Heat Engines and Refrigerators

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle first. It is a hot summer afternoon, and you are in a small closed room with a refrigerator. So you open the fridge door wide, to cool the whole room. The fridge makes things cold, so the room gets cooler, right? Hmm, are you sure about that? By the end of this video, you will know. Today: heat engines, refrigerators and heat pumps. The formulas are short, the traps are real, and questions on them come up often in JEE, so let's begin.

0:34Heat engine: heat in, work out

So, what is a heat engine? It is a device that works in a cycle, turning heat into work. Look at the figure. At the top, a hot reservoir at temperature T one: capital T, with a small one below. The engine takes heat Q one from it, the red arrow: capital Q with a small one below. Okay? Then it rejects heat Q two to the cold reservoir at temperature T two, the blue arrow. The orange arrow is the work delivered, labelled W equals Q one minus Q two. Let us see why, right? Now think. After one full cycle, the engine is back in its starting state. Delta U, with the capital Greek delta, means the change in internal energy. So what is delta U over one full cycle? Zero, exactly! Same state, same internal energy. Now the first law, with our sign convention: Q equals delta U plus W, where Q is the heat given to the system, and W is the work done by the system. Clear? So the net heat in equals the work out. Q one goes in and Q two comes out, so W equals Q one minus Q two, as the figure says. Here Q one, Q two and W are all positive amounts; the arrows show their directions. With me? Now, how good is an engine? We pay Q one as heat, and we get W as work, so we compare W with Q one. Can an engine ever give us all of Q one as work, so that W equals Q one? No, it cannot! The efficiency, the Greek letter eta, equals W over Q one, which is one minus Q two over Q one; the same formula sits at the top right of the figure. The less-than sign says it is always below one: some heat Q two always goes to the cold reservoir. The reason is the second law. That, and the Carnot line below it, one minus T two over T one, with temperatures in kelvin, come in the next lesson.

2:35Refrigerator: the engine run backwards

Okay? Now run the engine backwards, and you get a refrigerator. It takes heat Q two from the cold body, the inside of the fridge, okay? For that, it needs work W, done by its compressor. Right? Now, Q two goes into the machine, and the work W goes in too. Energy cannot just disappear. So where does all of it end up? In the hot body, the room outside the fridge, exactly! Q one equals Q two plus W: the heat delivered to the hot side is the heat removed, plus the work. Clear? Now, how good is a fridge? What we want is Q two removed, and what we pay is the work. So its coefficient of performance, the Greek letter beta, is Q two over W, got it? Now tell me. The efficiency of an engine is always less than one. Can beta be more than one? Yes, it can! The fridge does not turn its work into Q two. It only moves heat that is already there, from cold to hot. So a big beta is perfectly normal; just never call it an efficiency. Okay?

3:43Heat pump: the same device, used for heating

Now use the same device to heat a room in winter, and we call it a heat pump. This time, we want Q one, the heat delivered to the warm side. Right? So its COP, the coefficient of performance, is Q one over W, okay? And Q one is Q two plus W, remember? Can you write this using beta? Split the fraction: Q two over W, plus W over W, and that is beta plus one! So as a heat pump, the same machine has a COP one more than as a refrigerator. So always ask first: which heat do we want, Q two or Q one?

4:22Efficiency of a cycle: careful with the denominator

Clear? One last rule. Questions often give a cycle made of known processes, and ask for its efficiency. Then eta equals the net work, over the heat absorbed. Okay? Now, some steps of the cycle absorb heat, and some reject it. So for the heat absorbed, we just take the net heat of the whole cycle, right? No, no, careful here! Over a cycle, the net heat equals the net work, remember? So eta would be one, a perfect engine, every time. That is impossible. With me? So in the denominator, add only the positive heats, the steps where heat goes in. The heat rejected stays out.

5:04Example 1: fridge COP and why an open fridge warms a room

Got it? Alright, the first example. The refrigerator removes nine hundred joules from its cold compartment, using one hundred joules of work; capital J is joules. Find its COP, and the heat delivered to the room. Pause the video and try it first. Done? Okay, let's check. Beta is Q two over W: nine hundred over one hundred, which is nine. Much more than one, and perfectly fine. With me? Now the heat delivered to the room. The fridge removed nine hundred joules, so it delivers nine hundred joules to the room, right? Oh no, careful here: the work also ends up as heat. Q one equals Q two plus W: nine hundred plus one hundred, which is one thousand joules. Clear? Now, our puzzle from the start. Leave the door open in a closed room, and the inside of the fridge becomes part of the room. So, does the room get cooler, or warmer? Look. The fridge pulls nine hundred joules out of the room, through the open door. And it gives one thousand joules back to the same room. Okay? So the net heat into the room is one hundred joules: exactly the work W of the compressor. The room warms up, and that solves our mystery! An open fridge in a closed room is a heater. Got it?

6:30Example 2: 5 kW engine at 25%, heat absorbed and rejected

Now a slightly harder one. An engine of efficiency twenty five percent delivers five kilowatts; k W is kilowatts. Find the heat absorbed and rejected per second. Pause the video and try it first. Done? Okay, let's check. Five kilowatts means five kilojoules of work every second. So in one second, W is five kilojoules; k J is kilojoules. With me? Now Q one. The efficiency is twenty five percent, so Q one is zero point two five times five, which is one point two five kilojoules, right? No, no, that is the trap! The heat absorbed can never be less than the work it gives. Eta equals W over Q one, so Q one equals W over eta: five over zero point two five, which is twenty kilojoules per second. Okay? And the rest goes to the cold reservoir. Q two equals Q one minus W: twenty minus five, which is fifteen kilojoules per second. Clear? Quick check: one minus fifteen over twenty is zero point two five, which is twenty five percent, exactly as given, so everything matches. Very good work!

7:42Quick revision: efficiency, refrigerator COP, heat pump COP

So, quick revision. Engine: W equals Q one minus Q two, and eta equals W over Q one, one minus Q two over Q one, always less than one. Refrigerator: beta equals Q two over W, and Q one equals Q two plus the work; for a heat pump, COP equals beta plus one. For a cycle, eta is the net work over the heat absorbed, positive heats only. Now go and try the practice problems on this page. All the best!