Video: Calculating the Rate of Entropy Change in the Reservoirs of a Heat Engine

For the Carnot fcycle in the figure, consider the efficiency to be 0.65 and the heat consumed per cycle to be 300 J. What is the entropy change per cycle of the hot reservoir at 70Β°C? What is the entropy change per cycle of the cold reservoir?

08:52

Video Transcript

For the Carnot fcycle in the figure, consider the efficiency to be 0.65 and the heat consumed per cycle to be 300 joules. What is the entropy change per cycle of the hot reservoir at 70 degrees Celsius? What is the entropy change per cycle of the cold reservoir?

Okay, so in this question, we’ve got a Carnot cycle, which we can see in the diagram. And we’ve been told that the efficiency of this Carnot cycle is 0.65, and the heat consumed per cycle is 300 joules. Now this heat consumed per cycle seems to be 𝑄 sub β„Ž in the diagram. This is because the heat absorbed by the Carnot engine is 𝑄 sub β„Ž, and this heat gets converted to 𝑄 sub 𝑐 and the work done π‘Š.

Now, the first part of the question asks us to calculate the entropy change per cycle of the hot reservoir, which is at 70 degrees Celsius. To do this, we first need to recall how to calculate the entropy change of a process. The entropy change of a process Δ𝑆 is given by the integral of d𝑄 over 𝑇, where 𝑄 is the heat exchanged and 𝑇 is the temperature at which this process occurs. So we can use this equation to work out the entropy change per cycle of the hot reservoir.

Now, we need to be really careful here. Basically we know that the engine, the Carnot engine, is taking in a heat of 𝑄 sub β„Ž, and this heat is coming from the hot reservoir. Therefore, the hot reservoir itself is losing this heat 𝑄 sub β„Ž. So if we have to calculate the entropy change of the hot reservoir, then we know that this reservoir is losing heat. And this will become important in a second.

So we can say that the entropy change of the hot reservoir per cycle, which we’ve called Δ𝑆 sub β„Ž, is equal to the integral of d𝑄 over 𝑇 sub β„Ž, where 𝑄 is the heat exchanged by the reservoir and 𝑇 sub β„Ž is the temperature of the reservoir. Now we need to mention what the limits of the integral are. To do this, let’s first assume that the heat contained within the hot reservoir is 𝑄 sub res for the heat inside the reservoir. And then after the Carnot engine has drawn in the heat, 𝑄 sub β„Ž, the heat in the reservoir therefore is 𝑄 sub res, which is what it had initially, minus 𝑄 sub β„Ž, which is what it’s lost. And so those will form our limits. The integral is from 𝑄 sub res to 𝑄 sub res minus 𝑄 sub β„Ž.

Now, the other thing is is that the temperature of the hot reservoir 𝑇 sub β„Ž stays constant. So we can pull it out of the integral, and then we can evaluate it. So we can pull it out of the integral, and then we can evaluate the integral: the integral simply evaluated to 𝑄 between 𝑄 sub res and 𝑄 sub res minus 𝑄 sub β„Ž. Plugging in our limits, we find that the integral finally evaluates to negative 𝑄 sub β„Ž. And so we know that the entropy change of the hot reservoir is equal to negative 𝑄 sub β„Ž over 𝑇 sub β„Ž. Now at this point, we can start thinking about plugging in the numbers.

We know that the value of 𝑄 sub β„Ž is 300 joules. So we have negative 300 joules divided by the temperature of the hot reservoir, which is 70 degrees Celsius. But we need to convert this to Kelvin. So Δ𝑆 sub β„Ž ends up being negative 300 joules divided by 70 plus 273 Kelvin. Evaluating the fraction, we find that this is equal to negative 0.874 dot dot dot joules per Kelvin. To two significant figures, this value is negative 0.87 joules per Kelvin, which means that we can move on to finding out the entropy change per cycle of the cold reservoir. This time we are asked to work out the same quantity, the entropy change per cycle, but for the cold reservoir.

Now we could of course work this out, but there’s a quick way of doing this. We know that the Carnot cycle is a reversible cycle, which means that we can invoke Clausius’s theorem. Clausius’s theorem states that the integral over a closed loop of d𝑄 over 𝑇 is equal to zero, and this holds for a reversible cycle. Now as we saw earlier, the integral of d𝑄 over 𝑇 gives a change in entropy. But because we’ve got a reversible cycle here, we can integrate that over a closed loop. And the theorem tells us that this value is zero, and the theorem tells us that this integral is equal to zero. And once again, this is known as Clausius’s theorem.

