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What is the total entropy change of the system tea + air?

  1. Nov 26, 2007 #1
    1. The problem statement, all variables and given/known data
    You make tea with 0.250 kg of 89.0 °C water and let it cool to room temperature (20.0 °C) before drinking it.

    1)Calculate the entropy change of the water while it cools.
    2)The cooling process is essentially isothermal for the air in your kitchen. Calculate the change in entropy of the air while the tea cools, assuming that all the heat lost by the water goes into the air.
    3)What is the total entropy change of the system tea + air?

    2. Relevant equations

    m = 0.25kg
    T1 = 362K
    T2 = 293K

    3. The attempt at a solution

    1) I did m*c*ln(T2/T1) = -221 J/K And it's correct

    2) Now I'm not sure what to do here.

    Isothermal means that T = constant so is it 293K ?

    And what eq. am I suppose to use and why isn't the volume of the room a factor here?

    What I tried to do was to find the heat Q = m*L_f (L_f = 3.34*10^5 J/kg)

    So Q = 0.25*L_f = 83500 J

    Then (delta)S = Q/T = 83500J / 293 K = 289 J/K But thats not correct and I guess the entropy should me a small number but I'm not really sure since I'm not that good at understanding "High-Tech" english!

    3) If I can do 2) then 3 should not be a problem.

    Thank you.

    EDIT: I found out that I CAN use the eq. S = Q/T IF the process is isothermal so what am I doing wrong?
    ps. I've only got one attempt left and this is the last question of the year! (Apart from the test)

    EDIT2: Solution

    It was correct using (delta)S = Q/T using Q = 83500 J/K

    The T has to be the difference of temperature ((89°C-20°C)+273) = 342 K

    So Q/T = 244 J ! And it was correct.
    Last edited: Nov 26, 2007
  2. jcsd
  3. Nov 26, 2007 #2

    Andrew Mason

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    The flow of heat from the tea is [itex]\Delta Q = mc\Delta T_{tea}[/itex]. This has to be the amount of heat flowing into the air. So you don't have to know the heat capacity of the air.

    So the change in entropy of the air is just:

    [tex]\Delta S_{air} = \Delta Q /T_{air} = mc\Delta T_{tea}/T_{air}[/tex]

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