1. White Board Practice Problems © Mr. D. Scott; CHS

2. temperature added energy Review q =  H fus x grams q =  H vap x grams q = m x C gas x  t q = m x C liquid x  t q = m x C solid 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. © Mr. D. Scott; CHS

3. temperature added energy 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 77.0 °C q solid = ? q melt = ?q liquid = ? © Mr. D. Scott; CHS Problem 1

4. temperature added energy Next, calculate each step. -14.0°C 0 °C 77.0 °C q =  H fus x grams q= (333 J/g )(22.0 g ) q melt = 7326 J q = m x C solid x  t q = (22.0 g )(2.05 J/g∙°C )(14.0 C°) q ice = 631 J q = m x C solid x  t q = (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? © Mr. D. Scott; CHS Problem 1

5. temperature added energy q melt = 7326 J q ice = 631 J Finally, add the steps together. -14.0°C 0 °C 77.0 °C q 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? © Mr. D. Scott; CHS Problem 1

6. temperature 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 34.0 °C q freeze = ? q liquid = ? © Mr. D. Scott; CHS Problem 2

7. temperature added energy Next, calculate each step. 0 °C 34.0 °C q =  H fus x grams q= (333 J/g )(9.00 g ) q melt = -2997 J q = m x C liquid 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? © Mr. D. Scott; CHS Problem 2

8. temperature added energy 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 © Mr. D. Scott; CHS Problem 2

9. temperature added energy 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 100. °C q melt = ? q liquid = ? q boil = ? © Mr. D. Scott; CHS Problem 3

10. temperature added energy Next, calculate each step. 0 °C 100. °C q =  H fus x grams q= (333 J/g )(13.0 g ) q melt = 4329 J q = m x C liquid x  t q = (13.0 g )(4.184 J/g∙°C )(100.0 C°) q ice = 5439 J q =  H vap x grams q = (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? © Mr. D. Scott; CHS Problem 3

11. temperature added energy q melt = 4329 J q liquid = 54349 J Finally, add the steps together. 100.°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? © Mr. D. Scott; CHS Problem 3

12. temperature added energy 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 q steam = ? q liquid = ? q boil = ? © Mr. D. Scott; CHS Problem 4

13. temperature added energy Next, calculate each step. 0 °C 100. °C q = m x C gas x  t q = (11.5 g )(2.02 J/g∙°C )(45.0 C°) q gas = 1045 J q = m x C liquid x  t q = (11.5 g )(4.184 J/g∙°C )(100.0 C°) q liq = 4812 J q =  H vap x grams q = (2260 J/g)(11.5 g) q boil = 25,990 J 145 °C 100. °C © Mr. D. Scott; CHS Problem 4 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?

14. temperature 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 © Mr. D. Scott; CHS Problem 4 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?