Sections | |||
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Intro to Heat Engines | 12 mins | 0 completed | Learn |

Efficiency of Heat Engines | 14 mins | 0 completed | Learn |

Heat Engines & PV Diagrams | 19 mins | 0 completed | Learn |

Four Stroke Piston Engine | 44 mins | 0 completed | Learn |

Carnot Cycle | 8 mins | 0 completed | Learn |

Refrigerators | 54 mins | 0 completed | Learn |

Intro to Refrigerators | 27 mins | 0 completed | Learn |

Refrigerators & PV Diagrams | 27 mins | 0 completed | Learn |

Entropy and the Second Law | 46 mins | 0 completed | Learn |

Entropy & The Second Law | 27 mins | 0 completed | Learn |

Statistical Interpretation of Entropy | 12 mins | 0 completed | Learn |

Second Law for Ideal Gas | 7 mins | 0 completed | Learn |

Concept #1: Introduction to the Second Law of Thermodynamics

**Transcript**

Hey guys, in this video we're going to introduce the second law of thermodynamics in two forms known as the engine form and the refrigerator form. Alright let's get to it. Remember guys that real engines and real refrigerators have an efficiency. For an engine we call it the efficiency for a refrigerator we call it the coefficient performance but the efficiency is what determines how much work is going to be produced by the refrigerator sorry by the engine for a given heat flow from the hot to the cold reservoir. The coefficient of performance for a refrigerator determines how much heat is going to flow through the refrigerator against the normal flow for a given input of work.

Real engines can never have a 100 percent efficiency, real refrigerators can never have an infinite coefficient of performance and these are restricted by the first law sorry by the second law of thermodynamics. There are a lot of forms of the second law and we'll see a few of them but the first two very important forms are what we're going to discuss here. The one useful form is the engine form. The engine form of the second law says it's impossible for any cyclic system to convert heat absorbed from a reservoir completely into work. So this statement of the second law of thermodynamics is nothing new we already know this that no engine which does cyclic processes on gas, no cyclic system can entirely convert, can completely convert heat input from reservoir into work. That's impossible, we've already known this. Similarly there's a refrigerator form of the second law of thermodynamics which says that it's impossible for any system, this says specifically a cyclic system but this is for any system, it's impossible for system to transfer heat from a cold reservoir to a hot reservoir without work and this once again is nothing new we already know that this was impossible but there is a particular form of the second author of thermodynamics that says it's impossible. So why is it impossible? Because of the second law of thermodynamics specifically the refrigerator form. Both of these impossibilities can be described in terms of heat flow diagrams. We've seen these types of diagrams before we just didn't really name them but now I just got sick of writing diagrams that show the heat and whatever heat flow diagrams.

This is an impossible on the left, an impossible perfect engine. This engine takes however much heat is absorbed by it from the hot reservoir and converts it entirely into work with nothing sent to the cold reservoir. There's no heat that goes to the cold reservoir. This is absolutely impossible, it's a violation specifically of the engine form of the second law of thermodynamics. On the right we have a perfect refrigerator. This can transport heat from a cold reservoir into a hot reservoir without any work being put in. There's no work required in a perfect refrigerator. A refrigerator with a coefficient of performance of infinity and this is absolutely impossible given the refrigerator form of the second law of thermodynamics. So obviously from these two forms of the second law, we didn't really learn anything new we didn't gain any new insights into the underlying physics so we're going to remedy that with a more general form of the second law of thermodynamics but it's important to start here with the engine and refrigerator form all your books are going to show it your professors are probably talking about it in class so make sure that you know the refrigerator and the engine form. Alright guys that wraps up this introduction to the second law of thermodynamics. Thanks for watching.

Concept #2: Introduction to Entropy and the Second Law of Thermodynamics

**Transcript**

Hey guys, in this video we're going to define a new energy-like quantity called entropy and formulate a single and concise second law of thermodynamics in terms of entropy. Alright let's get to it. We want to formulate a concise single form of the second law of thermodynamics. We saw previously the engine and the refrigerator form but that didn't really tell us anything new plus those forms of the second law of thermodynamics didn't explain why that was true, it just said this is true. The new form will actually fully explain both the engine and the refrigerator forms and it'll give us new insight into thermal processes and specifically why do they occur and under what circumstances can thermal processes occur?

