Question

How many grams of ice at -12.1 ∘C can be completely converted to liquid at 15.2 ∘C if the available heat for this process is 4.49×103 kJ ? For ice, use a specific heat of 2.01 J/(g⋅∘C) and ΔHfus=6.01kJ/mol .

Answer 1

Answer:

The mass of ice is 1,06 kg

Explanation:

Formula for calorimetry:  Q = m . C . ΔT

Formula for phase transition, with latent heat: Q = L . m

m . C . ΔT + L . m + m . C . ΔT = 4490 kJ

m . 2,01 J/g°C (0° - (-12,1°C) + 334 J/g . m + m . 4,186 J/g°C . (15,2°C - 0°C)

Our unknown value is mass

You should look for latent heat of Ice and of course, specific heat of water.

Those are constant and they are known.

As the units are in J, and the heat for the process is in Kj we have to convert the final number into J

4490kJ . 1000 = 449000 J

m. 2,01 J/g°C (0°-(-12,1°C) + 334 J/g .m +m .4,186 J/g°C .(15,2°C - 0°C) = 449000 J

m. 24,321 J/g + 334 J/g .m + 63,62 J/g . m = 449000 J

421,94 J/g . m =  449000 J

m =  449000 J /421,94 g/J = 1064,1 g ---->1,06 kg