Carnot Heat Engines, Efficiency, Refrigerators, Pumps, Entropy, Thermodynamics - Second Law, Physics

in this video we're going to focus on the second law of thermodynamics we're going to talk about how to solve problems associated with heat engines efficiency the cardinal engine and refrigerators as well so let's begin now we know the first law is basically a statement of the conservation of energy the change in the internal energy is equal to q minus w so let's say if this box represents the system and if we add 300 joules of energy that's q that's the heat energy absorbed by the system and let's say the system performs 100 joules of work so it expends 100 joules of energy so the system gained 300 joules of energy it lost 100 so the change in the internal energy is a gain of 200 so energy is conserved that's the main idea behind the first law of thermodynamics the main idea behind a second law of thermodynamics is that irreversible processes tend to increase the disorder of a system let's talk about reversible and irreversible processes first let's say if you have a ball placed above ground level this ball naturally will fall down to the ground the ball is not going to go up by itself from the ground that just doesn't happen so this is an irreversible process the ball will naturally fall from a high position to a low position it doesn't naturally go from a low position to a high position the first process is a spontaneous process it happens naturally without any assistance the second process is non-spontaneous it doesn't happen on its own the only way to make it happen is to put energy into the system to lift the ball to a high position now what about a reversible process let's say if we have a level surface and if we have a ball the ball can easily roll towards the right just as easily as it can roll towards the left it can go in either direction because the surface is leveled so this is an idea of a reversible process it can occur in both directions now and let's not take friction into account now let's talk about heat let's say if we have a high temperature reservoir and a low temperature reservoir so let's say over here it's at 600 kelvin and the temperature on this side is 300 kelvin heat naturally flows from hot to cold it flows from a high temperature to a low temperature so in that direction it's spontaneous heat just doesn't flow from a low temperature to a high temperature it doesn't happen so the direction towards the right represents a spontaneous process and going back in the other direction is a non-spontaneous process so you don't need anything you don't have to do anything in order to get heat to flow from hot to cold it's going to happen naturally however if you want to cause heat to flow from let's say a cold object to a hot object since it doesn't happen naturally you have to put in energy into the system to pump heat from a low temperature environment to a hot temperature environment just the same way as you got to put energy to lift the ball from ground level to a higher position it takes energy to do that so basically this is one statement of the second law of thermodynamics heat always flows from hot to cold and never flows in reverse now let's say if the temperatures were approximately equal so in the last example where the temperatures were uh 600 and 300 kelvin that's an irreversible process heat flows in one direction on its own it doesn't flow in the other direction but let's say if the temperatures are approximately the same if they're both 300 kelvin then equilibrium will be established the amount of heat that flows towards the left is equal to the rate of heat that flows towards the right and so this is a reversible process an equilibrium is established in this case when you have an irreversible process like the one we had in the last example the entropy or the disorder tends to increase for an irreversible process things happen in such a way that the disorder goes up for example think of your room it's easier for your room to become messy than it is to uh keep it clean typically um it may you may find that your room naturally becomes disorderly but it takes effort it takes to work to put it back in order so for most natural processes the for irreversible natural processes the entropy tends to increase or the disorder now for reversible processes the total entropy can increase or if you have an isolated system at best it can remain constant the change in entropy could be zero the natural processes tend to lead towards disorder and that's another idea of the second law of thermodynamics now let's move on to heat engines so what exactly is a heat engine a heat engine is basically a device that converts heat energy into mechanical energy or work so let's say if we have a hot temperature reservoir let's say this is at 500 kelvin and if we have a low temperature reservoir which is at a 300 kelvin and let's say it is a big box in the middle is the system heat will naturally flow from hot to cold so the amount of heat energy that flows into the system or that is absorbed by the system is known as qh now the amount of heat energy expelled by the system into the environment or into the low temperature reservoir is qc now as heat flows from hot to cold the system can absorb some of that heat energy and use it to convert it to mechanical work so that's a heat engine now the heat engine is not able to take all of the heat energy that it absorbs and convert it to work it takes a fraction of that energy most of the energy is expelled into the environment but the heat engine takes some of it and converts it to mechanical work the work is equal to the difference between qh and ql qh tends to be positive because anytime the