- #1
bananabandana
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Homework Statement
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
Homework 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] $$ \Delta S_{Tea} = C_{P} ln\frac{T_{F}}{T_{I}} $$
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.