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.