system absorbs heat energy q is positive qc tends to be negative because heat energy is released by the system so that's why we have the absolute values now the efficiency is equal to the work divided by qh times 100 percent and sometimes you may need to use power power is equal to work divided by time and work is power multiplied by time keep in mind the unit of power is the watt one watt is one joule per second so power is the rate of energy transfer so if you have let's say 70 watts that means that in every in one second 70 joules of energy is being transferred every second now let's work on some problems a heat engine absorbs 2500 joules of heat and discards 2100 joules of heat calculate the work performed by this engine and its thermal efficiency so let's say this box represents the system now 25 joules of heat energy enters the system so this is qh and 21 joules of heat energy leaves the system so this is qc what is the work output of the system the work is basically the difference in these two values if 25 joules of energy enters the system for it to maintain its state 25 joules of energy must leave the system otherwise the temperature of the system will keep on going and that could be damaging so the work is going to be the difference between qh and qc in this case it's going to be 400 joules it's 2500 minus 2100 now what about the efficiency of this engine the efficiency is basically the work divided by qh times 100 so it's going to be 400 divided by 2500 times 100 percent now 400 divided by 2500 is point 16 times 100 so this uh device is 16 efficient so what that means is that 16 of the heat energy that travels through the system is converted to mechanical energy the other 84 percent is expelled to the environment so 16 travels this way and the other 84 escapes to the environment a jet engine releases 5000 joules of energy per cycle and performs 800 joules of work how much heat is absorbed by this engine per cycle so let's say this is the system and it releases five thousand joules of energy so this is equal to qc and the work output is 800 joules per cycle so how much heat energy is absorbed by the system qh is equal to w plus qc and of course we're dealing with absolute values so if a total of 5800 joules leaves a system per cycle then that same amount of energy must enter the cycle so that's a qh 5800 joules so that's how much heat energy is absorbed by the system per cycle now what about b what is the thermal efficiency the thermal efficiency is going to be the work performed by the machine per cycle divided by qh the heat energy absorbed by it times 100 and this is about 13.8 efficient so how much work can this engine perform in 50 cycles how would you answer part c the answer to this is straightforward if 800 joules of energy is produced per cycle then we simply need to multiply by 50 cycles notice that the unit cycles will cancel and this will give us the work output during those 50 cycles eight times five is forty and then after that we simply need to add the three zeros so forty thousand joules of work can be performed by this engine in 50 cycles now what about part d if the engine completes each cycle in point 20 seconds what is the power rating of the engine now we know power is the ratio of work in time so in one cycle the engine performs 800 joules of work and one cycle occurs in point 20 seconds so we simply need to divide the two eight hundred divided by point twenty is about four thousand joules per second so in one second this engine generates four thousand joules of work in ten seconds that would be forty thousand joules of work so you can say it's four thousand joules per second or simply four thousand watts 8000 joules of heat energy is absorbed per cycle by a diesel engine that is 15 efficient how much work does it perform per cycle let's find out so here's our system and 8 000 joules of energy enters into it per cycle to calculate the amount of work performed we need to realize that 15 percent of the energy travels this way and the other 85 percent is expelled to the environment so what is percent of eight thousand all you need to do is multiply eight thousand by point one five to convert fifteen percent into a decimal divided by a hundred or simply move the decimal point two units to the left so 8 000 times 0.15 is 1200 joules so that's the output work now if you take 8000 and multiply it by point eighty five that's going to give you the output work which is sixty eight hundred and you can check your work as you can see these two numbers add up to 8 000. so anytime you want to find the work from qh use this equation w is equal to the efficiency times qh divided by 100 and if you want to find qc just you can take 8000 subtract it by 1200 and that will give you qc to 6800 or you can multiply it by 85 percent both ways work an engine has a heat input of 175 kilowatts and a work output of 21 kilowatts what is the thermal efficiency so the thermal efficiency is the work divided by qh now this equation will work if you have the units of joules or if you have units of power such as watts so the work output is 21 kilowatts and qh is 175 kilowatts times 100 so the thermal efficiency in this problem is simply 12 percent now how much heat energy well i'm jumping to part c part b at what rate is heat discarded into the environment so we have 175 kilowatts of power being absorbed by the system 21 kilowatts is released well it's uh converted to mechanical work so the difference between these two 175 minus 21 represents the amount of heat or the rate of heat energy that it leaves into the environment so the answer is simply 154 now part c how much heat energy is released into the environment if this machine were to operate