Part A
Calculate the enthalpy change, ΔH, for the process in which 44.0 g of water is converted from liquid at 7.6 ∘C to vapor at 25.0 ∘C .
For water, ΔHvap = 44.0 kJ/mol at 25.0 ∘C and s = 4.18 J/(g⋅∘C) for H2O(l)
Express your answer numerically in kilojoules.
Part B
How many grams of ice at -11.0 ∘C can be completely converted to liquid at 9.4 ∘C if the available heat for this process is 5.66×103 kJ ?
For ice, use a specific heat of 2.01 J/(g⋅∘C) and ΔHfus = 6.01 kJ/mol .
Express your answer numerically in grams.
A)
the steps:
Q1 = liquid at 7.6°C to 25°C
Q2 = vaporization of liquid at 25°C
then
Q1 = m*C*(Tf-Ti) = 44*4.18*(25-7.6) = 3200.208 J
Q2 = m*LH = (44/18)(44000) = 107555.55 J
QTotal = 3200.208 + 107555.55 = 110755.758 J
Qtotal = 110.756 kJ
B)
mass of ice for:
Q1= ice to 0°C
Q2 = fusion of ice to water
Q3 = water at 0 to 9.4
Assume a basis of 1 gram f water
Q1 = 1*2.01*(0--11) = 22.11 J
Q2 = 1*6.01*1000/18 = 333.88J
Q3 = 1*4.18*(9.4-0) = 39.292 J
Qtotal = 22.11 +333.88 + 39.292
Qtotal = 395.282 J per gram...
Total E = 5.66*10^3 = 5660 J
Total mass = 5660 /39.292 = 144.0 g
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