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

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

1 Problem Start by planning how many steps are needed. 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. 77.0 °C Sice= 2.02 J/g∙°C Sliq= 4.184 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

1 Problem Next, calculate each step. How much heat is need to change 22.0 grams of water from -14.0 °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

1 Problem Finally, add the steps together. How much heat is need to change 22.0 grams of water from -14.0 °C to 77.0 °C? Finally, add the steps together. q total = q ice + q melt + q liquid q total = 631 J + 7326 J + 7088 J q total = 15045 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

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= 4.184 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

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)(34.0 C°) q liquid= -1280 J 34.0 °C temperature 0 °C 0 °C q = DHfus x moles q= (6.03 KJ/mol)(0.499 moles) q melt= -3.01 KJ added energy © Mr. D. Scott; CHS

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 = -1280 J + -3010 J q total = -4290 J = - 4.29 kJ q liquid= -1280 J temperature 0 °C 0 °C q freeze= -2710 J added energy © Mr. D. Scott; CHS

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

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

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 = 54349 J 100.°C q melt= 4329 J temperature q total = q melt + q liquid+ q boil q total = 4329 J + 5439 J + 29,380 J q total = 39148 J = 39.1 kJ 0 °C 0 °C added energy © Mr. D. Scott; CHS

4 Problem Start by planning how many steps are needed. 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? 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

4 Problem Next, calculate each step. 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? 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

4 Problem Finally, add the steps together. 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? 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 + 1045 J q total = 31847 J = 31.8 kJ 0 °C added energy © Mr. D. Scott; CHS 14