continuously for one day so qc divided by time is basically 154 kilowatts which is 154 kilojoules per second so we need to find out how many seconds are there in one day and then we can calculate the total amount of heat energy released into the environment so in one day this 24 hours per day and there's 60 minutes in a single hour and there's 60 seconds in a single minute and 154 kilojoules of energy is released every second so it's 24 times 60 times 60 times 154 so it's 13 million joules of energy a gasoline engine takes in 12 000 joules of heat energy per cycle and produces 2 400 joules of mechanical work so let's begin by drawing a picture so 12 000 joules of heat energy enters the system and only 24 joules of mechanical work leave the system so therefore qc is the difference between these two 12 000 minus 2400 is about 9 600 joules so that's the answer to part a now what about part b what is the thermal efficiency of this particular engine well we know that efficiency is the work divided by qh times 100 so w is 2400 qh is 12 000. and so this machine is about 20 percent efficient now part c what massive fuel is consumed in each cycle if the heat of combustion is 45 000 joules per gram so what we need to use is qh because the energy absorbed by the system comes directly from the gasoline so let's start with 12 000 joules and let's convert it to grams so this number tells us that one gram of gasoline releases 45 000 joules and you want to write it in such a way that the unit joules cancel so you get grams so this is going to be equal to 0.2667 grams now what about part d if the engine goes through 50 cycles per second what is the power rating in watts kilowatts and horsepower power is equal to the work divided by the time and we have the mechanical work it's 2400 joules per cycle so what we need to find is the time it takes to complete in one cycle now we're given 50 cycles per second that's the frequency which is the same as 50 hertz the period represents the time it takes to complete a single cycle and it's 1 divided by the frequency or 1 over 50. so in one second 50 cycles will be completed so if we take one and divide it by 50 this will give us 0.02 which is .02 seconds per cycle so in .02 seconds this device will generate 2 400 joules of mechanical work so if it can perform 2400 joules of mechanical work in 0.02 seconds then its power rating is 120 thousand watts now how much is that in kilowatts well let's convert it one kilowatt is equal to a thousand watts so 120 000 divided by a thousand is 120 kilowatts now what is the answer in horsepower well you need to know the conversion one horsepower is equal to 746 watts so if you take 120 000 and divide it by that number you're going to get 160.9 horsepower now let's talk about the carnot cycle the cardinal engine is basically the ideal model of a heat engine it was designed to represent or to obtain the maximum possible efficiency so it's simply an ideal model no other engine can have an efficiency that's greater than the cardinal engine at least as far as we know but let's draw a pv diagram for this so we're going to start at position a in the first step of the carnot cycle we have an isothermal expansion so going from a to b during this isothermal expansion heat is added to the system so we'll call this qh and the machine or the engine performs positive work anytime a gas expands the work done by the gas is positive now in the second step going from b to c we have an adiabatic expansion and during an adiabatic process no heat transfer occurs so q is zero however since the gas is still expanding the engine performs positive work during this process now going from c to d this process is known as an isothermal compression for any isothermal event the change in temperature is equal to zero but going from c to d heat is released from the engine into the environment and because the gas is being compressed the engine performs negative work which means that the surroundings do work on the engine so w is negative during this point it takes effort to compress the gas you got to put energy into it and going from c i mean from d back to a to complete the cycle that is an adiabatic compression q is zero there's no heat transfer and because the gas is being compressed the work performed on it is negative so that's the carnot cycle but there's two equations that you really need to know for this uh ideal engine the efficiency of the carnot engine is one minus tc divided by th where tc is the temperature of the cold reservoir and th is the temperature of the hot reservoir and also the ratio of qc and qh is equal to the ratio of tc and th a cardinal engine takes in 4 500 joules of heat energy from a reservoir at 800 kelvin how much heat energy will be released to a cold reservoir at 300 kelvin so we can use this equation qc divided by qh is equal to tc over th so the absolute value of qc we're looking for that and qh is 4500 joules that's the amount of heat energy that is absorbed by the system which comes from the hot reservoir and tc that's the low temperature or the temperature of the cold reservoir that's 300 kelvin th is the temperature of the hot reservoir that's 800 kelvin so if we cross multiply this is going to be 4500 times 300 and that equals 800 times qc 4500 times 300 is 1.35 million and if you divide that by 800 you'll see that qc is about 1687.5 joules so now let's draw a picture so let's say this is the system 500 joules of energy enters the system and 16 87.5 joules leaves the system so the temperature of the hot reservoir is 800 kelvin and the temperature of the low or the cold reservoir is 300 kelvin now once we have uh qh and qc we can calculate the work which is the difference