1. Heating & Cooling Curves
White Board Practice Problems
© Mr. D. Scott; CHS

2. GAS VAPORIZE LIQUID MELT SOLID
q = m x sgas x Dt
The heat quantity for each step is calculated separately from the rest.
GAS
q = DHvap x grams
VAPORIZE
LIQUID
q = DHfus x grams
q = m x sliquid x Dt
temperature
MELT
The total energy amount is found by adding the steps together.
q = m x ssolid x Dt
SOLID
added energy
© Mr. D. Scott; CHS

3. 1 Problem Start by planning how many steps are needed.
How much heat is need to change 22.0 grams of water from °C to 77.0 °C?
Start by planning how many steps are needed.
77.0 °C
Sice= 2.02 J/g∙°C
Sliq= J/g∙°C
Sgas= 2.02 J/g∙°C
DHfus =6.03 KJ/mol
DHvap = 2260 J/g
q melt= ?
q liquid= ?
temperature
0 °C
0 °C
q solid = ?
-14.0°C
added energy
© Mr. D. Scott; CHS

4. 1 Problem Next, calculate each step.
How much heat is need to change 22.0 grams of water from °C to 77.0 °C?
Next, calculate each step.
q = DHfus x moles
q= (6.03 KJ/mol)(1.22 mol)
q melt= 7.36 KJ
77.0 °C
q = m x ssolid x Dt
q = (22.0 g)(4.184 J/g∙°C)(77.0 C°)
q liquid= 7090 J
temperature
0 °C
0 °C
q = m x ssolid x Dt
q = (22.0 g)(2.02 J/g∙°C)(14.0 C°)
q ice= 622 J
-14.0°C
added energy
© Mr. D. Scott; CHS

5. 1 Problem Finally, add the steps together.
How much heat is need to change 22.0 grams of water from °C to 77.0 °C?
Finally, add the steps together.
q total = q ice + q melt + q liquid
q total = 631 J J J
q total = J = 15.0 kJ
77.0 °C
q melt= 7326 J
q liquid= 7088 J
temperature
0 °C
0 °C
q ice= 631 J
-14.0°C
added energy
© Mr. D. Scott; CHS

6. 2 Problem Start by planning how many steps are needed.
How much heat must be removed from 9.00 grams of 34.0 °C water to freeze it?
Start by planning how many steps are needed.
Sice= 2.02 J/g∙°C
Sliq= J/g∙°C
Sgas= 2.02 J/g∙°C
DHfus =6.03 KJ/mol
DHvap = 2260 J/g
34.0 °C
q freeze= ?
temperature
q liquid= ?
0 °C
0 °C
added energy
© Mr. D. Scott; CHS

7. 2 Problem Next, calculate each step.
How much heat must be removed from 9.00 grams of 34.0 °C water to freeze it?
Next, calculate each step.
q = m x sliquid x Dt
q = (9.00 g)(4.184 J/g∙°C)( C°)
q liquid= J
34.0 °C
temperature
0 °C
0 °C
q = DHfus x moles
q= (6.03 KJ/mol)(0.499 moles)
q melt= KJ
added energy
© Mr. D. Scott; CHS

8. 2 Problem Finally, add the steps together.
How much heat must be removed from 9.00 grams of 34.0 °C water to freeze it?
Finally, add the steps together.
q total = q liquid + q freeze
q total = J J
q total = J = kJ
q liquid= J
temperature
0 °C
0 °C
q freeze= J
added energy
© Mr. D. Scott; CHS

9. 3 Problem Start by planning how many steps are needed.
Starting with a 13.0 g ice cube at 0.00 °C, how much heat is needed to completely boil it?
Start by planning how many steps are needed.
100. °C
q liquid= ?
100. °C
q boil= ?
temperature
0 °C
0 °C
q melt = ?
added energy
© Mr. D. Scott; CHS

10. 3 Problem Next, calculate each step.
Starting with a 13.0 g ice cube at 0.00 °C, how much heat is needed to completely boil it?
Next, calculate each step.
100. °C
q = DHfus x grams
q= (333 J/g)(13.0 g)
q melt= 4329 J
100. °C
q = DHvap x grams
q = (2260 J/g)(13.0 g)
q liquid= 29,380 J
temperature
0 °C
0 °C
q = m x sliquid x Dt
q = (13.0 g)(4.184 J/g∙°C)(100.0 C°)
q ice= 5439 J
added energy
© Mr. D. Scott; CHS

11. 3 Problem Finally, add the steps together
Starting with a 13.0 g ice cube at 0.00 °C, how much heat is needed to completely boil it?
Finally, add the steps together
q boil = 29,380 J
100. °C
q liquid = J
100.°C
q melt= 4329 J
temperature
q total = q melt + q liquid+ q boil
q total = 4329 J J + 29,380 J
q total = J = kJ
0 °C
0 °C
added energy
© Mr. D. Scott; CHS

12. 4 Problem Start by planning how many steps are needed.
Starting with a 11.5 g liquid water at °C, how much heat is needed to raise it’s temperature to 145°C?
145 °C
Start by planning how many steps are needed.
q steam = ?
100. °C
q liquid= ?
100. °C
q boil= ?
temperature
0 °C
added energy
© Mr. D. Scott; CHS 12

13. 4 Problem Next, calculate each step.
Starting with a 11.5 g liquid water at °C, how much heat is needed to raise it’s temperature to 145°C?
145 °C
Next, calculate each step.
100. °C
q = m x sliquid x Dt
q = (11.5 g)(4.184 J/g∙°C)(100.0 C°)
q liq= 4812 J
100. °C
q = m x sgas x Dt
q = (11.5 g)(2.02 J/g∙°C)(45.0 C°)
q gas= 1045 J
temperature
0 °C
q = DHvap x grams
q = (2260 J/g)(11.5 g)
q boil= 25,990 J
added energy
© Mr. D. Scott; CHS 13

14. 4 Problem Finally, add the steps together.
Starting with a 11.5 g liquid water at °C, how much heat is needed to raise it’s temperature to 145°C?
145 °C
q gas= 1045 J
Finally, add the steps together.
q boil = 25,990 J
100. °C
q liquid = 4812 J
100.°C
temperature
q total = q liquid + q boil+ q gas
q total = 4812 J + 25,990 J J
q total = J = kJ
0 °C
added energy
© Mr. D. Scott; CHS 14