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309 SP 11-4 HEAT OF FUSION

309 SP 11-4 HEAT OF FUSION. HOW MANY GRAMS OF ICE AT 0°C AND 101.3KJ COULD BE MELTED BY THE ADDITION OF 2.25 KJ OF HEAT ?. H FUS. HEAT OF FUSION = 6.01 KJ / 1 MOLE. =. 1 MOLE OF WATER ( H 2 0) IS 1g + 1g + 16 g = 18g. 2.25 kj 1. 1 MOLE 6.01 kj. .37 MOLES. 18 g 1 MOLE. x. =.

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309 SP 11-4 HEAT OF FUSION

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  1. 309 SP 11-4 HEAT OF FUSION • HOW MANY GRAMS OF ICE AT 0°C AND 101.3KJ COULD BE MELTED BY THE ADDITION OF 2.25 KJ OF HEAT ? HFUS HEAT OF FUSION = 6.01 KJ / 1 MOLE = 1 MOLE OF WATER ( H20) IS 1g + 1g + 16 g = 18g 2.25 kj 1 1 MOLE 6.01 kj .37 MOLES 18 g 1 MOLE x = x = 6.74 g IT TAKE 6.01 KJ ( 6010) SMALL MATCHES TO MELT 18 g

  2. 309 SP 11-5 H of VAPORIZATON • HOW MUCH HEAT IN KJ, IS ABSORBED WHEN 24.8g OF WATER AT 100ºC IS CONVERTED TO STEAM AT 100 ºC ? H2O (L) + 40.7 KJ ----> H20 (g) IT TAKE 40.7 KJ TO CONVERT 1 MOLE OF LIQUID WATER AT 100 ºC TO STEAM AT 100 ºC . 24.8 g 1 1 MOLE 18 g = 40.7 kj 1 MOLE 1.38 MOLES 56.1 KJ =

  3. SP 11-6 HEAT OF SOLUTION • HOW MUCH HEAT ( IN kJ) IS RELEASED WHEN 2.5 MOLES OF Noah IS DISSOLVED IN WATER ? HEAT OF SOLUTION IS - 445.1 kj / MOLE HSoln = - 445 .1 kj / MOLE NaOH (s) ----> Na+1 (aq) + OH - (aq) WHEN 1 MOLE IS DISSOLVED IN WATER, 445 kj IS RELEASED 2.5 MOLES 1 - 445 kj 1 MOLE - 1113 kj =

  4. SP 11-6 HEAT OF SOLUTION HOW MUCH HEAT INKJ IS RELEASED WHEN 2.5 MOLES OF NaOH IS DISSOLVED IN WATER NaOH (s) -------> Na + + OH - + 445 KJ/ MOLE HEAT CHANGE IS - 445 KJ BECAUSE THIS IS EXOTHERMIC…. HEAT IS LOST TO THE WATER 2.5 MOLES 1 445 KJ 1 MOLE -113 KJ =

  5. START OF 11.1

  6. TERMS FOR TEST 11 • 1. THERMOCHEMICAL EQUATION • 2 JOULE • 3 ENDOTHERMIC PROCESS • 4 CHEMICAL POTENTIAL ENERGY • 5 HEAT OF COMBUSTION • 6 CALORIMETRY • 7 THERMOCHEMISTRY • 8 MOLAR HEAT OF FUSION

  7. SAMPLE PROBLEMS FOR TEST 11 • 11-1 FINDING SPECIFIC HEAT • 11-2 CALCULATING HEAT • 11-3 THERMOCHEMICAL REACTIONS • 11-4 HEAT OF FUSION • 11-5 SKIP • 11-6 HEAT OF SOLUTION

  8. 312 SP11-2 MOLAR HEAT OF SOLUTION IS THE AMOUNT OF HEAT CREATED BY DISSOLVING 1 MOLE OF A SUBSTANCE IN WATER. 1000g 52º 22º HEAT OF SOLUTION= g OF WATER x SH x TEMP. CHANGE 1000g 4.18 30º = 125. 4 kj

  9. CHAPTER 11: THERMOCHEMISTRY • 11.1 THE FLOW OF ENERGY and HEAT • 11.2 MEASURING AND EXPRESSING HEAT CHANGES • 11.3 HEAT IN CHANGES OF STATE • 11.4 CALCULATING HEAT CHANGES HEAT AND CHEMICAL CHANGE

