- #1
Flucky
- 95
- 1
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.
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.
Homework Statement
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.