between the two values it's qh minus qc so this engine performs 2812.5 joules of work so now we can calculate the efficiency so one way we can do that is using qh over w and that's going to be i mean i wrote it the other way it's supposed to be w over qh times 100 so 2812.5 divided by 4 500 times 100 is about 62.5 percent efficient now you can also get that same answer using another equation if you're dealing with a cardinal engine so we can use this equation to get the answer so it's 1 minus tc divided by th so this gives you a decimal answer 0.625 and if you multiply that by 100 it represents 62.5 a heat engine absorbs 4 000 joules of energy from a hot reservoir at 500 kelvin and ejects 3 000 joules of heat energy from a cold reservoir 300 kelvin so let's say this is the hot reservoir and it's going to put in 4 000 joules of energy into the system and 3 000 joules of heat energy gets ejected to a code reservoir 300 kelvin so how much work is performed by the engine we know work is qh minus qc so 4 000 minus three thousand the work performed is simply one thousand now what about b what is the efficiency of this engine the efficiency is simply the work divided by q h times a hundred percent so that's a thousand divided by four thousand which is point two five times a hundred that should be uh twenty-five percent so that's the efficiency of the the actual heat engine now what is the maximum theoretical efficiency so we need to find the efficiency of the cardinal engine so we're going to use the equation 1 minus tc over th so that's one minus 300 over 500 and that's going to be 0.4 so if we multiply that by hundred percent the maximum efficiency of this engine is forty percent that's as as high as it can go so 25 divided by 40 is 0.625 so this engine is operating at 62.5 percent of its cardinal capacity or its carnot efficiency now let's talk about refrigerators and air conditioners these devices work in the opposite direction compared to a heat engine in a typical heat engine heat flows from a hot reservoir to a cold reservoir let's say this box is the system so heat flows from hot to cold as that happens the heat engine can perform work so a heat engine converts heat into mechanical work but a refrigerator does the opposite in a refrigerator electricity is used to perform work on the device and to pump heat from a cold reservoir to a hot reservoir so this is qh and this is qc now just like the previous equations for refrigerators air conditioners and heat pumps the work is still the difference between qh and ql and qh is still the sum of qc and the work the only difference is you're using work to pump heat from a cold environment to a hot environment so how can we compare two refrigerators and know which one is more efficient is there a system in which we can measure the performance of a refrigerator there's something called the coefficient of performance some books might use the symbol k others might use cp but for refrigerator it's the ratio of qc and the work that you put into it to pump heat from the cold environment to the hot environment the purpose of a refrigerator is to cool something down and so the more heat that you can take away from the cold reservoir the more efficient the refrigerator is so defined as qc over w now for a heat pump the coefficient of performance is defined differently let's say if you have a house and you're up north and it's very cold outside it's winter and you want to use a heat pump to pump heat into your house to keep it warm you're concerned with qh rather than qc you want to know how much heat is being delivered into the house as opposed to just how much heat is being absorbed from the cold reservoir so for a heat pump the coefficient of performance is measured differently what matters is how much heat is coming into the house rather than how much heat is being removed from a cold reservoir which is important for a refrigerator or an ac unit so the equations are defined differently for these different devices so just keep that in mind now for a refrigerator calculate the car note coefficient of performance this is the the maximum ideal that you can get to it's equal to tc divided by th minus tc so that's another equation that you want to have with you a refrigerator uses 1200 joules of work to pump 3000 joules of heat from a cold reservoir at 275 kelvin to a hot reservoir at 320 kelvin what is the coefficient of performance so for a refrigerator the coefficient of performance is equal to uh qc divided by w qc that's the amount of heat that is pumped from the cold reservoir which is uh 3 000 joules and the amount of work that is used to accomplish that is 1200 joules so we simply have to divide these two numbers so the coefficient of performance is 2.5 now what is the maximum coefficient of performance so basically what is the coefficient of performance for the carnot device the equation is equal to tc divided by th minus tc so tc the temperature of the cold reservoir is 275 kelvin the temperature of the hot reservoir is 320 kelvin minus the 275 and if you type this in you should get an efficiency or rather a performance of 6.1 so that's the maximum coefficient of performance for this particular device but the actual performance is 2.5 now how much heat is delivered to the hot reservoir so let's draw a picture so we're putting in 1200 joules of work and at the same time let's say this is the code reservoir 275 kelvin we're pumping uh 3 000 joules of heat energy from the cold reservoir into the hot reservoir qh is basically the sum of w and qc so all we need to do is add twelve hundred and three thousand so about forty two hundred joules of energy is being pumped into the high temperature reservoir and