Entropy change of calorimetric process

In summary, the conversation is discussing a problem involving a 20 kg sample of solidified mercury and the determination of its original temperature and entropy change. The solution involves using calorimetry equations and knowledge of the melting point, heat of fusion, and specific heat of mercury. The correct approach for finding the entropy change is to use the equation ΔS = mCln(Tf/Ti) + L(fusion)*m / T(of fusion), where the second term should have a minus sign in front of it to account for the decrease in entropy during freezing.
  • #1
vetgirl1990
85
3

Homework Statement


A 20 kg sample of mercury is completely solidified and liberates 231.6 kJ of energy. What is the original temperature of the mercury? (The melting point of mercury is 234K, the heat of fusion of mercury is 11.3 kJ/kg, and the specific heat of mercury is 140 J/kg•K.) . What is the mercury entropy change ΔS in this process?

Homework Equations


q=mcΔT
ΔS = ∫Q/dt
For phase change: ΔS = (latent heat)m / T(of phase change)

The Attempt at a Solution


I found the original temperature of mercury to be 236K. A pretty straightforward calorimetric problem.

But having trouble finding the entropy change.
ΔS = mCln(Tf/Ti) + Lfusion*m/Tfreezing point
= (20)(0.14)ln(234/236) + (11.3)(20)/234
= 941.98J/K

The answer is -19819J/K
 
Physics news on Phys.org
  • #2
Must be an error in the answers.
Because ΔQ is 231.6 kJ and all of this is taking place at a temperature of around 234 K, ΔS is slightly less than 1 kJ/K.
 
  • #3
vetgirl1990 said:
But having trouble finding the entropy change.
ΔS = mCln(Tf/Ti) + Lfusion*m/Tfreezing point
= (20)(0.14)ln(234/236) + (11.3)(20)/234
= 941.98J/K

The answer is -19819J/K
That 11.3 should be negative. Are you sure about that initial temperature? I'm guessing the answer is supposed to be -1981.9, and the initial temperature is supposed to be 336.
 
  • Like
Likes vetgirl1990
  • #4
Chestermiller said:
That 11.3 should be negative. Are you sure about that initial temperature? I'm guessing the answer is supposed to be -1981.9, and the initial temperature is supposed to be 336.

Yes, it was a MC exam and there was no option for 336K and -1981.9J/K.
I'm not too concerned about the answer (as my prof has said himself sometimes the answers are wrong), I'm just more concerned with knowing how to do the question. I was able to calculate 236K (one of the answers) but there was no corresponding "correct" entropy change so I'm not sure if I'm using the wrong approach or if the answer was wrong.

ΔS = mCln(Tf/Ti) + L(fusion)*m / T(of fusion)
Can you verify whether that's the right approach to finding the entropy change of this process?
 
  • #5
vetgirl1990 said:
Yes, it was a MC exam and there was no option for 336K and -1981.9J/K.
I'm not too concerned about the answer (as my prof has said himself sometimes the answers are wrong), I'm just more concerned with knowing how to do the question. I was able to calculate 236K (one of the answers) but there was no corresponding "correct" entropy change so I'm not sure if I'm using the wrong approach or if the answer was wrong.

ΔS = mCln(Tf/Ti) + L(fusion)*m / T(of fusion)
Can you verify whether that's the right approach to finding the entropy change of this process?
As I said, that should be a minus sign in front of the second term. The entropy decreases during freezing.
 
  • Like
Likes vetgirl1990

What is entropy change?

Entropy change is a measure of the disorder or randomness of a system. In a calorimetric process, it refers to the change in the entropy of a substance as it undergoes a change in temperature.

How is entropy change related to heat transfer?

Entropy change is related to heat transfer through the second law of thermodynamics, which states that the total entropy of a closed system always increases with time. Heat transfer causes changes in the entropy of a system as it redistributes energy and increases the disorder of the system.

What factors affect the entropy change of a calorimetric process?

The entropy change of a calorimetric process is affected by the change in temperature, the specific heat capacity of the substance, and the amount of heat transferred. The more energy is transferred and the greater the change in temperature, the larger the entropy change will be.

How is entropy change measured in a calorimetric process?

The entropy change of a calorimetric process can be measured by using a calorimeter, which is a device that measures the heat transfer between a system and its surroundings. The change in temperature and the specific heat capacity of the substance can then be used to calculate the entropy change.

Can entropy change be negative?

Yes, entropy change can be negative if the process involves a decrease in disorder or randomness. This is often the case in chemical reactions where the products have lower entropy than the reactants. However, the overall entropy of a closed system will always increase due to the second law of thermodynamics.

Similar threads

  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
9
Views
2K
Replies
4
Views
923
  • Introductory Physics Homework Help
Replies
16
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
776
  • Introductory Physics Homework Help
Replies
7
Views
5K
  • Introductory Physics Homework Help
Replies
8
Views
4K
Back
Top