How much heat is released when 105 g of steam at 100.0°C is cooled to ice at -15.0°C? The enthalpy of vaporization of water is 40.67 kJ/mol, the enthalpy of fusion for water is 6.01 kJ/mol, the molar heat capacity of liquid water is 75.4 J/(mol • °C), and the molar heat capacity of ice is 36.4 J/(mol • °C).
|
A)347 kJ |
| B)54.8 kJ |
| C)319 kJ |
| D)273 kJ |
Ti = 100.0 oC
Tf = -15.0 oC
Hvap = 40.67KJ/mol =
40670J/mol
Lets convert mass to mol
Molar mass of H2O = 18.016 g/mol
number of mol
n= mass/molar mass
= 105.0/18.016
= 5.8282 mol
Heat released to convert gas to liquid at 100.0 oC
Q1 = n*Hvap
= 5.8282 mol *40670 J/mol
= 237030.9725 J
Cl = 75.4 J/mol.oC
Heat released to convert liquid from 100.0 oC to 0.0 oC
Q2 = n*Cl*(Ti-Tf)
= 5.8282 mol * 75.4 J/mol.oC *(100-0) oC
= 43944.2718 J
Hfus = 6.01KJ/mol =
6010J/mol
Heat released to convert liquid to solid at 0.0 oC
Q3 = n*Hfus
= 5.8282 mol *6010 J/mol
= 35027.198 J
Cs = 36.4 J/mol.oC
Heat released to convert solid from 0.0 oC to -15.0 oC
Q4 = n*Cs*(Ti-Tf)
= 5.8282 mol * 36.4 J/mol.oC *(0--15) oC
= 3182.1714 J
Total heat released = Q1 + Q2 + Q3 + Q4
= 237030.9725 J + 43944.2718 J + 35027.198 J + 3182.1714 J
= 319185 J
= 319 KJ
Answer: 319 KJ
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