I don't understand why, when our professor taught this, I left the class more clueless than when I walked in. in Less than 10 minutes I have a fundamental idea of this and now know how to calculate entropy change for a simple heat transfer. smh...
when we talk about entropy, it seems that we can only calculate the delta entropy in the physical change, how to calculate delta entropy in a chemical change? we always say that we can use the law of thermodynamics into chemistry, but what real problems of chemistry can the law of thermodynamics do?
If I use the method of integration for this problem I get delta S = 25.38J/K Is this answer technically more correct/accurate? I used: Delta S = Integral(dQ/T) = mc ln(Tf/Ti) to find the two changes in entropy and add them together. (as dQ = mc dT, where ln is log base e and Tf final temp, Ti initial temp.)
Thanks! I’m so horrible at math and wish I saw it as clear as you all. How are Celsius canceled out by Kevin? I always get confused how multiple units can be whittled down to a single x to y.
A Kelvin degree is the same size as a Celsius degree. So when we are considering changes in temperature it doesn't matter if we use K of C and one can cancel the other. You cannot do that if they are actual temperatures. 0 C = 273 K
It turns out that for relative small temperature changes that give you a good (and close) approximation, compared to integrating over the changing temperature. (We have some examples where we use the integration technique)
I always think of entropy as the dispersal of energy, and thus being unable to do work (because you need differentials to do work). So is the 25.2 J/K the energy that is dispersed (being the 2 cups being both 50°C)?
Note that the units involve both energy and temperature. Thus it is not only energy. Heat will naturally transfer from hot to cold (2nd law of thermodynamics). The greater the temperature gradient the faster the heat will transfer. Increasing entropy will decrease the difference in temperature, which will decrease the ability of heat to be transferred, or another way to look at it, there will be less heat available to transfer from hot to cold. The universe is like a big heat battery. Eventually it will run out of available heat.
@@MichelvanBiezen So... here's a question a little out of left field for you; is there a relationship between the action functional in Hamiltonian Mechanics and entropy?
@@MichelvanBiezen Based on units alone, the units of entropy are Joules per Kelvin, and the units of action are Joule - seconds. So the equivalence relation would have to be something like. DA / Dt = E (differential expression for Action) S = DE / DT (expression for entropy) dA = E dt dE = S dT E = integral ( S dT )_overT A ~ = integral( integral( S dT)_over-T dt)_over-t There's probably a concept missing here; like E in the action functional is supposed to be a the lagrangian and DE in the entropy expression is supposed to represent heat; so the expression probably isn't really complete. But based on the units in these expressions; maybe this form is roughly correct? Intuitively this would represent the sum of all probability density states as associated with temperature? Maybe?
@@historyisthebest5831The actual value of entropy is arbitrary, since there is no objective point at which entropy equals zero. All formulas for entropy are really formulas for change in entropy. There are conventions for where we define entropy to equal zero for a given substance, but these are arbitrary, and are just bookkeeping and not reality. For instance, ASHRAE uses the reference point of saturated liquid at -40C, as the zero point for thermodynamic potentials of refrigerants. Negative entropy ultimately doesn't mean anything. It's just like negative gravitational potential energy, that just means it is less than the reference datum. A negative change in entropy of the universe by contrast, indicates an impossible process.
Hey prof, can you check that my reasoning is correct? The formula used is only valid for reversible processes, and clearly the physical situation is irreversible. So, we can imagine taking each cup to a room seperately and making it reach the final temperatures of 50ºC by manipulating each of the room temperature, so that it is in equilibrium with the surroundings for all time (so it is an isothermal and therefore reversible process?) I.E for the cooler cup, take it to a room at 0ºC and increment the temperature very slowly so that the process can be reversed, and likewise for the warmer cup, take it to a room at 100ºC and decrement the tempreature very slowly so that the process can be reversed. The change in entropy of the SYSTEM (and system only) will be the same as entropy is a function of state.
at 3:49, can anyone tell me why Centigrade degrees is the same size as Kelvin degree? Why we do not multiply 273 to cancel out these two units (i.e., Kelvin degree and Centigrade degree)? In addition, the final outcome (25.2 J/K) means, we lose 25.2 J/K energy based on the concept of entropy?
