Entropy change of calorimetric process

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Homework Help Overview

The problem involves a calorimetric process where a 20 kg sample of mercury solidifies, releasing energy and requiring the calculation of the original temperature and the change in entropy. The melting point, heat of fusion, and specific heat of mercury are provided as context for the calculations.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of the original temperature of the mercury and the entropy change associated with the phase transition. There are attempts to apply the relevant equations for entropy change, with some participants questioning the correctness of their initial temperature and the sign of the heat of fusion.

Discussion Status

Multiple interpretations of the calculations are being explored, with some participants expressing uncertainty about the initial temperature and the resulting entropy change. There is a focus on verifying the approach to calculating entropy change, with concerns raised about the accuracy of the provided answers.

Contextual Notes

Participants note that the problem was part of a multiple-choice exam, and there were discrepancies between calculated values and provided answer options. There is an acknowledgment of potential errors in the answers given by the exam.

vetgirl1990
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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
 
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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.
 
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.
 
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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?
 
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.
 
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