How much heat is required to change 30.0 g of H2O(l) at 100 degrees
C to 30.0 g of H2O(g) at 100 degrees C?
Answer;
I got 91.5 kJ. Is that what you got? Here is how I solved the problem.
Solution:
In this type of problem the best approach is to determine each of the steps
along the heating curve required to effect the change;
Step 1) : H2O(s) at -10 degrees C to H2O(s) at 0 degrees
C
Step 2) : H2O(s) at 0 degrees C to H2O(l) at 0 degrees
C
Step3) : H2O(l) at 0 degrees C to H2O(l) at 100 degrees
C
Step 4) : H2O(l) at 100 degrees C to H2O(g) at 100 degrees
C
Step 5) : H2O(g) at 100 degrees C to H2O(g) at 110 degrees
C
Calculations;
Step 1) :
![](Ans2SolnP1.GIF)
Step 2) :
![](Ans2SolnP1.jpg)
Step3) :
![](Ans2SolnP3.GIF)
Step 4) :
![](Ans2SolnP4.GIF)
Step 5) :
![](Ans2SolnP5.GIF)
TOTAL 9.15 x 104 J (91.5 kJ)