That is possible, but it depends on a number of factors, the size and makeup of the conducting path, the contact area on each side, the circulation of the substance heated, etc. We can do relatively simple approximation calculations with the correct assumptions.
That would be difficult to explain like this as it would require integration. I will do some videos in the future to explain the time it takes to transfer heat.
Hi: You took out the confusion, thank you. You calculated the change of entropy for each cup and then added the two. Could you also please clarify the notion of entropy, the entropy being the maximum probable macrostate in equilibrium. What exactly is the maximum probable state for this two cup example with a number of particles known in each cup in equilibrium? Does it mean a total number of permutations of particles for each cup when the two systems or the two cup, in this case, have reached the maximum probable state and in a state of equilibrium? Could you please use a finite number of particles, may be 10, and actually calculate the maximum probable state using the permutation formula or counting principles? This will remove the confusion. Thanks. I l enjoy your lectures, they remove the doubts in mind. Great!
+mohamed zaki Since we are calculating the difference between 2 temperatures, you don't have to convert to K, because you will get the same result. But for the average temperature, you need the actual temperature in K
No, you are finding the difference of the entropy of the WHOLE system after the heat transfer and the entropy of the WHOLE system before the hear transfer.
Then you have to use the technique of calorimetry to find the final temperature. Look at that playlist. PHYSICS 23 CALORIMETRY th-cam.com/play/PLX2gX-ftPVXVFiZGjaL29aulIfiR-qY4C.html
Universe is an isolated system. No heat into or out of the universe. Does it means this increase is entropy is due to the entropy generation? And, the entropy in the previous problem is also due to entropy production? Thanks
i'm sorry but I think you are wrong with how you cancel out the different units in the problem, you cannot cancel units using subtraction in (50 c - 100 c) and even if you get the difference you cannot cancel kelvin with celcius without changing the celsius to kelvin first.
One of the best instructors in terms of simplicity and clarity. Thank you Mr. Biezen, I have learned and realized many things from your videos.
I don't understand why, when our professor taught this, I left the class more clueless than when I walked in. in Less than 10 minutes I have a fundamental idea of this and now know how to calculate entropy change for a simple heat transfer. smh...
Really awesome. My problem is solved finally.
when we talk about entropy, it seems that we can only calculate the delta entropy in the physical change, how to calculate delta entropy in a chemical change? we always say that we can use the law of thermodynamics into chemistry, but what real problems of chemistry can the law of thermodynamics do?
If I use the method of integration for this problem I get delta S = 25.38J/K
Is this answer technically more correct/accurate?
I used: Delta S = Integral(dQ/T) = mc ln(Tf/Ti) to find the two changes in entropy and add them together.
(as dQ = mc dT, where ln is log base e and Tf final temp, Ti initial temp.)
For small delta T it makes a small difference as you have seen.
Thanks! I’m so horrible at math and wish I saw it as clear as you all.
How are Celsius canceled out by Kevin?
I always get confused how multiple units can be whittled down to a single x to y.
A Kelvin degree is the same size as a Celsius degree. So when we are considering changes in temperature it doesn't matter if we use K of C and one can cancel the other. You cannot do that if they are actual temperatures. 0 C = 273 K
@@MichelvanBiezen makes sense . Thanks so much
Sir, why did you take average temperature between final and initial temperatures?
It turns out that for relative small temperature changes that give you a good (and close) approximation, compared to integrating over the changing temperature. (We have some examples where we use the integration technique)
@@MichelvanBiezen thankyoooou so much Sir, you are the world Best teacher.
Sir, is tha average temperature which is 75°C given? I can't figure out where 75°C coming from🙁. I need help, anyone?