so this is the answer for part c a cardinal freezer removes 1500 joules of heat energy from a reservoir at 250 kelvin and pumps it to a reservoir at 320 kelvin how much heat energy is transferred to the hot reservoir well since we have a cardinal device we can use this equation qc divided by qh is equal to tc divided by th so we have the value for qc that's the amount of heat energy being released from the cold reservoir 250 kelvin so qc is 1500 tc is 250 kelvin and th that's the temperature of the the hot reservoir that's at 320 kelvin so let's go ahead and calculate qh so let's cross multiply so it's 250 times qh and that equals 1500 times 320 1500 times 320 is 480 000 and if we divide that by 250 we'll see that qh is 1920. now that we have qh we can calculate w which is the difference between qh and qc so 1920 minus 1500 it's 420. so let's draw a picture so we're putting in 420 joules of work in each cycle and we're pumping 1500 joules of heat energy from the cold reservoir and so 19 20 joules of heat energy is going to be transferred to the hot temperature reservoir and so this is the answer for part a now what about part b what is the coefficient of performance for this device the coefficient of performance for a refrigerator or even a freezer is going to be qc divided by w qc is 1500 and w is 420 so the answer is about 3.57 now we can also calculate k using the other equation since this is a cardinal device so let's use this t tc divided by th minus tc so tc is 250 th is 320 and 320 minus 250 is 70. 250 divided by 70 gives us the same answer of 3.57 a heat pump uses joules of work to transfer heat energy how much heat energy is absorbed by the hot reservoir so for a heat pump k is equal to qh divided by w so to find qh is going to be k times w so it's going to be 3.0 times 2500 which is 7500 joules now once we have that we could find qc the amount of heat energy released by the cold reservoir so let's draw a picture so 25 joules i mean 2500 joules of work is being put into the system and 7 500 joules three times the value of w is being transferred to the hot reservoir so how much energy is being pulled from the cold reservoir well it has to be the difference between these two numbers 7 500 divided by or minus 2500 is 5 000 joules so that's the answer to part b and that is it for this question as you can see these problems are not too difficult you can solve them as long as you know the equations and understand how it works you can find whatever you need to look for now going back to heat engines i want to talk about the auto cycle if you haven't learned this in class feel free to skip over it but let's draw the pv diagram so we're going to start at position a in the first step we're going to have an adiabatic compression so we're going to go from a to b now during this adiabatic compression the volume decreases greatly we have volume on the x-axis pressure on the y-axis as we decrease the volume adiabatically the pressure is going to go up and the temperature will increase as well these conditions make it favorable to ignite an air fuel mixture the auto cycle is used for is the ideal model of a gasoline engine and so when you compress the air and the fuel together and as you increase the pressure you increase the collisions and as you increase the temperature you increase the energy of the molecules thus causing increase in the rate of reaction between the air and the fuel which will make it favorable to ignite now during this adiabatic compression q is zero no heat is transferred and in order to compress something you have to perform work on the gas you have to apply a force to compress it so the work done by the gas is negative because the surroundings is doing work on the gas so you have to put in energy during this process now in the second step we have an isochoric process we're going to go from b to c now during an isochoric process the change in volume is zero the volume is constant this is also known as an isovolumetric process now in order for the pressure to increase we need to increase the temperature and the pressure will go up and the only way to increase the temperature is to add heat so during this part heat is added in the cycle so let's call it qh so where does the heat come from well in the first step from a to b during the adiabatic compression once the air in the fuel ignites that reaction releases a lot of energy and so that energy basically is going to be used to heat up the surrounding gas increasing the temperature and increasing the pressure as well now once we've reached this state of high temperature and pressure the next step is an adiabatic expansion let me draw that better so c to d now during this adiabatic expansion q is zero but anytime a gas expands it performs positive work most of the work produced by the gas actually all of it in this case is from c to d so energy leaves a system as mechanical work now what about the last part that is d to a so d to a is a an isochoric process the change in volume is zero for any isochoric process w is zero now notice that we're going down in the pv diagram the pressure is decreasing which means that the temperature must be decreasing as well so that tells us that heat is being released from the system in order for the temperature to decrease so this is qc during the auto cycle now there's another equation that you need to know for the auto cycle the efficiency is equal to one minus one divided by r raised to the gamma minus one r is the compression ratio and this symbol is the gamma ratio which is cp over cv so sometimes you might have questions relating to this equation so i'll give you a few examples on that what is the efficiency of an auto cycle that has a