In order to formulate the single concise form though we need to define an energy-like quantity called entropy now you've probably heard the word entropy before, it's used a lot, thrown around in a lot of everyday conversations. Sometimes people talk about entropy and time, sometimes people talk about entropy and at the state of the universe but it's very common to talk about entropy in terms of evolution and specifically how it pertains to the formulation of complex cellular machines inside of living organisms but that's like a biology problem we're not going to talk about that but we will talk about what exactly entropy is. The entropy of a system is an energy-like measurement, very important understand not energy, energy-like, that measures the amount of disorder in a system. This may seem very vague and that's because it is but we're going to continue evolving our definition of entropy as these videos go on so don't worry if it doesn't seem to be very precise yet but conceptually the best way to think about energy is it tells us how disordered a system is. It's very very important understand entropy is not an energy it's energy-like and it's energy-like because has very similar properties as energy for instance if you have two systems you haven let me draw this box a little bigger. You have system A, you have system B. System A has an internal energy UA and an entropy SA. Entropy is always given by a capital S. System B has UB and SB. If you bring them together into thermal contact right now they are not in thermal contact then you'll see that the new system A plus B has an internal energy of UA plus UB and it has an entropy of SA plus SB. So this ability to add entropy is when you combine two systems together is why it acts like an energy but it's only energy-like. The easiest way to see that it's not an energy is it has units of joules per Kelvin, it doesn't have units of joules so it's not an energy, units of joules per Kelvin energy-like. The more disorder that a system has the higher the entropy of that system and that's what it means to be dependent on the amount of disorder. If you have two states one that has more disorder whatever that is than the other states that will have more entropy and it turns out that the absolute entropy of a system isn't really all that important what I mean by absolute entropy is like this number SA and SB or this number SA plus SB. That value of the actual amount of entropy that a state has isn't really all that important what's important is the change in the entropy between two states. This is what the second law of thermodynamics concerns itself with it wants to talk about the change in entropy and this is actually pretty common with energy specifically potential energies. Physics doesn't care how much potential energy a system has, it only cares about the change in potential energy from one state to another. Now our concise singular form of the second law of thermodynamics says a system can only undergo processes that lead to no decrease in entropy. No system can undergo a process that decreases the entropy. You can also write delta S for a system can never be less than zero or the entropy, change in entropy for a system has to be greater than or equal to zero. The change in entropy at a constant temperature is a very important equation to remember. It's just the amount of heat that is exchanged during this process divided by the temperature that the process is occurring at. This only works for constant temperature processes which we know as isothermal processes. So only isothermal processes this is applied to. Now read the second law of thermodynamics very carefully. The second law is typically misstated as saying that the entropy has to increase, the second law doesn't say the entropy must increase, it just says the entropy cannot decrease. The best situation you can hope for is one that results in no change in entropy. These are called the isentropic processes by the way. Isentropic means the same entropy. It's weird because all the other ones that mean constant something like isothermal, isobaric, isochoric they all are iso this one's isentropic. There's no O it's probably because there's two vowels but still weird. For an engine there are three components to every system so there are three entropy changes. You have the hot reservoir, you have the cold reservoir which is often referred to as simply the environment and you have the engine itself. Those three components of the engine each contribute a change in entropy to the change in entropy of the system. The change in entropy for the system is going to be the sum of the change in entropies for each of the components of the system. This is another reason why entropy is always called energy-like because changes in energies for a system are always the sum of the change in energy of each of the components of that system so this equation looks just like the equation for the amount of work done for a system, the amount of heat transfer into or out of a system, the total change in internal energy for a system, etc. Lastly something that's very very important to remember about entropy is entropy is a state function just like internal energy is a state function. Changes for state functions depend only on the states, only on changes in the state and that means that changes in state functions are what we call path independent. Just like the internal energy doesn't, the changes in internal energy doesn't depend upon the path taken from an initial state to a final state, the changes in entropy also doesn't depend upon the path taken between an initial and a final state.

