### Video Transcript

A Carnot engine working between two heat baths of temperature 600 kelvin and 273 kelvin completes each cycle in 5.00 seconds. In each cycle, the engine absorbs 10.0 kilojoules of heat. Find the output power of the engine.

Okay, so in the Carnot engine, which can be represented by this circle that we’ve drawn at the diagram here, what happens is that some amount of heat is taken from a hot reservoir. Now, this hot reservoir is a temperature, which we’ll call 𝑇 sub ℎ for temperature sub hot.

Now, the Carnot engine converts this heat that’s taken in from the hot reservoir into some form of work which we’ll call 𝑤 and it rejects the rest of the heat into the cold reservoir, which happens to be at a temperature which we’ll call 𝑇 sub 𝑐. As well as this, we’ll call the heat absorbed by the Carnot engine 𝑄 sub ℎ and the heat rejected by the Carnot engine 𝑄 sub 𝑐 because 𝑄 sub ℎ is the heat coming from the hot reservoir and 𝑄 sub 𝑐 is the heat rejected to the cold reservoir.

Now, the Carnot engine works in cycles. Therefore, every cycle the Carnot engine absorbs the heat 𝑄 sub ℎ, rejects the heat 𝑄 sub 𝑐, and produces work 𝑤. But then knowing the values of 𝑄 sub ℎ, 𝑄 sub 𝑐, and 𝑤, we can work out the efficiency of the Carnot engine.

Now, the efficiency 𝜂 is defined as the useful energy output divided by the total energy input, which in our case happens to be the work done by the Carnot engine divided by the total heat input into the Carnot engine because the useful energy out is the work that the Carnot engine does and the total energy in of course is the heat absorbed from the hot reservoir.

So we’ve got an expression for the efficiency of the Carnot engine. But why is this useful? Well, we can recall that for a Carnot engine, there’s another way to work out the efficiency. Specifically, for a Carnot engine, the efficiency can be found by finding one minus the cold reservoir’s temperature divided by the hot reservoir’s temperature.

In other words then, the definition that we use for efficiency here — the useful energy output divided by total energy input — is a very general definition of efficiency. And the expression that we’ve used here is very specific to a Carnot cycle. But then because on the right-hand side of this expression, we’ve used terms from the Carnot engine itself, we’ve, therefore, worked out another expression for the efficiency of the Carnot engine.

And so we can equate the right-hand side of this equation and the right-hand side of this equation because they’re both measuring the efficiency of the Carnot engine. So once we do equate the two, we can then see that within the question itself we’ve been given a value of 𝑄 sub ℎ.

The amount of energy absorbed by the Carnot engine per cycle is 10.0 kilojoules. As well as this, we know 𝑇 sub 𝑐, which happens to be 273 kelvin. And we know 𝑇 sub ℎ, which happens to be 600 kelvin. This means that there’s only one unknown variable in this equation: that’s the work done 𝑤. So we can rearrange to find this work.

So let’s multiply both sides of the equation by 𝑄 sub ℎ, which means it cancels on the left-hand side. And at this point, we have an expression for the energy output from the Carnot engine per cycle because we know that every cycle, a work 𝑤 is done by the Carnot engine. And this work 𝑤 is actually the output of the Carnot engine. However, what we’ve been asked to find is the output power of the engine.

So to work out the power, we can recall that this is equal to the rate of energy transfer, in other words, the amount of energy transferred per unit time. Now, what we’ve worked out is the output energy from the Carnot engine per cycle. But then, we know that each cycle takes five seconds to complete.

And so if we divide both sides of this equation by the time taken per cycle, which we will call 𝑡, then what we’ve got now on the left-hand side is the output power because this is equivalent to the output energy from the Carnot engine per cycle divided by the time taken to complete a cycle. And so we can replace the left-hand side of the equation with the output power 𝑃.

Now, at this point, all we have to do is the plug in the values on the right-hand side of the equation, which looks a bit like this: we’ve got 10.0 kilojoules, that’s the heat transferred to the Carnot engine per cycle, divided by the time taken per cycle, that’s 5.00 seconds, multiplied by one minus 273 kelvin, that’s 𝑇 sub 𝑐, divided by 600 kelvin, that’s 𝑇 sub ℎ. And when we evaluate the right-hand side of the equation, we get 1.090 kilojoules per second.

Now, at this point, we can actually convert this quantity into joules per second because we can recall that one joule transferred per second is equivalent to one watt of power. And so we convert this quantity into joules per second by remembering that one kilojoule is equivalent to 1000 joules. So we multiply 1.090 by 1000. And so we’re left with 1090 joules per second or 1090 watts. And that is our final answer.