compression ratio of 8.5 and a gamma ratio of 1.41 so using the equation e is equal to 1 minus 1 over r gamma minus 1. this is going to be 1 minus r is 8.5 gamma is 1.41 and 1.41 minus 1 is 0.41 8.5 raised to the 0.41 power is about 2.4047 so the final answer is about 0.584 which represents 58.4 percent what is the compression ratio of an auto cycle that has an efficiency of 46 percent and a gamma ratio of 1.4 so let's rearrange the equation to solve for r so the first thing we want to do is add this to both sides it's negative on the right side so if we move it to the left it's going to be positive and e is positive on the left side if we move it to the right it's going to be negative so we can rearrange the expression and write like this 1 over r raised to the gamma minus 1 is equal to one minus e which is the same as one minus e over one now what we're going to do is flip both fractions so r raised to the gamma minus one is equal to one over one minus e now to get r by itself we need to get rid of this exponent the only way to do that is to raise both sides to the reciprocal of gamma minus 1 which is 1 over gamma minus 1. so 1 over gamma minus 1 times square minus 1 those two will cancel and so we have this equation r is equal to 1 divided by 1 minus e raised to the 1 over gamma minus 1. so now that we have the equation for r let's go ahead and find it so the efficiency is 46 if we divide it by a hundred it's point four six as a decimal and gamma is one point four so let's take this one step at a time one minus point four six is 0.54 and 1.4 minus 1 is 0.4 now 1 divided by 0.54 is 1.8519 and 1 divided by 0.4 is about 2.5 so the final answer is about 4.67 so that's the compression ratio for this particular auto cycle an auto cycle has a maximum efficiency of 71 percent and a compression ratio of 8.9 what is the gamma ratio of the working substance so what exactly is the working substance well the working substance is the gas or fluid that is in the device so let's say if you have a steam engine the working substance is steam now let's start with the original equation e the maximum efficiency is equal to one minus one over r gamma minus one so just like we did before we're going to take this term which is negative on the right side let's move it to the left so it's positive and then we're going to take e which is positive on the left move it to the right and so it's going to be negative so 1 over r gamma minus 1 is equal to 1 minus e and we're going to flip both sides just like we did last time but this time because you want to solve for gamma we're going to take the natural log of both sides of the equation now a property of natural logs allows us to take the exponent and move it to the front so gamma minus 1 times the natural log of r is equal to the natural log of 1 over 1 minus e so now we can divide by l and r so gamma minus 1 is equal to ln 1 over 1 minus e divided by l and r and then the last thing we need to do is add one to both sides so here's the equation that we need so e is going to be 0.71 and r is 8.9 so 1 minus 0.71 is 0.29 and natural log of 1 divided by 0.29 is about 1.23787 the natural log of 8.9 is about 2.18605 so the final answer is going to be about 1.566 so that's how you can find a gamma ratio if you're given the maximum efficiency and the compression ratio now let's talk about entropy entropy is a measure of disorder and is represented by the symbol s whenever a gas expands whenever it increases its volume the entropy of a system increases entropy is higher for gases than it is for liquids and solids because gases they have more random motion they're more chaotic and disorderly so they have a higher entropy value whenever you increase the temperature of a gas you increase the kinetic energy of the molecules and they're moving faster in the more random motion so the entropy goes up with temperature the change in entropy is the difference between the entropy of the final state and the entropy of the initial state you can also calculate the change in entropy using this equation it's equal to q divided by t where q is the heat transferred into or out of the system now this equation applies for a reversible isothermal process now keep in mind whenever you have an isothermal process the change in temperature is zero so there's only one single constant temperature for an isothermal process therefore this equation applies in that instance calculate the change in entropy that occurs when seven kilograms of ice melts at zero degrees celsius and we're given the latent heat of fusion for ice so delta s is it positive or is it negative what would you say in order to melt ice we need to add heat to it typically when you add heat to something the entropy usually increases so delta s is positive in this case now this is an isothermal process when ice melts the temperature remains constant until all of the ice cubes have melted into liquid water so therefore the change in temperature is zero whenever the temperature is constant you can calculate delta s using this equation it's simply q divided by t and q is equal to m times l it's the mass times the latent heat of fusion so q is equal to seven kilograms times three point three four times ten to the five joules per kilogram so as you can see the unit kilograms will cancel and then we're going to take that result divided by t which is 0 degrees celsius but in kelvin that's 273 kelvin so if we multiply seven times three point three four times ten to the five you should get two million three hundred thirty eight thousand joules and then we'll divide that by 273 so in this case the change in entropy is 8 564 joules per kelvin so what can we do if the temperature is not constant what if