Physics 30 Entropy (3 of 5) Entropy and Heat Exchange: Example 2

[@bambo3267]

You are an awesome teacher i am now in my 4th year of mechanical engineering and i cant thank you for how many subjects you helped me remember or taught me from scratch when i didnt understand it

[@MichelvanBiezen]

We are glad we were able to help. Thanks for sharing.

[@questionman5]

For this entropy calculation, for the lake's Q, you added together the negative of (deltaS_ice + deltaS_water_heating), not the Q's of the ice melting and cold water heating up. Wouldn't you want (-mL - mcT)/283 ? Instead of (-mL/273 - mcT/278) /283 ? Thanks! Edit: Oh, I see you caught that toward the end of the video. :)

[@MichelvanBiezen]

Walter, At equilibrium, the melted ice will reach the same temperature as the rest of the lake (10 C)

[@Eugenewhatt]

You are awesome!!!! I'm getting an A for pre med!

[@andreguimaraes9347]

I'dbeso much more confortable if you used the actual integrate :/ If anybo0dy is curious dS=Int{dQ/T} But dQ = CdT (big C at constant pressute) So dS=int{CdT/T} from To to Tf dS = C*ln(Tf/To) = m*c*ln(Tf/To) It gives approximetly the same answer, but is more correct when dealing with larger gaps in temperatures

[@kidamaroo]

Usually, when I solve these types of problems, I calculate the system entropy and the surrounding entropy separately, then I subtract them at the end

[@recall660]

is this all the physics we need for the university level ,thanks

[@jackgarlick2611]

To get the latent heat from the ice melting you need Q = mL (the mass times the latent heat of fusion), but you have used Q = mcL (mass times specific heat times the latent heat of fusion). Do you have some good reason for doing this? or have you just not noticed this error. Also, why are you using two different units (both joules and calories in the same equation)? Even without the aforementioned error, this would give you an incorrect answer, beside the fact that in most reputable physics schools this is considered very poor practice.

[@MichelvanBiezen]

No error here. The 4186 J/kg is just a conversion factor from cal/g to J/kg. The video is correct. Thanks for checking.

[@ammarshazly2141]

Sir, why when calculating entropy for lake losing heat you considered its latent heat although there is no phase change?

[@MichelvanBiezen]

The amount of heat that is gained by the ice to melt the ice is the same amount of heat that is lost by the lake water. That is why we use the same expression.

[@ammarshazly2141]

@@MichelvanBiezen Got it, Thanks Sir

[@jillianhurst2415]

I love weater heating up

[@francohuacoto1541]

how would you change your solution if you have the amount of water, let's you put 1kg of ice in 5kg of water? Thanks!

[@philipjung4635]

Ive seen T = T_ave and T = T_hot reservoir. Is there a T that is supposed to be used instead of "this would be approximately accurate"

[@MichelvanBiezen]

You have to use the method of integration if you don't want to use the average T.

[@onielbalicoco1072]

How about if a 1 kg of water at 10 degrees C placed in thermal contact with a large thermal reservoir at exactly 0 degrees C? Thank you!

[@MichelvanBiezen]

What do you mean by "how about"? "How about" is such a general term that it does not make the question clear.

[@onielbalicoco1072]

@@MichelvanBiezen oh sorry sir what I mean is if process is reversed, the water is initially 10°C and it is putted in a cold large reservoir at 0°C, how am I going to calculate the total entropy? It is the question stated on the problem I am dealing right now. By the way thank you so much for your videos. It help me to answer my assignments.

[@MichelvanBiezen]

The 1 kg of water will cool down from 10 C to 0 C. The change in entropy will be: delta Q / T average = mc delta T / 5 C = - (1) (4186)(10) / 278 for the lake water change in entropy = + (1) (4186) (10) / 273

[@onielbalicoco1072]

How about for this question? "Find the change in the total entropy when 0.1 kg. of liquid water at 10 °C is placed in thermal contact with a large thermal reservoir at exactly 0 °C." Do I need to add in the Delta S of the water turning into ice (mL/T ice) to get the total entropy? Thank you sir.

[@MichelvanBiezen]

The question here becomes: "does the water, once it reached 0 C begin to freeze when it is in contact with a reservoir at 0 C ?" The answer is: "no". The reservoir needs to be below 0 C in order to draw heat away from the water. Note that at thermal equilibrium no heat will flow.

[@elyasigroupelyasi1437]

hello Pls let me know why you considered lake temprature constant as 283k ?

[@MichelvanBiezen]

Dumping a block of ice in a lake will have no measurable effect on the temperature of the lake.

[@EvuLYT]

For the change in entropy of the lake, don't we have to use the mass of the water of the lake? (which is not provided)?

[@MichelvanBiezen]

No because the temperature of the lake doesn't change.

[@gooddeedsleadto7499]

Hi Could u please show the difference between the entropies in these examples & entropy generation. Could u do the same example for entropy generation? I don't see any difference between the the two, Change of entropy in your examples & entropy generation. Thank you so much, please help remove my confusion.

