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

[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.

Homework Statement

The Attempt at a Solution

Part a)

Key
m = mass
c_s = specific heat bronze
c_m = specific heat molten bronze
T_fus = melting temperature bronze
T₁ = initial temperature (of the bronze coin)
T₂ = final temperature (of the bronze coin)
H_fus = latent heat fusion bronze

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

ΔS = m [c_sln(T_fus /T₁) + (H_fus /T_fus) + c_mln(T₂ /T_fus)]

plugging the numbers in:

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



Part b)

Key
m_w = mass of water (the beer)
m_m = mass of molten bronze coin
c_w = specific heat water
c_m = specific heat molten bronze
Q_w = heat transfer water
Q_m = heat transfer bronze coin

For this question I used the equation Q = mcΔT

Q_w = m_w x c_w x ΔT
Q_w = 0.5 x 4200 x (T - 288)

Q_m = m_m x c_m x ΔT
Q_m = 0.007 x 325 x (1473 - T)

At equilibrium Q_w = Q_m

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.

 

[Chestermiller]

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

 

[Flucky]

Hi Chet thanks for pointing that out, missed it completely