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## Homework Statement

A mug containing a volume of ta initially at 90

^{o}C is in thermal contact with the enivronment at 20

^{o}C.

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.