The Carnot Cycle and the Second Law

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Imagine a perfect engine: no friction, no leaks, nothing wasted. You give it one thousand joules of heat. How much work do you get back? All one thousand joules, right? Well, physics says no, not even for a perfect engine. Today we will see the best any engine can do, and the law behind that limit. This comes up often in JEE, so let's begin.

0:31The Carnot cycle: four reversible steps

So, what does the best possible engine look like? It is the Carnot engine, running on this cycle. In the figure, small p, the pressure, goes up, and capital V, the volume, goes across. The gas goes round the points capital A, B, C and D, and every step is reversible. With me? First, from A to B; the arrow means goes to. The gas touches the hot source at temperature T one: capital T, with a small one below. It expands at this constant temperature, the orange isothermal curve, and absorbs heat Q one, capital Q with a small one. Okay? Next, from B to C, the dark blue curve. No heat flows in or out, so this step is adiabatic. The gas keeps expanding and cools to T two, the cold sink's temperature. Right? Then, from C to D, the green curve. The gas touches the cold sink and is compressed at the constant temperature T two, an isothermal compression, so it rejects heat Q two. Clear? Now we are at D, at T two. We must get back to A, at T one, but there is no heat source in this last step. So how do we warm the gas up? Compress it adiabatically, yes! From D to A, the work done on the gas raises its temperature back to T one. And as the caption says, the enclosed area is the work done per cycle. Got it?

1:59Carnot efficiency: only the temperatures matter

Now, how good is this engine? Here Q one, Q two and the work W are all positive amounts, and W equals Q one minus Q two. The efficiency, the Greek letter eta, like an n with a long tail, is W over Q one, which is one minus Q two over Q one. Okay? Now the special thing about a Carnot engine. Its two heats are in the same ratio as the two temperatures: Q two over Q one equals T two over T one. This line is the heart of the Carnot cycle. With me? So, put this into our formula for eta. What is the efficiency of a Carnot engine? One minus T two over T one, exactly! The word Carnot below the eta just names it. Look, there is no heat and no work in it, only the two temperatures. Right? Now, questions often give the temperatures in degrees Celsius. Can we put those numbers straight into this formula? No, no, careful here! T one and T two must be absolute temperatures, in kelvin; the small circle before the C means degrees Celsius. So always add two hundred and seventy three first. Clear?

3:11Carnot refrigerator: the cycle run backwards

Now, a Carnot engine is reversible, so we can run it backwards. Then it is a refrigerator: it takes heat Q two from the cold body, uses work W, and gives Q one to the hot body. Its coefficient of performance, the Greek letter beta, is Q two over W, and W is Q one minus Q two. Okay? Now use the Carnot line again. Divide the top and the bottom by Q one. What does beta become? Beta equals T two over, T one minus T two. Very good! Again in kelvin, and T two is the cold one, the inside of the fridge. Got it?

3:48Carnot's theorem and the second law

Now, Carnot's theorem: no engine working between two given temperatures can beat a Carnot engine. So the Carnot efficiency is the ceiling; nothing can go above it. With me? And all reversible engines between the same two temperatures have exactly the same efficiency. Clear? Now, back to our puzzle. Our perfect engine rejects nothing, so Q two is zero, and eta is one, a hundred percent. Is that possible? No, it is not! This is the second law. The Kelvin Planck statement: no engine working in a cycle can convert all the heat it absorbs into work. Some heat must always go to a cold sink, so eta is always less than one. That is our puzzle solved, okay? And the second statement, from Clausius: heat cannot flow by itself from a colder body to a hotter one. That is why a refrigerator needs work. Two statements, one law. Got it?

4:45Example 1: Carnot efficiency and work, 500 K to 300 K

Alright, examples. Find the efficiency of a Carnot engine between five hundred kelvin and three hundred kelvin, and its work for one thousand joules absorbed; capital K is kelvin, and capital J is joules. Pause the video and try it first. Done? Okay, let's check. T one, the hot source, is five hundred kelvin, and T two, the cold sink, is three hundred kelvin. Right? So eta equals one minus three hundred over five hundred, that is one minus three fifths, which is two fifths: forty percent. Clear? The work is eta times Q one: zero point four times one thousand joules, which is four hundred joules. So even the best possible engine here turns our thousand joules into only four hundred joules of work. Okay? The rest, Q one minus W, one thousand minus four hundred, that is six hundred joules, goes to the cold sink. With me?

5:37Example 2: two Carnot engines in series, find middle T

Now a harder one. Two Carnot engines work in series between eight hundred kelvin and two hundred kelvin. The first rejects heat to a middle reservoir at temperature capital T, which feeds the second engine. Find T if, part a, their efficiencies are equal, and part b, their works are equal. Here is a hint: write each engine's efficiency, eta, first. Pause the video and try it now. Done? Okay, part a first. Equal efficiencies, so T should be exactly halfway, five hundred kelvin, right? Oh no, careful here, do not just guess! The first engine works between eight hundred and T, so eta one, with a small one below, is one minus T over eight hundred. The second works between T and two hundred, so eta two is one minus two hundred over T, okay? Set them equal: T over eight hundred equals two hundred over T, right? The double arrow means so: T squared is eight hundred times two hundred, one hundred and sixty thousand. So T is four hundred kelvin, not five hundred! Got it? Now part b, equal works. Use the Carnot line: heat over temperature is the same at every reservoir. Q one, from the hot source, over eight hundred, equals Q, into the middle reservoir, over T, equals Q two, to the cold sink, over two hundred. With me? So each heat is proportional to its temperature; this sign, like a stretched alpha, means proportional to. And each engine's work is heat in minus heat out, so it is proportional to the temperature drop. Clear? So equal works mean equal drops. Now tell me, what is T? Eight hundred minus T equals T minus two hundred, so two T is one thousand, and T is five hundred kelvin. Halfway is right for equal works, not for equal efficiencies. That is the trap in this question. Got it?

7:41Quick revision: Carnot efficiency, fridge COP, second law

So, quick revision. For a Carnot engine, Q two over Q one equals T two over T one, and eta equals one minus T two over T one, in kelvin. For a Carnot refrigerator, beta equals T two over, T one minus T two. And no engine beats Carnot; no cycle turns all the heat it absorbs into work. Now go and try the practice problems on this page. All the best!