The change in entropy for an ideal gas is going to depend only on changes in temperature and volume or changes in pressure. These are interchangeable technically they're all interchangeable but rarely do you describe the state of a ideal gas without discussing the temperature. It's most common to always discuss the temperature and to discuss one more, either volume or pressure dealer's choice. It's usually volume. Let's do a quick example. This is something that we want to show. Where will heat flow if you connect a hot reservoir at TH to a cold reservoir at TC with some thermal conductor? So we have TH, we have TC. Now the heat is going to follow, I know it's going to flow out of the hot reservoir and into the cold reservoir so I'm going to use that fact to help me show why this is true so that's QH, that's QC and because of conservation of energy because energy cannot be created or destroyed however much heat leaves the hot reservoir has to enter the cold reservoir so I know that the amount of heat entering the cold reservoir will be a positive number because heat entering a system is always positive that's going to be the negative of the amount of heat leaving the hot reservoir. That hot reservoir is going to be a negative number and I'll just call them Q. They're both gonna have the same magnitude Q, the cold reservoir QC is going to be positive Q and QH is going to be negative Q. This states right here that QH is -Q, if we consider Q to be positive. So what I want to see is is this process allowed by the second law of thermodynamics and more specifically is the reverse process forbidden? So what I'm interested in is how the entropy changes for the system. This is going to be the change in entropy for the hot reservoir plus the change in entropy for the cold reservoir and remember that the change in entropy at constant temperature is Q over T. A reservoir by definition is something that can give out an infinite amount of energy or take in an infinite amount of energy without changing temperatures. So energy exchanges between reservoirs are never going to change the temperature. They're always at constant temperature so we can absolutely use this equation for reservoirs.

So what's the change in entropy for the system? Well what's Q for the reservoir? QH. What's T for the reservoir? TH plus QC over TC. Now I want to use this definition right here from the conservation of energy. So this is going to be -Q for the hot reservoir where Q is a positive number and positive Q for the cold reservoir. Now I want to combine the denominators that means that I have to multiply TC up, I have to multiply TH up and then the denominator becomes TH times TC. So this is -QTC over TH plus QTH over, sorry, this is THTC in the denominator. THTC, THTC. And now I can combine them based on their denominators. Notice that both of them share Q over THTC, the only thing that's different is this guy has -TC, this guy has positive TH that's the only difference. So this is going to be Q over THTC times TH minus TC. I just reordered the addition of the positive TH and the -TC. This by the second law, by second law, has to be greater than or equal to zero. When is this inequality greater than or equal to zero? That's only true if TH is greater than TC and I started this process out by assuming that this was the hot reservoir and this was the cold reservoir and I showed that heat can flow in this direction heat can absolutely flow from TH to TC if and only if TH is greater than TC so only if TH is truly the hot reservoir and TC is truly the cold reservoir. Notice that if I say that TH is less than TC then the number is negative. That violates the second law of thermodynamics. So you can never ever transport heat without any work being done because that changes the entropy. You can never transfer heat from the cold reservoir to the hot reservoir by itself, that violates the second law of thermodynamics. That's why the refrigerator form of the second law exists as it exists. The refrigerator form says that this cannot happen, that cannot happen that's what the refrigerator law the refrigerator form of the second law says and the reason is because of this result right here. Alright guys, that wraps up this introduction to entropy and specifically how it applies to the second law of thermodynamics. Thanks for watching guys.

Example #1: Why Do Perfect Engines Violate the Second Law?

**Transcript**

Hey guys, lets do an example show that a perfect engine will in fact violate the second law of thermodynamics alright so here's a figure describing a perfect engine right the second law of thermodynamics says that the change in entropy for the system must be greater than 0 greater than or equal to 0 right that's the requirement for the second law where the change in entropy for the system is the change in entropy for the hot reservoir plus the change an entropy for the engine plus the change in entropy for the cold reservoir now right off the bat you only ever have a change in entropy when a heat is being transferred there is no heat transfer in the engine because however much heat it gains it loses, it loses either as work or loses as heat going into the cold reservoir so always the change an entropy for an engine is 0 another way that you can think about is engines are always what kind of processes, engines are always cyclic processes right what's the change in entropy for a cyclic process remember that entropy is a state function the change entropy for a cyclic process always zero because you start and end in the same state because you start in the same state the change in entropy is always going to be 0 for a cyclic process right that's because entropy is a state function so the only entropy changes that you are going to have in this case is the hot and the cold reservoir. So remembering our equation for the change in entropy at a constant temperature. If we say that the hot. Reservoir loses an amount of heat Q.H. which is what the diagram shows at the temperature H H sorry T H and the cold reservoir gains an amount of heat Q.C at a temperature T.C that's going to be the change in entropy for the system.

However in a perfect engine cold reservoir doesn't gain any heat right. That's the whole point of a perfect engine for a perfect engine whatever heat would normally go into the cold reservoir isn't there it can convert all of the heat into work what does this say though about the change in entropy for the system. That the change in entropy is negative and what is this, this violates the second law this violates the second law and therefore it cannot happen this perfect engine absolutely cannot exist because it violates the second law of thermodynamics the change in entropy for the system is negative and there is absolutely no way to make this positive unless some amount of heat is transferred to the cold reservoir. Alright guys that wraps up this problem. Thanks for watching.

