Entropy change for two masses of water mixed adiabatically

[LancsPhys14]

Homework Statement

Two quantities of water, of mass M and nM where n is constant, have temperatures T1 and T2. They are adiabatically mixed together and the pressure remains constant. What is the entropy change of the universe in this process?

Relevant Equations

dU=δW+δQ
ΔS=ΔSh+ΔSc
ΔS=Q/T
dS=δQ/T
Q=mcΔT

the entropy change for a reversible adiabatic process is zero as it remains constant. Is this a reversible process?

assuming T1>T2:
hot (h) water has mass M, temp T1
cold (c) water has mass nM, temp T2

let the final temperature be Tf

if δQ=0 as the process is adiabatic, |Qh|=|Qc| so Qh=-Qc

by Q=mcΔT, Qh=Mc(T1-Tf) and Qc=nMc(T2-Tf) hence Mc(T1-Tf)=-nMc(T2-Tf)

this give Tf=(T2+(n^-1)T1)/((n^-1)+1)

I am unsure where to go from here as I have the equation ΔS=mcln(Tf/Ti) but an unsure what to use for Ti

 

[Chestermiller]

Homework Statement:: Two quantities of water, of mass M and nM where n is constant, have temperatures T1 and T2. They are adiabatically mixed together and the pressure remains constant. What is the entropy change of the universe in this process?
Relevant Equations:: dU=δW+δQ
ΔS=ΔSh+ΔSc
ΔS=Q/T
dS=δQ/T
Q=mcΔT

the entropy change for a reversible adiabatic process is zero as it remains constant. Is this a reversible process?

No. It is spontaneous. So entropy is generated within the system.

assuming T1>T2:
hot (h) water has mass M, temp T1
cold (c) water has mass nM, temp T2

let the final temperature be Tf

if δQ=0 as the process is adiabatic, |Qh|=|Qc| so Qh=-Qc

by Q=mcΔT, Qh=Mc(T1-Tf) and Qc=nMc(T2-Tf) hence Mc(T1-Tf)=-nMc(T2-Tf)

this give Tf=(T2+(n^-1)T1)/((n^-1)+1)

I am unsure where to go from here as I have the equation ΔS=mcln(Tf/Ti) but an unsure what to use for Ti

What you do is use the equation to get the entropy change for each of the two quantities of water separately. You then add together the two entropy changes to get the total entropy change for the system.