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Heating & Cooling Curves. White Board Practice Problems. Review. q = m x C gas x D t. The heat quantity for each step is calculated separately from the rest. GAS. q = D H vap x grams. VAPORIZE. LIQUID. q = D H fus x grams. q = m x C liquid x D t. temperature. MELT. - PowerPoint PPT Presentation
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White Board Practice Problems
tem
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ture
added energy
Review
q = Hfus x grams
q = Hvap x grams
q = m x Cgas x t
q = m x Cliquid x t
q = m x Csolid x t
SOLID
MELT
LIQUID VAPORIZE
GAS
The heat quantity for each step is calculated separately from the rest.
The total energy amount is found by adding the steps together.
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How much heat is need to change 22.0 grams of water from -14.0 °C to 77.0 °C?
Start by planning how many steps are needed.
-14.0°C
0 °C 0 °C
77.0 °C
q solid = ?
q melt= ? q liquid= ?
Problem
1
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Next, calculate each step.
-14.0°C
0 °C 0 °C
77.0 °C
q = Hfus x gramsq= (333 J/g)(22.0 g)q melt= 7326 J
q = m x Csolid x tq = (22.0 g)(2.05 J/g∙°C)(14.0 C°)
q ice= 631 J
q = m x Csolid x tq = (22.0 g)(4.184 J/g∙°C)(77.0 C°)q liquid= 7088 J
How much heat is need to change 22.0 grams of water from -14.0 °C to 77.0 °C?
Problem
1
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q melt= 7326 J
q ice= 631 J
Finally, add the steps together.
-14.0°C
0 °C 0 °C
77.0 °Cq liquid= 7088 J
q total = q ice + q melt + q liquid
q total = 631 J + 7326 J + 7088 J
q total = 15045 J = 15.0 kJ
How much heat is need to change 22.0 grams of water from -14.0 °C to 77.0 °C?
Problem
1
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added energy
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.
0 °C 0 °C
34.0 °C
q freeze= ? q liquid= ?
Problem
2
tem
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added energy
Next, calculate each step.
0 °C
0 °C
34.0 °C
q = Hfus x grams
q= (333 J/g)(9.00 g)
q melt= -2997 J
q = m x Cliquid x t
q = (9.00 g)(4.184 J/g∙°C)(34.0 C°)
q liquid= -1280 J
How much heat must be removed from 9.00 grams of 34.0 °C water to freeze it?
Problem
2
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q freeze= -2997 J
Finally, add the steps together.
q liquid= -1280 J
q total = q liquid + q freeze
q total = -1280 J + -2997 J
q total = -4277 J = - 4.28 kJ
How much heat must be removed from 9.00 grams of 34.0 °C water to freeze it?
0 °C
0 °C
Problem
2
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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.
0 °C
0 °C
100. °C
100. °C
q melt = ?
q liquid= ? q boil= ?
Problem
3
tem
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added energy
Next, calculate each step.
0 °C
100. °Cq = Hfus x grams
q= (333 J/g)(13.0 g)q melt= 4329 J
q = m x Cliquid x tq = (13.0 g)(4.184 J/g∙°C)(100.0 C°)
q ice= 5439 J
q = Hvap x gramsq = (2260 J/g)(13.0 g)q liquid= 29,380 J
0 °C
100. °C
Starting with a 13.0 g ice cube at 0.00 °C, how much heat is needed to completely boil it?
Problem
3
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added energy
q melt= 4329 J
q liquid = 54349 J
Finally, add the steps together.
100.°C
0 °C 0 °C
100. °C
q boil = 29,380 J
q total = q melt + q liquid+ q boil
q total = 4329 J + 5439 J + 29,380 J
q total = 39148 J = 39.1 kJ
Starting with a 13.0 g ice cube at 0.00 °C, how much heat is needed to completely boil it?
Problem
3
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Starting with a 11.5 g liquid water at 0.00 °C, how much heat is needed to raise it’s temperature to 145°C?
Start by planning how many steps are needed.
145 °C
0 °C
100. °C
100. °C
q steam = ?
q liquid= ? q boil= ?
Problem
4
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Next, calculate each step.
0 °C
100. °C
q = m x Cgas x tq = (11.5 g)(2.02 J/g∙°C)(45.0 C°)
q gas= 1045 J
q = m x Cliquid x tq = (11.5 g)(4.184 J/g∙°C)(100.0 C°)
q liq= 4812 J
q = Hvap x gramsq = (2260 J/g)(11.5 g)q boil= 25,990 J
145 °C
100. °C
Problem
4Starting with a 11.5 g liquid water at 0.00 °C, how much heat is needed to raise it’s temperature to 145°C?
tem
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added energy
q gas= 1045 J
q liquid = 4812 J
Finally, add the steps together.
100.°C
145 °C
0 °C
100. °C
q boil = 25,990 J
q total = q liquid + q boil+ q gas
q total = 4812 J + 25,990 J + 1045 J
q total = 31847 J = 31.8 kJ
Problem
4Starting with a 11.5 g liquid water at 0.00 °C, how much heat is needed to raise it’s temperature to 145°C?