How Do You Calculate Entropy Change in Thermodynamics Problems?

In summary: Thank you for your input! I'll definitely check that out. :)In summary, the problem the student is having is trying to find the time that the gold is exposed to the air, and then calculating the change in entropy. They need to know the heat flow into the surroundings and the change in entropy of the gold.
  • #1
physninj
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Homework Statement


http://postimage.org/image/9artvc1h5/ see attached


Homework Equations


ΔS=Q/T


The Attempt at a Solution



The real issue I'm having is how to go about solving this, I think I'm going to need to use kinematics to find the time the gold is exposed to the air correct? but then I'm not sure how to use that time value or anything. My book has no problems like this. Any and all guidance will be greatly appreciated.
 

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  • #2
physninj said:

Homework Statement


http://postimage.org/image/9artvc1h5/ see attached

Homework Equations


ΔS=Q/T

The Attempt at a Solution



The real issue I'm having is how to go about solving this, I think I'm going to need to use kinematics to find the time the gold is exposed to the air correct? but then I'm not sure how to use that time value or anything. My book has no problems like this. Any and all guidance will be greatly appreciated.
You do not have to worry about the time the gold is exposed to the air. You just need to know the total heat flow into the surroundings.

When everything reaches equilibrium, has the temperature of the air or ice changed? So if you know the amount of heat flow into the surroundings, what is its change in entropy?

What is the heat flow into the surroundings generated by the cooling of the gold? (hint: you need an expression for ΔQ as a function of ΔT).

What is the change in entropy of the gold? (hint: ΔS = ∫dQrev/T).

What is the heat flow into the surroundings generated by the kinetic energy of the gold? Assume that energy is absorbed by the ice and air.

What is the total amount of heat flow into surroundings? So what is the change in entropy (hint: that part is easy).

Add the changes in entropy together.

AM
 
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  • #3
So I calculated the Q for the gold to reach -20 degrees using the specific heat of gold (wasn't provided for practice test??), I'm not sure if that's the right move. I assume because the surroundings are at -20 that there is no real change in the temp of the air and ice overall, otherwise I wouldn't know what to do.

Q=m(spH)ΔT

Q=(1500 g)(.126 J/gK)(-20-1500)= -287280 J

For the entropy change of the gold

ΔS=∫dQ/T=c∫dT/T=(m(spH))ln(T2/T1)=(189)ln(253/1773)= -367.99 J/K

Now, when you say kinetic energy of the gold, do you mean from its speed falling? If so I used the change in gravitational potential to calculate that part. I don't know if I can just say its Q2 but yeah.

ΔUg=mghf-mghi=0-(1.5 kg)(9.8 m/s^2)(150 m)=-2205 J=Q2??

total heat flow into surroundings 289485 J

change in entropy of surroundings ΔS=Q/T=(289485 J)/(253 K)= 1144.21 J/K

adding changes in entropy gives 776.219 J/K, is this change in entropy of the universe then?

Not very confident in all of this so please let me know wherre I'm wrong. Thank you :)
 
  • #4
Does my work look alright? I don't want to let the thread die quite yet. If it goes down again I suppose I can take it to mean I'm all good.
 
  • #5
physninj said:
Does my work look alright? I don't want to let the thread die quite yet. If it goes down again I suppose I can take it to mean I'm all good.
Everything looks good and is well explained. Well done.

AM
 
  • #6
Thank you very much for your help.

For reference I believe I forgot a piece, the latent heat of fusion for the gold which solidifies at 1064 °C

Q_fusion=-m*Lh=-(1.5 kg)(63000 j/kg)= -94500 J
ΔS=Q/T=-94500/(1064+273)= -70.68 J/K

Making total change in entropy of the gold -438.67 J/K

making heat flow into surroundings 383985 J and entropy change 1517.727 J/K

change in entropy of universe becomes 1079.057 J/K
 
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  • #7
physninj said:
Thank you very much for your help.

For reference I believe I forgot a piece, the latent heat of fusion for the gold which solidifies at 1064 °C

Q_fusion=-m*Lh=-(1.5 kg)(63000 j/kg)= -94500 J
ΔS=Q/T=-94500/(1064+273)= -70.68 J/K

Making total change in entropy of the gold -438.67 J/K

making heat flow into surroundings 383985 J and entropy change 1517.727 J/K

change in entropy of universe becomes 1079.057 J/K
You are thinking! Since the gold is a liquid at 1500C you might want to check the heat capacity for liquid gold as it may be higher than for solid gold - not that it will make a huge difference here.

AM
 

FAQ: How Do You Calculate Entropy Change in Thermodynamics Problems?

Question 1: What is the definition of entropy in thermodynamics?

The definition of entropy in thermodynamics is a measure of the disorder or randomness in a system.

Question 2: How is entropy related to the Second Law of Thermodynamics?

The Second Law of Thermodynamics states that the total entropy of a closed system will always increase over time. This means that the disorder or randomness in a system will continuously increase, leading to a decrease in the available energy for work.

Question 3: Why is the increase in entropy inevitable?

The increase in entropy is inevitable because it is a natural tendency for systems to move towards a state of greater disorder. This is due to the fact that there are a larger number of possible disordered states compared to ordered states, making it more likely for a system to move towards a higher entropy state.

Question 4: Can entropy be decreased in a system?

Yes, entropy can be decreased in a system, but it can only occur with an input of energy. This energy input is required to reverse the disorder and create a more ordered state, resulting in a decrease in entropy.

Question 5: How is entropy calculated in a system?

Entropy is calculated using the formula S = k ln W, where S is the entropy, k is the Boltzmann constant, and W is the number of microstates that correspond to a given macrostate. The higher the number of microstates, the higher the entropy of the system.

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