Δ𝑇 sub 𝑖 and Δ𝑇 sub 𝑤. And so it looks like we’ve got one equation with two variables which we can’t solve for.
Except we have another equation that tells us the relationship between Δ𝑇 sub 𝑤 and Δ𝑇 sub 𝑖. So let’s rearrange this equation in order to solve for either Δ𝑇 sub 𝑖 or Δ𝑇 sub 𝑤 and then substitute that back into this equation. So let’s say we solve for Δ𝑇 sub 𝑖. Well then, the right-hand side of that equation becomes 𝑇 sub 𝑤 plus Δ𝑇 sub 𝑤 minus 𝑇 sub 𝑖. Then let’s take the right-hand side of this expression and sub it in to Δ𝑇 sub 𝑖 over here, at which point the only variable we now don’t know when this equation is Δ𝑇 sub 𝑤. So let’s rearrange to solve for Δ𝑇 sub 𝑤. When we do we should see that Δ𝑇 sub 𝑤 is equal to here we go: 𝑚𝑖𝐶𝑖 multiplied by 𝑇𝑖 minus 𝑇𝑤 minus 𝐿𝑤 𝑚𝑤 all divided by 𝑚𝑤 𝐶𝑤 plus 𝑚𝑖𝐶𝑖.
Now at this point, we can substitute in all the quantities on the right-hand side of this equation. When we do, it looks something like this and, remember, we need to convert the latent heat of vaporization of water from kilojoules per kilogram into joules per kilogram. We do this by multiplying by a factor of 10 to the power of three. And then at this point, we can find a value for Δ𝑇 sub 𝑤 by evaluating the right-hand side of this equation. When we do, we find that Δ𝑇 sub 𝑤 is equal to 6.077 dot dot dot degrees Celsius. But because the lowest number of significant figures that we were given a quantity to in the question is two significant figures, that’s 𝑚 sub 𝑤, then we need to round our answer to two significant figures as well.
Now let’s be careful here, because the temperature of the water we’ve said is zero degrees Celsius. Isn’t that one significant figure? Well yes it is, but in the question all we were told was that the ice was at its melting point. So we could have very easily written this as 0.00000 degrees Celsius. And so the quantity with the lowest number of significant figures in the question was 𝑚 sub 𝑤. So anyway, Δ𝑇 sub 𝑤 becomes 6.1 degrees Celsius to two significant figures. Now we’ve just had to do a lot of maths here. So let’s take a step back and think about what Δ𝑇 sub 𝑤 actually means.
Well Δ𝑇 sub 𝑤 is the change in temperature of the ice or water once all of the heat exchange is complete between the iron and the ice and the two are in thermal equilibrium. Now because this value is positive, this means that the temperature of the ice by the end of the process will have increased by 6.1 degrees Celsius. Remember we said earlier that the final temperature of the water was equal to the initial temperature of the water plus the change in temperature of the water. Well, the initial temperature of the water was zero degrees Celsius, and we just worked out Δ𝑇 sub 𝑤.
Therefore zero degrees Celsius plus 6.1 degrees Celsius give us a value of 6.1 degrees Celsius for the final temperature of the water. Now we very easily could’ve worked out the value of Δ𝑇 sub 𝑖 as well. And if we did that, we’d find it to be negative. Moreover, we’d also find that the final temperature of the iron cylinder was also 6.1 degrees Celsius. But then this shouldn’t surprise us, because this is what we stated was a condition of thermal equilibrium. But anyway, at this point, we have our final answer: the equilibrium temperature is 6.1 degrees Celsius.