molar heat of fusion
A mass of 45.0 g of molten (liquid) silver at its melting point is poured into 750 mL of calorimeter water at 25.0°C. The final temperature reached by both is 30.0°C. If the specific heat capacity of silver is 0.235 kJ/kg°C, calculate the molar heat of solidification of silver.
Assuming no heat losses to the environment:
the increase in the water energy = the decrease in the silver energy
Water (1g = 1cm3)
E = mcΔT
E = 0.75 x 4.18 x 5 = 15.675 kJ
Silver
The energy uptake by the water is the sum of the energy released by the silver solidification process + the energy of the silver cooling.
Enthalpy of solidification = mass/RMM x molar enthalpy of solidification
E = 45/107.86 x molar enthalpy of solidification + (0.045 x 0.235 x ΔT)
E = [0.4172 x ΔH] + [0.045 x 0.235 x ΔT]
To proceed further you must know the melting point of silver in order to show ΔT otherwise there are two unknowns.
From the data book Ag melting point = 1234K
Therefore ΔT = 1234 – 303 = 931 degrees
15.675 = 0.4172 x ΔH + [0.045 x 0.235 x 932]
(15.675 – 9.856)/ 0.4172 = ΔH
ΔH = 13.95 kJmol-1
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