How Does Mixing Water at Different Temperatures Affect Entropy?

[toothpaste666]

Homework Statement

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.

Homework Equations

ΔS=∫dQ/T.
dQ = mcTdT

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

[itex] Q_{tot} = 0 [/itex]

[itex] \int dQ_1 + \int dQ_2 = 0 [/itex][itex] dQ_1 = mcTdT [/itex]

[itex] \int dQ_1 = mc \int_{273}^{T_f} TdT [/itex]

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

[itex] \int dQ_1 = \frac{1}{2}mc (T_f^2 - 273^2) [/itex]

[itex] dQ_2 = mcTdT [/itex]

[itex] \int dQ_2 = mc \int_{326}^{T_f} TdT [/itex]

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

[itex] \int dQ_2 = \frac{1}{2}mc (T_f^2 - 326^2) [/itex]

plugging back in

[itex] \int dQ_1 + \int dQ_2 = 0 [/itex]

[itex] \frac{1}{2}mc (T_f^2 - 273^2) + \frac{1}{2}mc (T_f^2 - 326^2) = 0 [/itex]

[itex] (T_f^2 - 273^2) + (T_f^2 - 326^2) = 0 [/itex]

[itex] T_f^2 - 74529 + T_f^2 - 106276 = 0 [/itex]

[itex] 2T_f^2 = 180805 [/itex]

[itex] T_f^2 = 90402.5 [/itex]

[itex] T_f = 301 [/itex]

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

[itex]ds = \frac{dQ}{T} [/itex]

[itex]ds = \frac{mcTdT}{T} [/itex]

[itex]ds = mcdT [/itex]

[itex]s_1 = \int_{273}^{301} mcdT [/itex]

and

[itex]s_2 = \int_{326}^{301} mcdT [/itex]

and then

[itex] s_1 + s_2 = s_{sys} [/itex]

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

 

[rude man]

Homework Statement

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.

Homework Equations

dQ = mcTdT

whence this?

 

[toothpaste666]

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

 

[rude man]

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

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.

 

[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 don't tell me how the temp is changing as a function so I am not sure what to plug in there

 

[rude man]

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 don't tell me how the temp is changing as a function so I am not sure what to plug in there

"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.

 

[toothpaste666]

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

 

[rude man]

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

Yer' welcome!

(Except I got the units wrong. Cross out the "s" (for seconds).
http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/spht.html

 

[cabon7969]

please solve
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 .

 

[Chestermiller]

"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.

J/(kgK)

 

[Chestermiller]

please solve
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 .

Please do not hijack someone else's thread to seek help with your own problem. Please start a separate thread using the required template, and show some effort.

Chet