Ch 24: The Second Law of ThermodynamicsWorksheetSee all chapters
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Ch 01: Intro to Physics; Units
Ch 02: 1D Motion / Kinematics
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Ch 07: Friction, Inclines, Systems
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Ch 22: Kinetic Theory of Ideal Gasses
Ch 23: The First Law of Thermodynamics
Ch 24: The Second Law of Thermodynamics
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Ch 39: Atomic Structure
Ch 40: Nuclear Physics
Ch 41: Quantum Mechanics
Intro to Heat Engines
Efficiency of Heat Engines
Heat Engines & PV Diagrams
Four Stroke Piston Engine
Carnot Cycle
Intro to Refrigerators
Refrigerators & PV Diagrams
Entropy and the Second Law
Entropy & The Second Law
Statistical Interpretation of Entropy
Second Law for Ideal Gas

Concept #1: Heat Engine Efficiency


Hey guys, in this video we're going to talk about the efficiency of heat engines. Now heat engines absolutely cannot convert all of the heat available to them from a hot reservoir into usable work. This would violate the second law of thermodynamics. So by physical law, engines can only convert so much heat into work and therefore we can define an efficiency as how good that engine is at converting heat into work. Let's get to it. Now remember guys the energy used in a heat engine is heat flow from a hot reservoir to a cold reservoir. Heat flows from the hot into the engine, from the engine into the cold reservoir and that heat flow is what powers the engine and allows the engine to output work. The heat emitted from a hot reservoir can never never be converted completely into usable work, never, violation of the second law. This is a consequation of the second law of thermodynamics which as I said we'll cover later.

Different engines have different amounts of usable work for a particular heat flow. So given let's say that one hot, one reservoir is at 300 Kelvin, one reservoir is at 100 Kelvin. One engine might output X amount of work. For the same reservoirs, one engine might output 2X of work. For the same reservoirs, one engine my output 1 half X of work. Different engines are gonna output different amounts of work for the same heat flow right and all of the examples is the same temperature of the hot reservoir, same temperature for the cold reservoir but you should have outputs of different work because there are different engines. Now how good, in quotes, how good an engine is at converting heat into work is known as the efficiency of the engine and the efficiency is very easily defined, it's defined as the amount of work output by the engine divided by the heat input into the engine so the heat from the hot reservoir. Very very straightforward. If somehow an engine could produce no work, no matter how much heat you put into it. You could have a reservoir that's at a billion degrees 1 times 10 to the 9 Kelvin and a cold reservoir or at 1 Kelvin this is a huge temperature difference but even for that 0 joules of work is output, this is an engine that would have a 0 efficiency which would be the quintessential bad engine, the worst possible engine you could think of because it doesn't produce any work no matter how big the temperature gradient is for the reservoirs. If an engine could, I all CAPS-ed it because this is a hypothetical it's mathematically possible but it's not possible in physics it's physically barren by the second law of thermodynamics. If an engine could somehow convert all of the heat input into work, it would have a one hundred percent efficiency. If the work equals the amount of heat input so we would have some hot reservoir, we would have an engine, we would have work and then the cold reservoir would receive nothing and work equaled this. If this engine were possible, which this is absolutely not possible, if it were possible it would have a one hundred percent efficiency and this would be the quintessential perfect engine. Theoretically the most efficient engine is known as the carnot engine. This is the theoretically, the engine with the maximum efficiency for a given temperature gradient. So when between a hot reservoir at TH and a cold reservoir at TC, this efficiency is one minus TH over TC. No engine could possibly be more efficient than this for a given hot reservoir and given a cold reservoir. So our 10 to the 9 Kelvin and 1 Kelvin reservoir, no engine that you could ever put in between those two reservoirs could ever get more efficient than a carnot engine.