Practice: An engine operates with the following heat flow diagram. How much entropy is generated in each cycle?

Concept #3: Microstates and Macrostates of a System

**Transcript**

Hey guys, in this video we're going to introduce the concept of a macrostate and a microstate of a system and how it pertains to the entropy of the system. Alright let's get to it. So far whenever we talk about the state of the system we've been referring to simply the state we've just been saying the word state. That state that we've been talking about is more properly referred to as a macrostate. Macro meaning macroscopic this means pressure, volume, temperature, internal energy, entropy etc. and any of these macroscopic measurements that you can make on a system those define its macrostate. The macrostate is defined by its measurable thermodynamic properties just like I listed. For ideal gases the pressure and the volume and therefore the temperature because they're related by the gas law define system's, this should say, macrostate that is a typo or an autocorrect in word that appeared. Something important to realize is that a particular macrostate is not actually unique to a system. A system can rearrange itself to make the same macrostate multiple times. There can be multiple arrangements of the system internally to make a particular macrostate. It can have many many different microscopic arrangements that produce the same macrostate for instance two samples of a gas can have the same temperature but they could have different positions of the gas particles that make up that gas that lead to the same temperature that's a very very easy one. You can think about all the different positions in fact a gas's particles are constantly changing position even if the gas is in thermal equilibrium and its temperature isn't changing.

A microstate of a system is just a single microscopic arrangement of a system that leads to a particular macrostate. Macrostates typically have multiple microstates but it has to be at least one. At least one microscopic arrangement has to exist for a macrostate otherwise what's the point of even considering that a macrostate because it's not possible for the system to arrange itself in that manner. The number of microstates for a particular macro state is called its multiplicity which is given by the capital Greek letter omega. So a multiplicity is particular to a macrostate so you could have a macrostate of one, a macrostate of two, macrostate three, macrostate four and they could all have different multiplicities, different numbers of microstates available to them. Let's do an example. Consider 4 coins. A particular macrostate for the system could be 2 coins heads up, and this should be 2 because I cannot add, 2 coins heads down. How many different microstates are there for this particular macrostate? So let's just do it. We can have heads up, heads up, heads down, heads down. Heads up, down, up, down. Up, down, down, up. Down, up, down, up. Down, up, up, down. Or down, down, up, up. Those are all the possible arrangements for this macrostate. How many microstates are there? Each of these is a micro state that still results in 2 coins heads up. There is 1, 2, 3, 4, 5, 6. So the multiplicity of this particular macrostate is 6. There are six different microstates available for the system in this particular macrostate.

Mathematically the entropy is best defined in terms of how many microstates are available to a particular macrostate. So in terms of the multiplicity, the entropy is best defined as the Boltzmann constant which we've seen before, 1.38 times 10 to the -23, the natural logarithm LN of the multiplicity. That's the entropy of a particular macrostate. Remember different macrostates have the same, sorry, different microstates with one macrostate have the same entropy. Entropy is a thermodynamic measurable at the macroscopic level. What's the entropy of the system of coins in the previous problem with a two heads up macrostate? Remember that the multiplicity of that macrostate is 6 so the entropy of that macrostate which is KBLN of omega is just going to be 1.38 times 10 to the -23 times LN of 6 which if you plug this into a calculator ends up being 2.47 times 10 to the -23 joules per Kelvin.

Now based on this definition the entropy is never going to be negative. The entropy will always be positive because you can never have fewer microstates than one. Remember all macrostates have to have at least one microstate. So this is mathematically the best possible way to define entropy and as the disorder of a system increases the number of available microstates increases for a particular macrostate and so the entropy also increases as the multiplicity goes up, as the number of available microstates goes up, the entropy goes up. So that's why systems tend to move towards more disorder more available microstates, more disorder, more entropy, that's where the second law of thermodynamics says the system wants to go. Okay guys, that wraps up this discussion on microstates and macrostates and specifically how we can define entropy in terms of them. Thanks for watching guys.

