Thermodynamics question, entropy Example

In summary, in this problem we determine the final temperature of a mixture of water in a well insulated container after a mixing process. The final temperature is found to be 44.375 degrees celsius. We also calculate the change in entropy of the water as a result of the mixing process, which is 0.246 kJ/kg-K.
  • #1
abs123456
15
1

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
 
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  • #2
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.
 
  • #3
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..
 
  • #4
I think that should be correct.

For the entropy part, you just need to apply an entropy balance in the same way.
 
  • #5
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
 
  • #6
You'd need to do

[tex]\Delta S = \frac{Q_{cold}}{T_{cold}} - \frac{Q_{hot}}{T_{hot}}[/tex]
 
  • #7
how do u work out delta q?
 
  • #8
You have the Q=mc(T2-T1) for each of the hot and cold fluids.
 
  • #9
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
 

FAQ: Thermodynamics question, entropy Example

What is the definition of thermodynamics?

Thermodynamics is a branch of physics that studies the relationships and behavior of energy and heat within a system.

What is entropy in thermodynamics?

Entropy is a measure of the disorder or randomness in a system. In thermodynamics, it is a measure of the amount of energy that is unavailable for work.

Can you provide an example of entropy in action?

One example of entropy in action is the expansion of a gas. When a gas expands, it becomes more disordered and its entropy increases.

How is entropy related to the second law of thermodynamics?

The second law of thermodynamics states that the total entropy of an isolated system always increases over time. This means that as energy is transferred or transformed, the overall disorder or randomness of the system increases.

How does entropy affect the efficiency of engines?

Entropy plays a role in the efficiency of engines, as it represents the amount of energy that is unavailable for work. In a perfect engine, all of the energy would be converted into work, but in reality, some energy is lost as heat due to entropy. This limits the efficiency of engines.

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