Now what does this mean for us? Well it means that the entropy change of a Carnot cycle, which is a reversible cycle, is zero if we consider the entire cycle. In other words, the entropy change of the hot reservoir must be cancelled out by the entropy change of the cold reservoir. Because in one cycle of the Carnot cycle, the total entropy change must be zero. Therefore, the entropy change per cycle of the cold reservoir must be positive 0.87 joules per Kelvin. This is in order to cancel out the entropy change of the hot reservoir. However, let’s not just take my word for it. Let’s actually calculate what Δ𝑆 sub 𝑐 ends up being.

As we saw earlier, Δ𝑆 sub 𝑐 will be the integral of d𝑄 over 𝑇 sub 𝑐, where 𝑇 sub 𝑐 is the temperature of the cold reservoir. Now this time, let’s say that the cold reservoir initially started off with a heat of 𝑄 sub 𝑐 res for cold reservoir, and then it gains a heat 𝑄 sub 𝑐. So the final heat is 𝑄 sub 𝑐 res plus 𝑄 sub 𝑐 because, remember, the Carnot cycle dumps heat into this cold reservoir. Therefore, the heat of the cold reservoir increases. So let’s put the limit into our integral. Now once we’ve done that, we can see that the integral will evaluate in a similar way to what we did earlier.

We find out that when we put the limits in, this is what we end up getting. And so the integral just simplifies to 𝑄 sub 𝑐. Therefore, the entropy change of the cold reservoir is 𝑄 sub 𝑐 over 𝑇 sub 𝑐. Unfortunately for us though, we don’t know either of these quantities here. We don’t know what 𝑄 sub 𝑐 is, and we don’t know what 𝑇 sub 𝑐 is. Fortunately for us, we’ve got enough information to work it out. Let’s start by working out what 𝑄 sub 𝑐 is. We can do this by utilizing the efficiency of the Carnot engine. The efficiency of an engine is defined as the useful energy output divided by the total energy input. In the case of our Carnot cycle, the efficiency happens to be the useful energy output, which is the work done, because it’s an engine it is meant to do work, divided by the total energy input, which is 𝑄 sub β„Ž, because that’s how much heat is taking in.

Now we already know that the efficiency is 0.65, and we also know what 𝑄 sub β„Ž is. We know that it’s 300 joules. So we can work out what the work done by the Carnot engine is. The work done is 0.65 times 300 joules, which ends up being 195 joules. We can use this to calculate what 𝑄 sub 𝑐 is because we know that the Carnot engine converts the heat 𝑄 sub β„Ž into the heat 𝑄 sub 𝑐 and the work π‘Š. In other words, 𝑄 sub β„Ž is equal to 𝑄 sub 𝑐 plus π‘Š. And so 𝑄 sub 𝑐 is equal to 𝑄 sub β„Ž minus π‘Š. This is 300 joules minus 195 joules, which ends up being 105 joules. So we can plug this into our equation here, 105 joules, divided byβ€” we still need to find out what 𝑇 sub 𝑐 is.

To find out the value of 𝑇 sub 𝑐, we need to remember what the efficiency of a Carnot engine specifically is defined as. To find out the value of 𝑇 sub 𝑐, we need to remember how we calculate the efficiency specifically for a Carnot engine. For a Carnot engine, the efficiency is one minus 𝑇 sub 𝑐 over 𝑇 sub β„Ž. And once again, we know the efficiency is 0.65. And we know that this must be equal to one minus 𝑇 sub 𝑐 over 𝑇 sub β„Ž, which happens to be 70 plus 273 Kelvin as we saw earlier. We can rearrange this to give us 𝑇 sub 𝑐 is equal to 70 plus 273 kelvin multiplied by one minus 0.65. And this ends up being 120.05 Kelvin. So we found the value of 𝑇 sub 𝑐 and we can plug it into our equation.

At this point we’ve got all the information we need to find out what Δ𝑆 sub 𝑐 is, the entropy change per cycle of the cold reservoir. Evaluating the fraction gives us an entropy change of 0.874 joules per Kelvin, which to two significant figures is 0.87 joules per Kelvin just as we expected. Now before we go, a very final point: we’ve been asked to find the entropy change per cycle in both cases. So have we ignored the per cycle bit? Have we not taken into account that this is per cycle? Well no, don’t worry, we have accounted for this because the heats 𝑄 sub β„Ž and 𝑄 sub 𝑐 are the heat taken from the hot reservoir and the heat passed onto the cold reservoir respectively per cycle. In other words, every cycle, the Carnot engine takes in a heat 𝑄 sub β„Ž and gives out a heat 𝑄 sub 𝑐. So by using these values, we’ve already worked out the entropy change per cycle. So just to recap, the entropy change per cycle of the hot reservoir is negative 0.87 joules per Kelvin and the entropy change per cycle of the cold reservoir is positive 0.87 joules per Kelvin.

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