# Thermal physics - entropy change (latent heat, specific heat etc)

1. Oct 30, 2013

### Flucky

Hi all, could somebody have a look over my answers for this question please? The value I got for the second part seems quite feeble.

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

3. The attempt at a solution

Part a)

Key
m = mass
cs = specific heat bronze
cm = specific heat molten bronze
Tfus = melting temperature bronze
T1 = initial temperature (of the bronze coin)
T2 = final temperature (of the bronze coin)
Hfus = latent heat fusion bronze

Ok so to find the entropy change I used the following equation:

ΔS = m [csln(Tfus /T1) + (Hfus /Tfus) + cmln(T2 /Tfus)]

plugging the numbers in:

ΔS = 0.007 [377ln(1223/300) + (168000/1223) + 325ln(1473/1223)]
∴ ΔS = 5.09 JK-1

Part b)

Key
mw = mass of water (the beer)
mm = mass of molten bronze coin
cw = specific heat water
cm = specific heat molten bronze
Qw = heat transfer water
Qm = heat transfer bronze coin

For this question I used the equation Q = mcΔT

Qw = mw x cw x ΔT
Qw = 0.5 x 4200 x (T - 288)

Qm = mm x cm x ΔT
Qm = 0.007 x 325 x (1473 - T)

At equilibrium Qw = Qm

So 0.5 x 4200 x (T - 288) = 0.007 x 325 x (1473 - T)

∴ T = 289.4 K which is about 16 °c, a mere 1 °c rise.

2. Oct 30, 2013

### Staff: Mentor

It looks like part A was done correctly. In part B, why didn't you take into account the latent heat to freeze the bronze, and the sensible heat to cool the solid bronze which forms? Also, in part b, you have to also calculate the change in entropy for the beer between its initial temperature and its final temperature.

Chet

3. Nov 1, 2013

### Flucky

Hi Chet thanks for pointing that out, missed it completely