If a cup of water starts at 100 C and then cools down to 50 C, the average = (100 + 50) / 2 = 75
I always think of entropy as the dispersal of energy, and thus being unable to do work (because you need differentials to do work). So is the 25.2 J/K the energy that is dispersed (being the 2 cups being both 50°C)?
Note that the units involve both energy and temperature. Thus it is not only energy. Heat will naturally transfer from hot to cold (2nd law of thermodynamics). The greater the temperature gradient the faster the heat will transfer. Increasing entropy will decrease the difference in temperature, which will decrease the ability of heat to be transferred, or another way to look at it, there will be less heat available to transfer from hot to cold. The universe is like a big heat battery. Eventually it will run out of available heat.
@@MichelvanBiezen Yeah okay. The universe will run out of available heat, because temperature/energy has been equally distributed, right?
That is correct.
@@MichelvanBiezen So... here's a question a little out of left field for you; is there a relationship between the action functional in Hamiltonian Mechanics and entropy?
@@MichelvanBiezen Based on units alone, the units of entropy are Joules per Kelvin, and the units of action are Joule - seconds. So the equivalence relation would have to be something like.
DA / Dt = E (differential expression for Action)
S = DE / DT (expression for entropy)
dA = E dt
dE = S dT
E = integral ( S dT )_overT
A ~ = integral( integral( S dT)_over-T dt)_over-t
There's probably a concept missing here; like E in the action functional is supposed to be a the lagrangian and DE in the entropy expression is supposed to represent heat; so the expression probably isn't really complete. But based on the units in these expressions; maybe this form is roughly correct? Intuitively this would represent the sum of all probability density states as associated with temperature? Maybe?
Hey is C in the delta S formula a constant 1 cal/g•k ?
In the real world, it is not a constant. But for our purpose (and since the change is minor) it makes sense to just call it a constant 1 cal/g K
The total entropy is always positive even though the entropy might increase or decrease during the processes.
The change in entropy is always positive
@@MichelvanBiezen Yeah right! Total change in entropy.
@@historyisthebest5831The actual value of entropy is arbitrary, since there is no objective point at which entropy equals zero. All formulas for entropy are really formulas for change in entropy.
There are conventions for where we define entropy to equal zero for a given substance, but these are arbitrary, and are just bookkeeping and not reality. For instance, ASHRAE uses the reference point of saturated liquid at -40C, as the zero point for thermodynamic potentials of refrigerants.
Negative entropy ultimately doesn't mean anything. It's just like negative gravitational potential energy, that just means it is less than the reference datum. A negative change in entropy of the universe by contrast, indicates an impossible process.
Hey prof, can you check that my reasoning is correct?
The formula used is only valid for reversible processes, and clearly the physical situation is irreversible. So, we can imagine taking each cup to a room seperately and making it reach the final temperatures of 50ºC by manipulating each of the room temperature, so that it is in equilibrium with the surroundings for all time (so it is an isothermal and therefore reversible process?) I.E for the cooler cup, take it to a room at 0ºC and increment the temperature very slowly so that the process can be reversed, and likewise for the warmer cup, take it to a room at 100ºC and decrement the tempreature very slowly so that the process can be reversed. The change in entropy of the SYSTEM (and system only) will be the same as entropy is a function of state.
at 3:49, can anyone tell me why Centigrade degrees is the same size as Kelvin degree? Why we do not multiply 273 to cancel out these two units (i.e., Kelvin degree and Centigrade degree)? In addition, the final outcome (25.2 J/K) means, we lose 25.2 J/K energy based on the concept of entropy?
One centigrade degree = 1 Kelvin degree. (the size of each degree is the same) But 1 degree C = 274 degrees K
@@MichelvanBiezen I see: change degree is same size. I will keep in mind that degree centigrade is not equal to centigrade degree. Thank you!
How do we find the Tavg?
(Q final + Q initial) / 2
(100 C + 50 C) / 2 = 75 C
(50 C + 0 C ) / 2 = 25 C
Du Pham thnxx
hi, great explaining, but why dont you use the integral of dQ/T?
