C ice = J/g°CC water = J/g°C C steam = J/g°C ∆H fus = kJ/mol, ∆H vap = kJ/mol What is the change in enthalpy when g.

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Presentation transcript:

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

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!

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

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 = J/g°CC water = J/g°C C steam = J/g°C ∆H fus = kJ/mol, ∆H vap = kJ/mol

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 ,800J =-646,800J Or 43.04kJ kJ = kJ