the temperature changes how can we estimate or calculate the change in entropy we're going to do it two ways we're going to use a simple method to ballpark the answer and we're going to use calculus to get the exact answer so first let's estimate the answer you can still use the equation delta s is equal to q divided by t but you need to use the average temperature in kelvin not in celsius so first let's calculate q we can use the equation q is equal to mcat since we have a temperature change in the last problem we had a phase change when ice melts into liquid water the temperature is not changing so you want to use q equals ml but in this problem the temperature is changing so you want to use q equals m c delta c the mass is 4 kilograms the specific heat capacity is 4 184 joules per kilogram per celsius delta t the change in temperature could be in celsius or kelvin the difference will still be a hundred so as you can see the unit celsius will cancel and kilograms cancels as well so we're going to get q in joules so 4 times 4184 times 100 gives us a q value of 1 673 and 600 joules now let's divide that by the average temperature the average celsius temperature between zero and 100 is 50 celsius to convert that to kelvin add 273 to 50 which gives you 323 kelvin so now let's divide these two numbers so the change in entropy is approximately equal to 5181 joules per kelvin so let's round that to 5200 joules per kelvin now let's find the exact answer delta s is equal to the integral from a to b or the initial temperature to the final temperature of dq divided by t so that's the equation that you need now we know that q is equal to mcat so we could say that dq which is the differential of q is equal to mc which are constants times the differential of t that is the differential of the temperature so let's replace dq with mcdt m and c are constants so let's move that in front of the integral symbol the antiderivative of 1 over x is ln x therefore the antiderivative of 1 over t must be lnt so this is m c natural log of t evaluated from t initial to t final so final minus initial if we plug in those two values it's going to be m c natural log t final minus m c natural log teen issue a property of natural logs allows us to condense two single logs two logs into a single log so ln a minus lnb is the same as ln a over b so knowing that this expression is now equivalent to mc ln t final divided by t initial this is the equation that you want to use to calculate the change in entropy whenever you have a temperature change due to heating or cooling an object so let's go ahead and use that so delta s is going to be the mass which is four kilograms times the specific heat capacity of 4184 times the natural log of the final temperature which has to be in kelvin keep in mind delta t could be in celsius or kelvin but if you see t without the triangle it has to be in kelvin 100 plus 273 is 373 that's the final temperature the initial temperature which is zero celsius is the same as 273 kelvin so it's 4 times 4184 times natural log 373 divided by 273 and this is equal to 521 joules per kelvin the other answer was like 5181 joules per kelvin as you can see if you round both of them they're approximately equal to 5200 joules per kelvin but this is the exact answer this is the more accurate answer the other one is simply an approximation so this is the equation that you want to write down if you need to get the exact answer two moles of an ideal monatomic gas expands adiabatically from 300 kelvin to 250 kelvin the molar heat capacity of this gas is that number find the change in entropy now whenever a gas expands adiabatically the volume increases greatly and it increases at a factor that's greater than the decrease in pressure it turns out that adiabatic expansion cools the gas the temperature goes down as well now how can we calculate the change in entropy can we use this equation and if we can what is q now typically you might be thinking q equals mcat or some variation of this equation q equals n c times delta t if you have the molar heat capacity you can find q that way but it turns out that q is equal to zero for an adiabatic expansion or compression by definition whenever you have an adiabatic process there's no exchange of heat heat is not absorbed by the gas or released by the gas so this equation really doesn't apply and because q is zero for an adiabatic process the change in entropy is therefore zero so watch out for this question let's try this problem a cardinal engine absorbs 4 000 joules of heat energy from a hot reservoir at 800 kelvin how much heat energy does it release into the cold reservoir 300 kelvin so for cardinal engines you can use this equation qc divided by qh is equal to tc over th so qc we're looking for that qh is 4000 that's how much energy it absorbs from the hot reservoir tc is 300 kelvin and th is 800 kelvin so to find qc if we cross multiply it's going to be 4 000 times 300 divided by 800. so qc is 1500 joules so let's say this is the engine here is the hot reservoir and here's the cold reservoir so the engine absorbs 4000 joules from the hot reservoir and then it releases 1500 joules into the code of reservoir and this happens all within one cycle so if this information how can we calculate the net change in entropy of the engine so let's find out the entropy change when the engine absorbs energy from the hot reservoir it's going to be q divided by t and the temperature at the hot reservoir is constant it's 800 so we don't you need to use like the natural log function in this case so q is going to be 4 000 divided by 800. so delta s is positive 5 joules per kelvin now what about the change in entropy of the engine when it