Thermodynamics question, entropy Example

[abs123456]

Homework Statement

A well insulated container of negligible thermal capacity contains 30 kg of water at
90oc. A mixing process then takes place in which 50 kg of liquid water at 17oc is
added to the tank. The mixing process is continued until thermal equilibrium
is established.

Determine:
a) The final temperature of the mixture.
b) The change in entropy of the water as a result of the mixing process
(For liquid water, take Cp:4.18 kJ/kgK

Homework Equations

h = cpT

The Attempt at a Solution

Could someone please give some tips about how i should go about tackilng this question.
Thank you

 

[rock.freak667]

Applying a heat balance on the entire system would give the final temperature I believe.

The initial water will lose heat while the second cooler water will gain heat.

 

[berko1]

I believe this is the way to the answer:

Q=M*Cp*(t2-t1)
Q=0 - it is an isolated container, this means no heat is added or lost to the surroundings.
If we take into consideration both liquids:

0=30*Cp*(t2-90)+50*Cp*(t2-17)
Cp drops.
t2 is the final temp and it is equal for both of them.
==>t2=44.375 degrees c.

I hope this is right and helpful its been a while since i tried thermo questions..

 

[rock.freak667]

I think that should be correct.

For the entropy part, you just need to apply an entropy balance in the same way.

 

[abs123456]

Thanks for the replys, however when i am doing the entropy balance i seem to be getting it wrong...

Do i use this equation :

S= M x Cp x ΔT... for both waters and then take their differnece?
I get a different answer though, it should be 2.02kj/k

 

[rock.freak667]

You'd need to do

\Delta S = \frac{Q_{cold}}{T_{cold}} - \frac{Q_{hot}}{T_{hot}}

 

[abs123456]

how do u work out delta q?

 

[rock.freak667]

You have the Q=mc(T2-T1) for each of the hot and cold fluids.

 

[jadericdawson]

The energy of the system does not change. Therefore the energies of each "water" is the same as collective mass and final temperature energy.

m1*Cp*T1 + m2*Cp*T2 = (m1+m2)*Cp*Tf

where

Tf = (m1*T1 + m2*T2)/(m1+m2)

now the collective entropy change of each portion of water is
Δs = Cp*ln(Tf/T1) + Cp*ln(Tf/T2) = Cp*ln(Tf^2/(T1*T2))

Δs = 0.246 kJ/kg-K
Tf = 413K