Practice: The macrostate of a set of coins is given by the number of coins that are heads-up. If you have 100 coins, initially with 20 heads-up, what is Δ𝑆 when the system is changed to have 50 heads-up? Note that the multiplicity of k coins which are heads-up, out of N total coins, is Ω = 𝑁!/𝑘!(𝑁−𝑘)! . Does this change in macrostate satisfy the second law of thermodynamics?

Concept #4: The Second Law of Thermodynamics for an Ideal Gas

**Transcript**

Hey guys, in this video we're going to talk about the second law of thermodynamics as it specifically applies to an ideal gas like pretty much everything we do we want to talk in general and we want to apply it specifically on ideal gases because that's mainly what we deal with in thermodynamics let's get to it. Remember the entropy is a state function it only depends on the state of the gas and there are a bunch of consequences to that this means that a change in entropy most importantly is path independent it only depends on the change in the state of the gas so when you go from one state to another from some initial state to some final state it doesn't matter how you get there because all that matters is what the initial state is and what the final state is so the path doesn't matter. The state of an ideal gas is determined by two of the following three properties pressure volume or temperature any two of them will define the state property it's common to represent the state of a gas as a combination of pressure and volume right because we use PV diagrams so commonly on a PV diagram for a particular combination of pressure and volume you define state but for entropy it's actually more useful to represent the state of an ideal gas with temperature and volume just because that's how most people represent this equation that we're going to get to in a second we can use the first law of thermodynamics right that delta U equals Q plus W to find the change in entropy for an ideal gas it will be written in terms of the state given by the temperature and the pressure now how to actually convert delta U equals Q plus W that first law into this representation of the second law I'm not going to get into your book will probably do it your professor might do it if you do a lot of derivations in class but it's not worth doing here the final result is that the number sorry the change in entropy for the gas is the moles of the gas times the specific heat at constant volume times the logarithm of the final temperature divided by the initial temperature plus the number of moles times the ideal gas constant times a the logarithm of the final volume divided by the initial volume right and the second law of thermodynamics says this has to be greater than or equal to 0.

If the change in entropy of the gas is greater than 0 the process can occur on its own right if the change in entropy for the gas is less than zero the process cannot occur on its own I'm not saying you cannot occur at all because the gas is not the only thing that a system can be made up of right a gas can be part of an engine and that engine can be a part of a hot or cold reservoir and those reservoirs can influence a change in entropy for the system but it cannot occur without help. There needs to be heat flow from a hot reservoir to a cold reservoir that's always going to lead to a change in entropy of the reservoirs greater than 0 this is something that we addressed when we first introduced entropy that whenever heat flows from a hot reservoir to a cold reservoir the change entropy for those reservoirs always greater than 0 no matter the temperatures if the heat flow is large enough then when you add the change in entropy of a gas to the change in entropy of a reservoir you're going to get a number greater than 0 and the second will be satisfied right this is the basic idea of how an engine works that even though. The change in entropy of the gas is negative yeah even though the change in entropy for the gas is negative that the gas decreases it's disorder that the gas decreases it's entropy which seems to violate the second law It actually doesn't the reason is that the total change in entropy for the system which is a combination of both the change in entropy of the gas and the change in entropy the reservoirs that does increase that does become greater than the initial state so the second law is satisfied even though the change in entropy for the gas is negative this is the basic idea with heat engines that you can overcome certain processes that won't occur on their own by adding in reservoirs.

Let's do an example 4 moles of a monoatomic ideal gas is expanded isothermally from some initial volume to some final volume what's the change in entropy does it satisfy the second law of thermodynamics right the change in entropy for a gas is always going to be the number moles times the specific heat at constant volume times a logarithm of the final temperature over the initial temperature plus the number of moles times R the ideal gas constant times the logarithm of the final volume over the initial volume, something very very important to note is that the logarithm of one is always zero that means that if temperature final equals temperature initial. Then you get the logarithm of one which is 0. That's true for an isothermal process right so this whole term is 0 because the final temperature equals the initial temperature so you're going to get a logarithm of 1 so all we're left with is this term so we have 4 moles the ideal gas constant is 8.314 and we take the logarithm of the final volume 0.008 divided by the initial volume 0.005 plugging this into a calculator we see that this is 15.6 Joules per Kelvin is the second law of thermodynamics satisfied, yes just by the gas alone the second law of thermodynamics now is satisfied and this process can occur without any help without any heat flow from reservoirs this process can occur on its own. Alright guys that wraps up this discussion on the second law of thermodynamics and how it specifically applies to ideal gases. Thanks for watching you guys.

