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Stirring a Cup of Tea

  1. May 23, 2016 #1
    1. The problem statement, all variables and given/known data
    A mug containing a volume of ta initially at 90oC is in thermal contact with the enivronment at 20oC.

    Density of tea = ##10^{3}kgm^{-3}##
    Constant pressure specific heat capacity of tea: ##4.20 \times 10^{3} Jkg^{-1}K^{-1}##

    b) After reaching equilibrium, 0.01 J of work is done on the tea by stirring it after which it settles into equilibrium again. Calculate the resulting entropy chagne of the tea and of the universe
    2. Relevant equations
    Already have found out that:

    [1] $$ \Delta S_{Univ} = C_{P} \bigg( \frac{T_{0}-T_{F}}{T_{R}} + ln \frac{T_{I}}{T_{F}} \bigg) $$
    ##T_{I}## and ##T_{F}## are the inital and final temperatures respectively.

    [2] $$ \Delta S_{Tea} = C_{P} ln\frac{T_{F}}{T_{I}} $$

    3. The attempt at a solution
    Let ##T_{eq}## be the equilibrium temperature of the tea (i.e ##293K##/##20^{o}C##)

    (As an aside - we've done work on the tea, but it's volume hasn't increased - how does that make sense? Or is it meant to very subtly expand?)

    What I'd like to do is to say that ##dU = 0 \implies dQ = \Delta W ## - but that doesn't seem a very sensible statement to make - unless we make some argument that the tea must lose all of the energy it just gained as heat when it returns to equilibrium?

    If I do say that, then it follows that

    [3] $$ C_{P}dT = \Delta W \implies \Delta T = \frac{\Delta W}{C_{P}} $$

    From which it is then trivial to get a value for the entropy for tea and universe by substituting into the equations [1] and [2] above...

    Really appreciate the help! Thanks.
  2. jcsd
  3. May 23, 2016 #2
    Do they tell you how much tea there is?

    Before you get to part (b), you should focus on the changes in entropy of the system and the surroundings if part (a). So, how much does the entropy of the system change in part (a) and how much does the entropy of the reservoir (environment) change?

    You asked about the work in part (b). For your edification, pressure-volume work is not the only kind of work than can be applied to a system. In this case, stirring the tea does work on the liquid by deforming the liquid; the work is associated with viscous stresses that develop as a result of the deformation.
  4. May 24, 2016 #3
    Yes, volume is ##2.5 \times 10^{-4}m^{3}## - I've found the general expression for the change in the entropy of the universe and the tea in eqns (1) and (2),above? [These were the results given to derive, so I assume they're right!] To get a numerical answer, I would just stick in ## T_{I}=90^{0}C## and ##T_{F}=20^{o}C##,right?

    (Oh - and yep, that makes sense - but how come we don't consider these in the first law: ## dU = TdS - PdV## - is that due to the implicit assumption of reversibility, which including viscosity won't allow? )
  5. May 24, 2016 #4
    The first law does apply to this situation. For the system in this problem, ##\Delta U = Q-W##. The equation you wrote, ## dU = TdS - PdV## is not the first law. It just tells us the relationship between the changes in U, S, and V between two closely neighboring (i.e., differentially separated) thermodynamic equilibrium states.

    In part (b) of your problem, the initial temperature of the tea is 20 C. What is the final equilibrium temperature of the tea (that is in contact with a reservoir at 20 C)? What is the change in entropy of the tea in part (b), given that entropy is a physical property of the tea that is a function only of its initial and final states?
  6. May 24, 2016 #5
    Ah, I see.

    I'm guessing you're wanting me to say the ##dS=0## - since the temperature of the tea at the end will be the same as the temperature at the beginning. I guess that makes sense for the tea - like I said earlier, it seems sensible that the tea loses all the extra internal energy it gains. But the surroundings will have to take in that heat. But assuming that the surroundings are a large heat bath, that difference is negligible, so- do we say..
    $$ \Delta S_{Tea} = 0, \Delta S_{Surround} \approx 0 $$
    $$ \Delta S_{Univ} = 0 $$

    Though that does seem to me a bit odd...
  7. May 24, 2016 #6
    Also - regarding the point about entropy and state variables. Is there an intuitive justification of the fact that any given state variable ##S,U##etc., can be written explicitly in terms of only two variables i.e ## S=S(T,P)=S(V,N)## or do I need to start doing some more theoretical thermodynamics/stat physics to understand this? e.g phase space, Liouville..
  8. May 24, 2016 #7
    You said that 0.01 J of work was done in stirring the tea (this seems awfully low. Did you mean 0.01 kJ?). Since the internal energy of the tea didn't change, where did that work go?
  9. May 24, 2016 #8
    That's what they put in the question! I'm assuming that the work went out as heat into the surroundings when the tea cooled?
  10. May 24, 2016 #9
    Is the phase rule at a sufficiently fundamental level for you? What are the degrees of freedom for a single phase substance of constant composition?

    Or, for a single phase substance, if entropy per unit mass is a physical property, it can depend on T, P, and V, where V is the specific volume. But, the equation of state eliminates one of these. Does that work for you?
  11. May 24, 2016 #10
    This is correct. So, what is the change in entropy of the surroundings?
  12. May 25, 2016 #11
    The change in entropy of the surroundings is zero? Like I said above? Or, from equation (1) since the surroundings are a heat bath, there is no change in temperature, therefore no change in entropy...?

    Phase rule makes sense, had to look it up -## F= C-P+2## - in this case, ## C=1## (assuming we can model tea as pure water) ##P=1, \therefore F=2##
  13. May 25, 2016 #12
    The change in entropy of the heat bath is 0.01/293 J/K. This is the amount that the bath receives (ideally reversibly) from the tea. So the entropy is generated in the tea, but transferred to the bath.
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