Part A Calculate the enthalpy change, H , for the process in which 44.0 g of wat

ID: 543179 • Letter: P

Question

Part A

Calculate the enthalpy change, H, 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, Hvap = 44.0 kJ/mol at 25.0 C and s = 4.18 J/(gC) 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/(gC) and Hfus = 6.01 kJ/mol .

Express your answer numerically in grams.

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

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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