 ## Physics

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23. The Second Law of Thermodynamics

# Entropy and the Second Law of Thermodynamics

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concept

## Intro to Entropy 7m
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Problem

3 moles of an ideal gas are compressed isothermally at 20°C. During this compression, 1850 J of work is done on the gas. What is the change of entropy of the gas?

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Problem

You have a block of ice at 0°C. Heat is added to the ice, causing an increase in entropy of 120J/K. How much ice melts into water in this process?

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example

## Entropy Increase When Braking 3m
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everybody. So let's take a look at our problem. So we have a car that's driving along the road, it's gonna suddenly slam on the brakes. And we want to calculate what's the change in entropy of the air. So in other words, what we want to calculate here is delta S. And I want to say delta S. For the air. But before we get started, I want to sort of draw out what's going on here. Right? So we've got this little road, I'm gonna draw this little car like this, don't make fun of my car drawings like that's and what happens is it's chugging along initially at some initial speed, which is 20 m per second. Then what happens is over some time it's gonna slam to a stop. So eventually what happens is that the final is going to be zero. Alright, so what's going on here? If you want to calculate the change in the entropy, we're gonna have to use the change in entropy equation now. Remember we can only use this equation Q over t as long as the temperature remains constant. But it does because we're told in this problem that when you slam on the brakes, the air which is like so massive and large, Remains at a constant temperature of 20°C. So what happens is if you add a little bit of heat to it, the air is obviously so massive that it doesn't really change temperature. So we have delta S for the air is the heat that's added to the air divided by that temperature, right? So we have what the temperature is. So the question here is, well, how much heat is actually added to the air? We don't have an equation for this, right? We can't use like Q equals M. Cat or something like that because we don't have the mass of the air. We have the specific heat. So, what's really going on? But what happens here is that we have to realize that this initial velocity for the car means that this car has some kinetic energy. So, in other words, this kinetic energy, I'm gonna I'm gonna say K initial is equal to one half M. V. Naught squared. Actually, I'll call this K. Not, so I'll call this K zero. And eventually what happens is that when you slow to a stop, right? You slam on the brakes eventually when you get to this point here, you're K. Final is equal to zero. So in other words, you had some kinetic energy and then you lost it because you slammed on the brakes. So then where did all of this energy go? Remember that energy can only change forms. It can't be created or destroyed. It doesn't go to potential energy because it's still staying at the same height. So basically what happens here is that all of the K. That gets transferred, basically goes into friction. It goes into the friction of the brakes and that gets dissipated as heat. Remember friction. When you rub something, it's really just heat transfer. So what happens here is that the delta K. The change in the kinetic energy really just becomes Q. It becomes the cue that gets given off to the air. Alright, so that's what's going on. So basically what we can do here is we can say well the Q air really comes from the change in the kinetic energy. Now. Really, what happens is this change is really just the one half Mv not squared because the K final is zero so we don't have to do final minus initial. So this really just becomes 1/2 and this is going to be M V not square. We have the mass of the car and the initial speed so that we can go ahead and do that. So we're gonna have one half of the mass which is 1500 times the initial velocity which is 20 squared. So now you divide that by the initial temperature. Now the temperature is 20 degrees Celsius, but we have to convert that to kelvin. So this just becomes 293 Kelvin and that's what we plug into our denominator here. So if you work this out, what you're gonna get is that the change in the entropy for the air is equal to 1.02 times 10 to the third. And this is going to be jewels per Kelvin. Alright, so that's how much entropy gets, sort of dissipated. That's how much energy you've now dissipated. The energy is a little bit more random and therefore the energy of the universe increases. Alright, so that's it for this one. Guys, let me know if you have any questions.
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concept

## Calculating Entropy Changes for Systems of Objects 6m
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Problem

A non-Carnot heat engine operates between a hot reservoir at 610K and a cold reservoir at 320K. In a cycle, it takes in 6400 J of heat and does 2200 J of work. What is the total change in entropy of the universe over the cycle?

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example

## Entropy of Carnot Engine 3m
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