[@tavernofheart]

Sir I have an unsolved problem and I hope you can help me, I need to find entropy of closed system consist of water 100 g 8°C and ice 0°C 10g. I calculate q of water and ice and it's both 800 cal, does this mean the q of system is 0 and the entropy is also 0? But if the entropy is 0,​ it means that it's reversable but I don't think it's reversable. I'm really confused

[@MichelvanBiezen]

Whenever there is heat exchanged, there MUST be in INCREASE in entropy. First you need to find the final state of the system. It will take 800 cal to melt all the ice and it will require the removal of 800 cal from the water to bring it down to 0 C. Thus the final state is water at 0 C. Change in S = - m c delta T / T ave + m Lf / T = - (100) (1) (4.186) (10)/278 + (10)(80)(4.186)/273 (The units are J/K)

[@tavernofheart]

@@MichelvanBiezen thank you so much sir for taking your time to answer. It's really helpful.​

[@Akcakes2024]

Prof. the second ds represents wat?.initially v r throwing ice into lake after tat it starts melting tat s 1st ds .ten 2nd..?

[@MichelvanBiezen]

After all the ice melts, what is left over is the same amount of water at zero degrees C. Then the second ds represents the cold water heating up until it is the same temperature as the lake. (10 degrees C)

[@Akcakes2024]

Sorry prof

[@magida2578]

how did you get 80 from the ice ? what is that

[@MichelvanBiezen]

To melt ice, it requires 80 calories (335 joules) for every gram.

[@bollywood-wanna-be-dancer1827]

@@MichelvanBiezen Dear Prof, how can I calculate the 80 cal? Best greetings from Germany

[@recall660]

I have a Question prof ,is this all the physics we need up to university level or there is more to it ,thanks

[@messi102882]

That doesn't make sense; since heat is being added to the ice to melt it, the lake must be cooling down slightly. At thermal equilibrium the final temperature of the system is not 10C, it will be less than that.

[@zioshi2]

+Dhruval Shah he told that the assuming the lake is very large this change can be neglected.

[@gooddeedsleadto7499]

what is 80 you multiplied?

[@MichelvanBiezen]

That is the latent heat of fusion of water in calories per gram. (the amount of heat required to melt ice)

[@gooddeedsleadto7499]

Michel van Biezen 1kg x 4186 j/kcal x 80/1000 x kcal/kg x 1000. Is it 4186 j/ kcal & not 4186 j/kg? Thanks

[@Akcakes2024]

Prof. I need simple explanation for entropy?

[@MichelvanBiezen]

Have you watched this video? Physics - Thermodynamics: (1 of 5) Entropy - Basic Definition

[@Akcakes2024]

I saw 1 st .now I can understand

[@anhho7464]

I do not understand how you calculate deta S lake losing heat. Can you tell more about this? Thank's!

[@MichelvanBiezen]

delta S = (heat transferred in or out of the substance) / (temperature at which it occurs) In this case heat is transferred from the lake to the ice

[@icoolio123]

Why does the Ice that melted heat all the way up to 10C I thought it would be less

[@younique9710]

I have a question. I know why 4186J/kg is added (to cancel out the unit, "Kg," or convert Kg into J). But at 3:49, the professor uses 4186 J/Kg K to cancel out Kg and also Kelvin. I wonder if we can simply remove a unit (e.g., Kelvin) to get the final unit (J/K). In the previous video ( 5 out of 137), the professor even used 4186 J/kg C (C = centigrade degree) to get the final outcome unit, Centigrade degree. I don't know if freely deleting a unit does not affect a final outcome.

[@MichelvanBiezen]

I am having trouble finding your question in this comment. Also one cannot convert kg into J. (kg is a unit of mass and J is a unit of energy). If you could just ask the question, we can answer it.

[@younique9710]

@@MichelvanBiezen I am sorry about making you confused. My question is, you sometimes use 4186 J/kg (at 2:58), sometimes use 4186 J/kg K (at 3:40), and sometimes use 4186 J/kg C (in your previous video). I don't know why the unit of 4186 keeps changed. Doesn't that change matter for the final outcome?

[@MichelvanBiezen]

Again, the way the question is asked it is difficult to answer. Everything should be presented in the form of an equation. For example the specific heat of water is 4186 J/kg C = 4186 J/kg K (c (for water) = 4.186 J/kg C) Which means that it takes 4186 J to raise the temperature of 1 kg of water by 1 degree C (or 1 degree K)

[@younique9710]

@@MichelvanBiezen Thank you for your response. I understood "it takes 4186 J to raise the temperature of 1 kg of water by 1 degree C (or 1 degree K)." That's depending on what the form of an equation. For example, "[(1kg) (4186 J/kg K) (10K) ] / 278K = 150.58 J/K" is equal to "[(1kg) (4.186 J/kg C) (-263C) ] / 5C = -1,096 J/C." But, I am still confused with: how is 4186 J/kg C (degree C) equal to 4186 J/kg K (degree K)? For example of a speed, 10 meter/sec is not equal to 10 meter/min.

[@MichelvanBiezen]

1 Centigrade degree is equal to 1 Kelvin degree but 1 degree C = 274 K When we are dealing with changes in the temperature we use the "SIZE" of the degrees (for example 1 centigrade degree = (9/5) Fahrenheit degrees)

[@recall660]

I have a Question prof ,is this all the physics we need up to university level or there is more to it ,thanks