1. C ice = 2.108 J/g°CC water = 4.186 J/g°C C steam = 1.996 J/g°C ∆H fus = 6.020 kJ/mol, ∆H vap = 40.70 kJ/mol What is the change in enthalpy when 100.0 g of steam at 160.0°C is condensed into a liquid, then frozen into a solid, and left at 0.000  C? (Hint: Draw out the graph first)

2. Hints Remember that this is cooling so your  T and  H  are negative! (we are going down the graph) When you add all the values together make sure everything is in the same units!

3. Answer -The change in energy for the steam changing to ice is -689,800J or -689.8kJ.

4. If later the 100.0g of ice is left out in the room and heats up to 23  C how much energy did that take? What was the net change of energy from cooling the steam to ice, followed by melting and heating the ice? C ice = 2.108 J/g°CC water = 4.186 J/g°C C steam = 1.996 J/g°C ∆H fus = 6.020 kJ/mol, ∆H vap = 40.70 kJ/mol

5. Answer Part 2 For the ice to heat up and melt it takes 43,040J or 43.04kJ of energy! To find the net energy change you just add the 2 answers together 43,040J + -689,800J =-646,800J Or 43.04kJ + -689.8kJ = -646.8kJ