# Entropy problem

1. Oct 26, 2014

### toothpaste666

1. The problem statement, all variables and given/known data
An aluminum can, with negligible heat capacity, is filled with 150g of water at 0 ∘C and then is brought into thermal contact with a similar can filled with 150g of water at 53∘C.
Find the change in entropy of the system if no heat is allowed to exchange with the surroundings. Use ΔS=∫dQ/T.

2. Relevant equations
ΔS=∫dQ/T.
dQ = mcTdT

3. The attempt at a solution
cup1 T = 273 K cup2: T = 53 + 273 = 326 K
first I need to find the final temp they will both be at

$Q_{tot} = 0$

$\int dQ_1 + \int dQ_2 = 0$

$dQ_1 = mcTdT$

$\int dQ_1 = mc \int_{273}^{T_f} TdT$

$\int dQ_1 = mc ( \frac{1}{2} \left.T^2 \right|_{273}^{T_f} )$

$\int dQ_1 = \frac{1}{2}mc (T_f^2 - 273^2)$

$dQ_2 = mcTdT$

$\int dQ_2 = mc \int_{326}^{T_f} TdT$

$\int dQ_2 = mc ( \frac{1}{2} \left.T^2 \right|_{326}^{T_f} )$

$\int dQ_2 = \frac{1}{2}mc (T_f^2 - 326^2)$

plugging back in

$\int dQ_1 + \int dQ_2 = 0$

$\frac{1}{2}mc (T_f^2 - 273^2) + \frac{1}{2}mc (T_f^2 - 326^2) = 0$

$(T_f^2 - 273^2) + (T_f^2 - 326^2) = 0$

$T_f^2 - 74529 + T_f^2 - 106276 = 0$

$2T_f^2 = 180805$

$T_f^2 = 90402.5$

$T_f = 301$

now i am stuck when it comes to finding the actual entropy. I tried using

$ds = \frac{dQ}{T}$

$ds = \frac{mcTdT}{T}$

$ds = mcdT$

$s_1 = \int_{273}^{301} mcdT$

and

$s_2 = \int_{326}^{301} mcdT$

and then

$s_1 + s_2 = s_{sys}$

but it didnt give me the right answer which makes me question my entire method. where did i go wrong?

2. Oct 26, 2014

### rude man

whence this???

3. Oct 26, 2014

### toothpaste666

its specific heat. my book gives it as Q = mc(deltaT) and the differential form as
dQ = mc(T)dT

4. Oct 26, 2014

### rude man

dQ = mc(T) dT is not the same as dQ = mcT dT!. And then you carried this mistake all the way thru your computations.

It should be apparent that, assuming constant c between the two temperatures, that the mix temperature will be the average of the two.

Strangely enough, your mix temperature wasn't off by much.

EDIT: that goes for your entropy change calculations also, for which you again used the wrong formula.

5. Oct 26, 2014

### toothpaste666

I am confused... I guess the difference between dQ = mcTdT and dQ = mc(T)dT is that (T) should be replaced by a function of temperature but they dont tell me how the temp is changing as a function so I am not sure what to plug in there

6. Oct 26, 2014

### rude man

"mc(T)" says "mass times specific heat as a function of temperature". The units are J/s or W.
"mcT" says "mass times specific heat times temperature". Units are J-K/s or W-K.
The two are completely different entities.
In your case there is no difference between c and c(T) since c is assumed constant between your two temperatures. So forget c(T) and use c only:
dQ = mc dT
dS = dQ/T = mc dT/T.
c is specific heat in J/s/kg.

7. Oct 26, 2014

### toothpaste666

yeah the whole problem makes a lot more sense now. thank you!

8. Oct 27, 2014

### rude man

9. Jul 30, 2015

### cabon7969

Two vessels A and B each of volume 3m^3 are connected by tube if negligible volume. Vessel A contains air at 0.7 MPa , 95C while vessel B contains air at 0.35MPa , 205C. Find the change in entropy of system assuming the mixing to be complete and adiabatic .

10. Jul 30, 2015

### Staff: Mentor

J/(kgK)

11. Jul 30, 2015