The variables in the integral must be the same. So you have to convert Q in terms of T
Great videos! Thank you very much!!!
Sir , can we find out the total time taken in transferring heat from 100C to 0C jar of water??
That is possible, but it depends on a number of factors, the size and makeup of the conducting path, the contact area on each side, the circulation of the substance heated, etc. We can do relatively simple approximation calculations with the correct assumptions.
Please can u just tell me that one concept on which this ques can be solved which i have just asked you??
That would be difficult to explain like this as it would require integration. I will do some videos in the future to explain the time it takes to transfer heat.
Thank you so much sir!! Your videos are very help full for me.
Each video is very use full.. :)
Please can anyone help me how do you find the average of the first cup is 75 instead of 25.
The first cup starts at 100 C and then drops down to 50 C To find the average of two numbers A and B: average = (A + B) / 2
@@MichelvanBiezen thanks I was not getting it too.
Hi: You took out the confusion, thank you. You calculated the change of entropy for each cup and then added the two.
Could you also please clarify the notion of entropy, the entropy being the maximum probable macrostate in equilibrium.
What exactly is the maximum probable state for this two cup example with a number of particles known in each cup in equilibrium?
Does it mean a total number of permutations of particles for each cup when the two systems or the two cup, in this case, have reached the maximum probable state and in a state of equilibrium?
Could you please use a finite number of particles, may be 10, and actually calculate the maximum probable state using the permutation formula or counting principles? This will remove the confusion.
Thanks. I l enjoy your lectures, they remove the doubts in mind. Great!
Have you seen this playlist: PHYSICS 32.5 - STATISTICAL THERMODYNAMICS th-cam.com/users/ilectureonlineplaylists?sort=dd&view=50&shelf_id=4
Thank you 🙏💯🙏
No problem 👍
why we aren`t transfer ( 50c- 0c ) or ( 50c -100c ) into K
such as : Taverage
+mohamed zaki
Since we are calculating the difference between 2 temperatures, you don't have to convert to K, because you will get the same result. But for the average temperature, you need the actual temperature in K
If you're finding the difference in entropies why are you adding the two, why don't you subtract the delta S2 from delta S1?
You have to add the entropy of both before the heat transfer and then add the entropy of both after the heat transfer and compare.
So you're not finding the difference in entropy as in the deltaS you're finding the sum of entropies?
No, you are finding the difference of the entropy of the WHOLE system after the heat transfer and the entropy of the WHOLE system before the hear transfer.
Prof. Here ds positive if it negative wat...?
delta S of a closed system is always positive
Wow! Thank you Sir!
What if the two cups didn't have equal amount of water - how do we find the final temperature?
Then you have to use the technique of calorimetry to find the final temperature. Look at that playlist. PHYSICS 23 CALORIMETRY th-cam.com/play/PLX2gX-ftPVXVFiZGjaL29aulIfiR-qY4C.html
how is the average temperature 75
ok i got it
Yes same problem with i didn't understand why he get 75 instead of 25 ???!!
Could anyone help me here??? please 🙏🙏🙏
Universe is an isolated system. No heat into or out of the universe. Does it means this increase is entropy is due to the entropy generation? And, the entropy in the previous problem is also due to entropy production?
Thanks
Someone been playing with a nerf gun? I see a dart in the corner lol
Yes, we have kids armed with nerf guns.
Thank u sir
why is t average 75C and not 50C?
+1234av8tor For the hot container the average is 75C (100C + 50C) / 2 For the cold container the average is 25 C (0C + 50C) / 2
i'm sorry but I think you are wrong with how you cancel out the different units in the problem, you cannot cancel units using subtraction in (50 c - 100 c) and even if you get the difference you cannot cancel kelvin with celcius without changing the celsius to kelvin first.
Since we are considering the difference in temperature, centigrade degrees are equal to Kelvin degrees.
so, what if the given temperature is in kelvin would you still cancel it or change it to Celsius?