So let's do an example on this. You read about a proposed engine that only wastes 30 percent of the energy input when connected to a hot reservoir of 1000 Kelvin and a cold reservoir of 200 Kelvin. Do you think such an engine could exist? Not is it practically possible, not is it feasible given production costs but could it be allowed by physical law? For a hot reservoir at 1000 Kelvin and a cold reservoir at 200 Kelvin, the maximum allowed efficiency by the second law of thermodynamics is 1 minus T hot over T cold, sorry this by the way is totally backwards, this is T cold over T hot otherwise it wouldn't make any sense it's the smaller number divided by the bigger number that's my bad. T cold over T hot which is 1 minus 200 over 1000 which is 1 minus 0.2 which is 0.8 which is 80 percent. So 80 percent of all of the heat added into this engine is converted into work. That means that the minimum, sorry, theoretical wasted energy is 20 percent wasted. That's the theoretical minimum. If eighty percent is used and converted to work then 20 percent is still wasted that's heat that has to go back to the cold reservoir. The proposed amount was 30 and this is less than 30 percent the proposed amount so this engine is absolutely theoretically possible. It's possible because the proposed efficiency of the engine is not greater than the carnot efficiency. If this engine uses, sorry, loses 30 percent of energy as radiated heat then it can only convert 70 percent of energy into work so it's 70 percent efficient which is lower than the carnot efficiency so absolutely, this engine does not violate the second law of thermodynamics and is physically possible. Alright guys, that wraps up this discussion on heat engines and their efficiencies. Thanks for watching.

Example #1: Four-Stroke Internal Combustion Engine


Hey guys, let's do an example. A four-stroke internal combustion engine is a type of engine found commonly in cars. A hypothetical car engine can output 300 horsepowers of power if it's operating at 6000 cycles per minute. So 6000 times a minute it, produces work and it produces work at a rate of 300 horsepowers. If the efficiency of the engine is 40 percent, how much heat needs to be input in each cycle of the engine? How much heat is released in the cycle? Note that one horsepower is 746 watts.

So what we're given is the power and that's the amount of work done by the engine per unit time and this is 30, sorry, 300 horsepower but we want to convert this so that we can get it in SI units of watts because watts are joules per second. So that will tell us if we can figure out how long each cycle lasts, how long each completed cycle of the heat engine takes, we can figure out how much work is released by each cycle of the engine. Now we're told that one horsepower equals 746 watts so if you notice the horsepower is cancelled here and we're left with watts, the unit that we want. Plugging this is into a calculator we get 223800 watts which is 223.8 and a kilowatt is just a kilojoules per second. So the question we need to answer is how long does each cycle take? If we can figure out how long each cycle takes, then the work released in each cycle is just the power times the duration of each cycle. Now what we're told is not how long cycle last but how frequent the cycles are. We're told the frequency is 6000 cycles per minute. Now we can convert this into cycles per second which would be an SI unit by saying one minute goes into 60 seconds and then the minutes will cancel and it's going to be cycles per second which is absolutely an SI unit and this is 100, right 6000 divided by 60, cycles per second. Now the thing to remember here is that the period which is how long each cycle takes which is what we're looking for right this is by definition the period is one over the frequency. So this is 1 over 60 cycles per second sorry it's 100 cycles per second, it was 6000 cycles per minute which is 0.01 seconds that's how long each cycle lasts. This is by definition the time duration for each cycle. Knowing that we can say then that the work due to each cycle is just the power times delta T of each cycle, this equation which we already said, and that's going to be 223.8 kilojoules per second so remember our answer is going to be in kilojoules time 0.01 seconds which is 2.24 kilojoules. Don't forget the answer is in kilojoules not joules. So now we know how much work is produced by this engine per cycle it's 2.24 kilojoules, we are told that the efficiency of the engine is 40 percent or 0.4 and that equals the work done divided by the amount of heat that's input. So the amount of heat that's input per cycle is going to be 0.4 times the work released per cycle which is 0.4 times 2.24 kilojoules which is 5.6 kilojoules.

That's how much heat is released every cycle of the engine and remember guys the diagram for an engine, we have some heat flowing in from the hot reservoir into the engine, it converts some of it into work but some of it has to return to the cold reservoir and this tells us that the hot reservoir magnitude, the heat from the hot magnitude is equal to the heat from the cold magnitude plus W. So we know the heat returning to the cold reservoir which is as the question put it the amount of heat released by the engine, this is going to be Q hot minus W which is 5.6 minus 2.24 kilojoules and that is 3.36 kilojoules. That's how much heat is released by this engine every cycle. So every cycle the hot reservoir puts 5.6 kilojoules of heat into the engine, it converts 2.24 kilojoules of that into work and returns 3.36 kilojoules of that back into the cold reservoir releases that as heat and it does this 6000 times per minute to produce the 300 horsepowers of the car engine. Alright guys, that wraps up this problem. Thanks for watching.