releases energy into the cold reservoir because it releases energy q is going to be negative 1500 and not positive and the temperature is 300 kelvin so negative 1500 divided by 300 is negative 5 joules per kelvin now the net entropy change is the sum of these two individual entropy changes that occurs at the cold and hot reservoir for the engine so it's going to be positive five plus negative five which is zero so what does this all mean for any cyclic reversible process the change in entropy will always be zero because if you start from position a go to b c d and come back at a you're going to end up in the same location so the entropy for such a cyclic process doesn't change the change in entropy is zero and as you can see we have gotten an answer so that's something that you just need to know when you're taking your tests now if you have an irreversible process the change in entropy doesn't have to be zero for most for irreversible processes the change in entropy is usually greater than zero but for a cyclic process for an ideal cyclic process the change in entropy at best is constant it's equal to zero five kilograms of water at 80 degrees celsius is mixed with five kilograms of water at 20 degrees celsius what is the total change in entropy now let's solve this using the average temperature process that we used earlier and also using the natural log function that we came up with but first what is the final temperature of the mixture because the mass of water in the two samples are identical the final temperature is going to be the average of these two temperatures we have the same substance the same quantity of water so then therefore the change in the temperature of these two samples of water should be the same so the average or the final temperature is going to be 50 degrees celsius so that the change in temperature is the same for these equal masses of water now let's focus on the hot water let's calculate the entropy change for that sample it's going to be the heat that is released divided by t for the hot water the temperature changes from 80 to 50. since the temperature is decreasing q has to be negative and using the equation q is equal to mcat the mass is 5 the specific heat capacity is 4184 for water that's joules per kilogram per celsius and the change in temperature is negative 30 as it goes from 80 to 50 degrees celsius now we know we need to use the average temperature in kelvin what is the average temperature well if you average 80 and 50 you're going to get 65 and then convert this celsius temperature into kelvin so 65 plus 273 is equal to 338 kelvin 5 times 4184 times negative 30 divided by 338 is equal to negative 1856.8 joules per kelvin so this is the change in the entropy for the the hot water sample now what about the cold water sample we're going to use the same equation qc over the average temperature so it's going to be 5 times 4184 times this time the temperature is going from 20 to 50. so the change in temperature is positive 30 instead of negative 30. so q is positive because heat is being absorbed now the average temperature the average of 2050 if you add them up 20 plus 50 is 70 divided by 2 is 35 celsius now let's convert this into a kelvin temperature 35 plus 273 is 308 kelvin so if we multiply 5 4184 and 30 and then divided by 308 this is going to give us positive 2037.7 joules per kelvin so now that we have the change in entropy for the cold water sample and for the hot water sample the change the total change is basically the sum of these two values so if we add them together we're going to get 180.9 joules per kelvin so this is our estimation it's around 181 but now let's use the other equation and see what answer we come up with to calculate the entropy change whenever the temperature is not constant you can use the equation m c natural log t final divided by t initial so let's use it for the hot temperature first so m is 5 c is 4184 and for the hot water sample it goes from 80 degrees celsius to 50. but we need to convert both of these values into kelvin so 80 plus 273 is 353 kelvin and 50 plus 273 is 323 kelvin so t final is 323 and t initial is 353 kelvin so if we plug this in we're going to get negative 1858.0 joules per kelvin now let's do the same thing for the cold water sample so it's going to be m times c times natural log now this time the temperature is changing from 20 to 50 degrees celsius now we know the kelvin temperature is 323 at 50 degrees celsius 20 plus 273 is 293 kelvin so now we have our initial and final temperatures in kelvin so the final is 323 the initial is 293 and if we plug this in this is going to equal positive 2039.3 joules per kelvin so now let's add these two values to calculate the total change in entropy so this will give you positive 181.3 joules per kelvin the other answer was positive 180.9 as you can see the difference is not that great so you can use either technique to get a good estimation of the entropy but this answer is more accurate though so that is it for this video hopefully you found it to be helpful so do well on your next exam and thanks for watching

Channel: The Organic Chemistry Tutor

Views: 344,234

Keywords: carnot engine thermodynamics, carnot engine efficiency, heat engines and refrigerators, heat engines and second law of thermodynamics, coefficient of performance, refrigerators, heat pumps, heat engines, thermodynamics, efficiency, work, otto cycle, compression ratio, gamma ratio, heat, entropy, physics, carnot engines, adiabatic, isothermal, carnot cycle, ice, delta S, disorder, carnot refrigerator, power, reversible, spontaneous, second law of thermodynamics

Id: QBd2zraOe2k

Length: 78min 26sec (4706 seconds)

Published: Sun Dec 18 2016