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### Homework 8.87

Posted: Sun Jan 22, 2017 9:38 pm
The question asks: "How much heat is required to convert a 42.30g block of ice at -5.042 degrees C into water vapor at 150.35 degrees C?"

I know that I have to factor in the change in enthalpy due to the two phase changes. I converted the grams of H2O to moles and multipled that to the deltaH of fusion as well as vaporization to the kJ absorbed during the phase changes.

Then I used the equation q=nC$\Delta T$ three times (one for each phase since the specific heat changes). I then added all my energy together to get 111.014kJ. However, the answer in the back of the book is 132.0kJ. Help?

### Re: Homework 8.87

Posted: Sun Jan 22, 2017 10:32 pm
This process involves 5 steps and to solve this problem we are going to use the information given in table 8.2 (pg 269 in textbook) which shows specific heat capacities (for the steps 1,3 and 5)
And Table 8.3 (pg 284 in textbook) which shows the standard enthalpies of physical change (for steps 2 and 4)

1.heat you need the raise the temperature of the ice from -5.042 degrees celsius to 0.00 degrees celsius
ΔH= (42.30g)x(2.03J/gºC)x(0.00ºC - (-5.042ºC))= 0.433 kJ

2.then ice melts at 0.00ºC
ΔH= (42.30g/18.02 g/mol) x (6.01kJ/mol) = 14.1kJ

3.heat needed to raise the temperature of liquid water from 0.00ºC to 100.00ºC
ΔH= (42.30g)x(4.18J/gºC)x(100.00ºC - 0.00ºC)= 17.7 kJ

4.then at 100.00ºC, water vaporize
ΔH= (42.30g/18.02 g/mol) x (40.7kJ/mol) = 95.5kJ

5.Lastly, heat needed to raise the temperature of water vapour from 100.00ºC to 150.35ºC is:
ΔH= (42.30g)x(2.01J/gºC)x(150.35ºC - 100.00ºC)= 4.3 kJ

add the heat needed in all of the steps to calculate the total heat required to raise the temperature from -5.042ºC to 150.35ºC
0.4kJ + 14.1kJ + 17.7 kJ + 95.5kJ + 4.3 kJ = 132.0kJ

### Re: Homework 8.87

Posted: Mon Jan 23, 2017 6:19 am
That makes perfect sense. Thank you.