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Concept #1: Introduction to the Second Law of Thermodynamics

Concept #2: Introduction to Entropy and the Second Law of Th...

Example #1: Why Do Perfect Engines Violate the Second Law?

Practice #1: Entropy Generated by an Engine

Concept #3: Microstates and Macrostates of a System

Practice #2: Entropy Produced by a Change in Macrostates of ...

Concept #4: The Second Law of Thermodynamics for an Ideal Gas

n moles of an ideal gas expands while the temperature of the gas remains constant. In this process the entropy of the gas
(a) remains constant
(b) decreases
(c) increases

During an adiabatic change of state for an ideal gas 20 J of work are done on the surroundings. During this change, the temperature decreases from 400 K to 380 K. What is ΔU, ΔS and Q for this change of state?
A. ΔU = 20J, ΔS = 0, Q = 0
B. ΔU = -20J, ΔS = 0, Q = 0
C. ΔU = 20J, ΔS = 20/390 J/K, Q = 0
D. ΔU = -20J, ΔS = 20/390 J/K, Q = 0
E. ΔU = 20J, ΔS = 20/390 J/K, Q = 20 J

A system takes in 200 J of heat while undergoing an isothermal change of state at 500 K. What is ΔU, ΔS, W, and Q for this change of state?
A. ΔU = 0, ΔS = 1/2 J/K, W = 200 J, Q = 200 J
B. ΔU = 0, ΔS = 2/5 J/K, W = -200 J, Q = -200 J
C. ΔU = 200 J, ΔS = 1/2 J/K, W = -0, Q = 200 J
D. ΔU = 0, ΔS = 2/5 J/K, W = -200 J, Q = 200 J

Using the Einstein model of a solid, what is the change in entropy when adding two quanta of energy to a system of 5 atoms that already has 5 quanta of energy stored in it? (In an Einstein solid each atom corresponds to three independent oscillators.)
1. ΔS = kB ln (21! /14! • 7!)
2. ΔS = kB ln (21/7 • 5)
3. ΔS = kB ln (17/6)
4. ΔS = kB ln (10/7)
5. ΔS = kB ln (10! /7!)
6. ΔS = kB ln (20! /14! • 5!)
7. ΔS = kB ln (20! /7! • 5!)
8. ΔS = kB ln (10)
9. ΔS = kB ln (21! /7! • 5!)
10. ΔS = kB ln (16!7! • 7)

An engineer claims to have built four engines. When operating between two reservoirs at 400 K and 300 K, the engines have the following characteristics:1) QH = 200 J, QC = -175 J, and W = 40 J2) QH = 500 J, QC = -200 J, and W = 400 J3) QH = 600 J, QC = -200 J, and W = 400 J4) QH = 100 J, QC = -90 J, and W = 10 JDo any of these engines violate either the first or the second law of thermodynamics?

A glass of water is left outside overnight to freeze. During this process, which of the following statements about entropy is true?
A) The entropy of the surrounding air doesn't change overnight
B) The entropy of the water changes more than the surrounding air overnight
C) The entropy of the surrounding air changes more than the water overnight
D) The change in entropy of the water is the same as the change in entropy for the surrounding air, with opposite signs

An ideal gas is taken through the cycle shown in the diagram below. The process is carried out with 150 mg of Helium, which is a monoatomic gas with an atomic mass of 4 g/mol. First, the gas undergoes an adiabatic expansion (from point A to point C), followed by an increase in pressure at constant volume (from point C to point B). Finally, the gas is taken back to the initial state by an isothermal compression (from point B to point A). What is the change in entropy of the gas during the isothermal process (from point B to point A)? Note that the work during an isothermal process is W = nRT ln(Vi / Vf).

Which of the following violates the second law of thermodynamics? (TH > TC)
A. Heat taken from a body at TH is converted completely into work.
B. Heat flows from a body at TH to another body at TC.
C. Work is done to remove heat from a body at TC and exhaust it at TH.
D. Heat taken from a body at TH is converted partly into work.

Blocks A and B have 300 and 200 oscillators, respectively. The blocks are placed in contact with each other in an insulated environment. At the instant the blocks touch, block A has 99 quanta of energy while block B has 1 quanta of energy. What will happen sometime after contact?
I. The entropy of block A will increase.
II. The entropy of block B will increase.
III. The entropy of the two-block system will increase.
1. None
2. I, II only
3. II only
4. I, II, III
5. I, III only
6. II, III only
7. I only
8. III only

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