Entropy change of system two substances

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  • #1
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Homework Statement


A 3.00-kg block of silicon at 60.0°C is immersed in 6.00 kg of mercury at 20.0°C. What is the entropy increase of this system as it moves to equilibrium? The specific heat of silicon is 0.17 cal/(g·K) and the specific heat of mercury is 0.033 cal/(g·K).

Homework Equations


Q = mCΔT
ΔS = ΔSH + ΔSC
ΔS = Q/T

The Attempt at a Solution


mSi = 3.00 Kg
mHg = 6.00 Kg
TSi = 60.0 °C = 333.15 K
THg = 20.0 °C = 293.15 K
CSi = 0.17 cal/(g⋅K) = 711.62 J/(Kg⋅K)
CHg = 0.033 cal/(g⋅K) = 138.138 J/(Kg⋅K)
ΔT = TH - TC = 333.15 K - 293.15 K = 40 K

Q = mCΔT
QSi = (3.00 Kg)(711.62 J/(Kg⋅K))(40 K)
= 85394.4 J

QHg = (6.00 Kg)(138.138 J/(Kg⋅K))(40 K)
= 33153.12 J

ΔS = Q/T
(Heat leaves the silicon block, so its change in entropy is negative)
ΔSSi = -(85394.4 J / 333.15 K)
= -256.324 J/K

(Heat enters the Mercury, so its change in entropy is positive)
ΔSHg = 33153.12 J / 293.15 K
= 113.093 J/K

ΔS = ΔSH + ΔSC
ΔS = -256.324 J/K + 113.093 J/K
= -143.231

This is a multiple choice question and my answer does not match any of the five possible choices. However, the question also states that we can use our own answer if we explain why we believe it to be correct, but I do not feel confident that my answer is correct.
 

Answers and Replies

  • #2
21,375
4,797
Before you can determine the change of entropy for the system, you need to determine the final equilibrium temperature. Do you know how to do that? (Hint: The final temperature will be somewhere between 20 C and 60 C). After you have determined the final equilibrium temperature, I will help you find the change in entropy of each substance.

Chet
 
  • #3
18
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I assume it is (20°C + 60°C) / 2 = 40°C (313.15 K), being that no energy left the system in the process.
 
  • #4
21,375
4,797
I assume it is (20°C + 60°C) / 2 = 40°C (313.15 K), being that no energy left the system in the process.
Suppose the silicon block were 3000 kg instead of 3 kg. Would the final temperature still be (20°C + 60°C) / 2 = 40°C? Have you learned about the concept of internal energy U?

Chet
 
  • #5
18
0
Q = mSi * CSi * (333.15 K - HeatLost)
Q = mHg * CHg * (HeatGained - 293.15 K)

mSi * CSi * (333.15 K - HeatLost) = mHg * CHg * (HeatGained - 293.15 K)
(3.00 Kg) * (711.62 J/(Kg⋅K)) * (333.15 K - x) = (6.00 Kg) * (138.138 J/(Kg⋅K)) * (x - 293.15 K)
(2134.86 J/K) * (333.15 K - x) = (828.828 J/K) * (x - 293.15 K)
x = 321.964 K = 48.814 °C

I believe that is correct (for final temperature)?
 
  • #6
18
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ΔS = ΔSH + ΔSL = -Q/THM + Q/TLM

ΔSSi = - (85394.4 J)/(321.96 K) = -265.23 J/K
ΔSHg = (33153.12 J)/(321.96 K) = 102.97 J/K

ΔS = -265.23 J/K + 102.97 J/K = -162.26

Am I supposed to use the middle heat (321.96 K) to calculate the ΔQ for each substance (ΔTSi = 333.15 K - 321.96 K = 11.19 K and ΔTHg = 321.96 - 293.15 K = 28.81 K)?
 
  • #7
21,375
4,797
Q = mSi * CSi * (333.15 K - HeatLost)
Q = mHg * CHg * (HeatGained - 293.15 K)

mSi * CSi * (333.15 K - HeatLost) = mHg * CHg * (HeatGained - 293.15 K)
(3.00 Kg) * (711.62 J/(Kg⋅K)) * (333.15 K - x) = (6.00 Kg) * (138.138 J/(Kg⋅K)) * (x - 293.15 K)
(2134.86 J/K) * (333.15 K - x) = (828.828 J/K) * (x - 293.15 K)
x = 321.964 K = 48.814 °C

I believe that is correct (for final temperature)?
Yes. What you called HLost and HGained is really the final temperature x.

Chet
 
  • #8
21,375
4,797
ΔS = ΔSH + ΔSL = -Q/THM + Q/TLM

ΔSSi = - (85394.4 J)/(321.96 K) = -265.23 J/K
ΔSHg = (33153.12 J)/(321.96 K) = 102.97 J/K

ΔS = -265.23 J/K + 102.97 J/K = -162.26

Am I supposed to use the middle heat (321.96 K) to calculate the ΔQ for each substance (ΔTSi = 333.15 K - 321.96 K = 11.19 K and ΔTHg = 321.96 - 293.15 K = 28.81 K)?
If you use 321.96 for the final temperature, you get the same heat lost by the Silicon as the heat gained by the Hg. How many Joules is this.

To calculate the change in entropy for each of the substances, you need to dream up a reversible path for each, to bring it from its initial temperature to its final temperature. This path will be different from the actual path that each of the substances takes in the actual process. In a reversible path, each substance passes through a continuous sequence of thermodynamic equilibrium states. For this particular kind of system, you get dQrev=mCdT, where C is the heat capacity. So, for the reversible path

$$dS=mC\frac{dT}{T}$$

The integral of this equation from the initial state to the final state is

$$ΔS=mC\ln\left(\frac{T_{final}}{T_{Initial}}\right)$$

This is the entropy change for each of the substances individually.

What does this equation give you for ##ΔS_{Si}## and ##ΔS_{Hg}##?

I think it would be very helpful for you to read my recent Physics Forums Insights article on Entropy and the 2nd Law of Thermodynamics at the following link: https://www.physicsforums.com/insights/understanding-entropy-2nd-law-thermodynamics/

Chet
 
  • #9
18
0
I calculate that ΔSSi = -72.939 J/K and ΔSHg = 77.697 J/K.

Added them together, I got a net entropy change of 4.758 J/K.
 
  • #10
21,375
4,797
I calculate that ΔSSi = -72.939 J/K and ΔSHg = 77.697 J/K.

Added them together, I got a net entropy change of 4.758 J/K.
It is good that you were able to take the data and plug into the equations that I presented. But it is much more important that you understand the fundamentals, and how these equations were obtained. I'm hoping that you have done the more important part and read/digested my Insights article.

Chet
 

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