# Entropy change of calorimetric process

1. Dec 6, 2015

### vetgirl1990

1. The problem statement, all variables and given/known data
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?

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

3. 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

2. Dec 7, 2015

### PietKuip

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. Dec 7, 2015

### Staff: Mentor

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.

4. Dec 7, 2015

### vetgirl1990

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. Dec 7, 2015

### Staff: Mentor

As I said, that should be a minus sign in front of the second term. The entropy decreases during freezing.