Heating / Cooling Curve Calculations EQ: Why is an ideal heating curve not a straight line graph?

Ideal Heating/Cooling Curve 1) Heating a solid 2) Change from a solid to liquid 3) Heating a liquid 4) Change from a liquid to a gas 5) Heating a gas gas solid & liquid liquid liquid & gas solid present

On the back of you notes write:

1) Diagonal lines: KE changes Flat lines: Potential Energy changes ∆Hvap > ∆Hfus because it requires more energy

y= mx + b q= mc∆T The slope of the diagonal line is determined by the specific heat

Example How much heat is needed to convert 50.0 grams of water from a temperature of 70.0 degrees to 110.0 degree steam? C ice = 2.03 J/gºC C water = 4.184 J/gºC C steam = 2.01 J/gºC Hfus = 6.01 kJ/mol Hvap = 40.7 kJ/mol

1st: sketch part of the graph you need 110.0- 2 100.0- ∆T= 10.0C 1 3 70.0- ∆T= 30.0C Heating: Draw graph going up 0.0-

2nd: Perform calculation for each line segment 1) Diagonal (q = m c DT) q = 50.0 g x 4.184 J/goC x (30.0oC ) 1 KJ = 6.28 kJ q = 6280 J x 1000 J

1st: sketch part of the graph you need 110.0- 2 100.0- ∆T= 10.0C 1 3 70.0- ∆T= 30.0C 0.0-

2) Flat line (Hvap = 40.7 kJ/mol) 1 mol H2O 40.7 kJ = 114 kJ 50.0 g H2O x x 18.02 g H2O 1 mol H2O

1st: sketch part of the graph you need 110.0- 2 100.0- ∆T= 10.0C 1 3 70.0- ∆T= 30.0C 0.0-

3) Diagonal (q= mc∆T) q = (50.0 g)(2.01 J/goC)(10.0oC) q = 1010 J 1 KJ = 1.01 kJ x 1000 J

3rd : Add them up + + 1.01 kJ = 121 kJ 6.28 kJ 114 kJ

Summary Use H (Flat lines) Use q = m x c x DT (Diagonal lines)

When dealing with a cooling curve… draw the curve backwards ∆H solidification & condensation are negative

Notebook Problems Calculate the amount of heat needed to convert 10.0 grams of ice from a temperature of -23.0oC to water at 27.0oC. Calculate the amount of heat released when 50.0 grams of steam at a temperature of 123.0oC cools into water at 77.0oC.

Answers 1) 4.94 kJ 2) 120. kJ