1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Change of entropy

  1. Nov 13, 2015 #1
    1. The problem statement, all variables and given/known data
    to illustrate an aspect of the second law of thermodynamics, we regard entropy changes in two cases.
    We have a vessel with 1 kg of warm water with T = 80 C and another vessel with 2 kg of cold water with T = 10 C and we mix them.

    Case 1:
    we pour hot water in the vessel with the cold water. The total entropy change becomes S1

    Case 2:
    we have a third empty vessel. We mix warm water with 1 kg of the cold water in the vessel. The contents of the third container is then poured with the remainder (1 kg) of cold water. The total entropy change becomes S2

    calculate S2 / S1

    2. Relevant equations
    ΔS = m*c*dT/T
    ΔS(tot)= ΔS(hot water)+ΔS(cold water)
    c=4190 [J/kg K]
    T (hot water) = 80 C = 353 K
    T (cold water)= 10 C = 283 K
    3. The attempt at a solution
    The answer is 1 that means S2 = S1 but I dont get it! I have problem with Case 2

    My solution for Case 1 is :

    ΔS(hot) = 1 * 4190 * ln(283/353)= -926 J/K
    ΔS(cold)=1*4190 * (353-283)/283 = 1036.4 J/K
    ΔS1 = 110,4 J/K


    Case 2 :
    We say that we mix first 1kg warm water with 1 kg cold water:
    ΔS (hot) = 1 * 4190 * ln(283/353)= -926 J/K (same !)
    ΔS (cold) = (1*4190 * (353-283)/283) = 1036.4 J/K (same !)

    Then I will mix the other cold water what I think that when we mixed first 1kg of warm water with cold water, we got a new temperature, when we think of Q (missing) = Q (gain)
    when the new value is 45 C, which is 318 K.
    Now we have 2 kg of water with temperature 318 K in a third vessel and we want to pour 1 kg of water with temperature 283 K. but when I calculate I get it:
    ΔS(remainder cold) = 1 * 4190 * ln(318/283) = 488 J/K !

    what is the problem ?! how could I think ?! As I said the solution is S2/S1 = 1 ! But I get a bigger S2
     
  2. jcsd
  3. Nov 13, 2015 #2

    DrClaude

    User Avatar

    Staff: Mentor

    What is the final temperature of the water?
     
  4. Nov 13, 2015 #3
    In case 2:
    When we first mix 1kg warm and 1 kg cold water
    Q(warm water) = Q (cold water)
    1 * 4190 (T(warm) - T(average)) = 1* 4190 (T(average)- T(cold))
    T average = 45 C
     
  5. Nov 13, 2015 #4

    DrClaude

    User Avatar

    Staff: Mentor

    And for case 1? You don't seem to have taken the final temperature into account anywhere.
     
  6. Nov 13, 2015 #5
    Should I do that for case 1 too ?!
    Ok
    we mix 1 kg hot water with 2 kg cold water

    1 * 4190 * (80 - T(av)) = 2* 4190 *(T(av)-10)
    T(av) = 100/ 3 C= 33.33 C

    This is the final temperature in case 1...
     
  7. Nov 13, 2015 #6

    DrClaude

    User Avatar

    Staff: Mentor

    How can you not do it? To calculate the change in entropy, you need to know the change in temperature.
     
  8. Nov 13, 2015 #7

    ok now I think I understand what you mean
    the first case I used this formula:
    m * 4190 * ln (T2 / T1)

    but I could also do so here
    S = mcdT / T (average)

    T (average) is = (C 80 + C 10) / 2 = 45 C
    and we must write it in Kelvin when 45 +273 = 318 K

    S (heating water) = 1 * 4190 * (10-80) / 318 = -922 J / K

    Was it that you meant?!
     
  9. Nov 13, 2015 #8

    DrClaude

    User Avatar

    Staff: Mentor

    The first equation for ΔS is correct, not the second one. What is incorrect is the value of T2 you were using.

    This is not the final temperature. You need to take into account the amount of water.
     
  10. Nov 13, 2015 #9
    OK thanks! :smile:
    but what is the value of T(2) ?!
    And my important question is why there is no difference between case 1 and case 2 ?!

    the case 1 is clear but the second case , when I first calculate with 1 kg cold water I get the same answer, but when I add1 kg cold water further, I do not get the same response as the case first. I know the new 1 kg cold water will mix with 1 kg cold & 1 kg warm water which are already mixed, that made me dizzy!:frown:
     
  11. Nov 13, 2015 #10
    please can somebody help me ?
     
  12. Nov 13, 2015 #11
    If s0 is the entropy of 1 kg of water at 273 K, then the entropy of 1 kg of water at temperature T is:
    $$s=s_0+c\ln{(T/273)}$$
    where c is the heat capacity. Does that make sense to you so far?

    Chet
     
  13. Nov 14, 2015 #12
    thanks but why 273 ?! the cold water is 283 k, in case two for the last process, we have a vessel with 2 kg water at 318 K and now we want to pure 1 kg cold water with T = 283 K in that vessel ... This part make me dizzy
     
  14. Nov 14, 2015 #13
    I just arbitrarily chose 273 as a reference temperature, with an entropy at this reference temperature of so. This choice will not affect the final answer.
    Please be patient. I will help you work out the answer.

    So, I ask again. Are you comfortable with this so far?

    Chet
     
  15. Nov 14, 2015 #14
    Yes I think I do ... I know that is the change in entropy...:smile:
     
  16. Nov 14, 2015 #15
    You don't seem too sure. But OK, here we go.

    We are going to focus for now exclusively on Case 1. After we successfully do Case 1, we will analyze Case 2.

    CASE 1:

    Initial thermodynamic equilibrium state of system:
    1 kg water at 80 C (353 K)
    2 kg water at 10 C (283 K)


    Final thermodynamic equilibrium state of system:
    3 kg water at temperature ??


    What is the temperature of the water in the final equilibrium state of the system?

    In terms of s0, using the equation in post #11 (which applies to 1 kg of water), what is the entropy of the system in the initial thermodynamic equilibrium state?

    In terms of s0, what is the entropy of the system in the final thermodynamic equilibrium state?
     
  17. Nov 14, 2015 #16

    Bingo !

    I think I solved the problem !

    well for case one when we mix 2 kg cold water with 1 kg hot water we have:
    1 (353 -T)=2(T-283)
    and T = 306.33 K
    S(hot)= 1 * 4190*Ln(306.33/353)= -598.68 J/K
    S(cold)= 2*4190*Ln(306.33/283)= 655 J/K
    S1=56 J/K

    And Case 2
    When we mix 1 kg cold and hot water our new T will be 318 K
    and
    4190*ln(318/353) + 4190 *ln(318/280)= 51,57 J/K

    when we add 1 kg cold water again T will be 306
    2*4190*ln(306/318) + 1*4190*ln(306/283) = 5.2
    S2 = 56 J/K

    Now I understand it ! because I'm beginner with entropy and I started with Ice in a big lake problem I thought that I should do the same algorithm but I see the algorithm depends on the situation !
    Thanks for help!:wink::smile:
     
  18. Nov 14, 2015 #17
    OK. So you have no interest in trying the problem by the alternate approach I was going to suggest. Fine.

    Chet
     
  19. Nov 14, 2015 #18
    what's the alternative method ?!
     
  20. Nov 14, 2015 #19
    The alternative method is answering the questions I asked.

    Chet
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Change of entropy
  1. Entropy change (Replies: 0)

  2. Entropy change (Replies: 1)

  3. Change in entropy (Replies: 1)

  4. Change in entropy (Replies: 11)

Loading...