  10. 293 TERM for 11.l • 11. CALORIE • 12. JOULE • 13. HEAT CAPACITY • 14. SPECIFIC HEAT • CAPACITY • 15. SPECIFIC HEAT • 1.THERMOCHEMISTRY • 2,ENERGY • 3.CHEMICAL POTENTIAL ENERGY • 4.HEAT • 5.SYSTEM • 6. SURROUNDING • 7. UNIVERSE • 8. LAW OF CONSERVATION OF ENERGY • 9. ENDOTHERMIC • 10. EXOTHERMIC

  11. 294 EXOTHERMIC PROCESS HOT COPPER PLACED IN COLD WATER 10 -c 50 -c 50-c 100 -c HEAT aka CALORIE HEAT IN WATER = 2 HEAT IN COPPER = 2 TOTAL HEAT = 4 HEAT IN WATER = 3 HEAT IN COPPER = 1 TOTAL HEAT = 4 • EXOTHERMIC PROCESS • THE WATER (SURROUNDINGS) GET WARMER • HEAT (CALORIES) MOVE INTO THE WATER • THE SYSTEM ( COPPER) GETS COLDER

  12. 294 EXOTHERMIC PROCESS HOT COPPER PLACED IN COLD WATER 10 -c 50 -c 50-c 100 -c HEAT aka CALORIE HEAT IN WATER = 2 HEAT IN COPPER = 2 TOTAL HEAT = 4 HEAT IN WATER = 3 HEAT IN COPPER = 1 TOTAL HEAT = 4 • EXOTHERMIC PROCESS • THE WATER (SURROUNDINGS) GET WARMER • HEAT (CALORIES) MOVE INTO THE WATER • THE SYSTEM ( COPPER) GETS COLDER

  13. 299 SP 11-1 SPECIFIC HEAT THE TEMPERATURE OF A PIECE OF Cu WITH A MASS OF 95.4 g INCREASES FROM 25 -C TO 48 -C WHEN THE METAL ABSORBS 849 J OF HEAT. WHAT IS THE SPECIFIC HEAT ? FORMULA HEAT in J or C SPECIFIC HEAT = x MASS TEMP CHANGE 849 J 849j .387 j 1g x -C 95.4 g 2194 gxC 23 - C x

  14. 294 EXOTHERMIC PROCESS HOT COPPER PLACED IN COLD WATER 10 -c 50 -c 50-c 100 -c HEAT aka CALORIE HEAT IN WATER = 2 HEAT IN COPPER = 2 TOTAL HEAT = 4 HEAT IN WATER = 3 HEAT IN COPPER = 1 TOTAL HEAT = 4 • EXOTHERMIC PROCESS • THE WATER (SURROUNDINGS) GET WARMER • HEAT (CALORIES) MOVE INTO THE WATER • THE SYSTEM ( COPPER) GETS COLDER

  15. 294 ENDOTHERMIC PROCESS OPPOSITE OF EXOTHERMIC PROCESS HEAT FROM THE SURROUNDINGS IS LOST TO THE SYSTEM 100 -c 50 -c 50-c 10 -c HEAT IN WATER = 1 HEAT IN COPPER = 3 TOTAL HEAT = 4 HEAT IN WATER = 2 HEAT IN COPPER = 2 TOTAL HEAT = 4 • ENDOTHERMIC PROCESS • THE WATER (SURROUNDINGS) GET COLDER • HEAT (CALORIES) MOVE FROM THE WATER • THE SYSTEM ( COPPER) GET WARMER

  16. 1. THE STUDY OF HEAT CHANGES IN CHEMISTRY. 293 TERMS 1-5 THERMOCHEMISTY 2. THE CAPACITY TO DO WORK OR SUPPLY HEAT ENERGY 3. ENERGY STORES IN CHEMICAL SUBSTANCES CHEMICAL POTENTIAL ENERGY 4. THE ENERGY TRANSFERED FROM ONE OBJECT TO ANOTHER BECAUSE OF TEMPERATURE DIFFERENCES. HEAT 5. PART OF THE UNIVERSE WE FOCUS ON SYSTEM

  17. 1. IMMEDIATE VICINITY OF SYSTEM SURROUNDINGS 293 TERMS 6-10 2. SYSTEM + SURROUNDING UNIVERSE 3. ENERGY IS NEIHER CREATED OR DESTROYED LAW OF CONV. OF ENERGY 4. ABSORBS HEAT FROM THE SURROUNDINNGS ENDOTHERMIC 5. RELEASES HEAT TO ITS SURROUNDINGS EXOTHERMIC

  18. 1. HEAT THAT RAISES 1 g OF WATER 1 -C CALORIE 293 TERMS 11-15 2. SI UNIT OF HEAT …… ALSO CALORIE = 4.18 OF THESE JOULE 3. NEEDED TO RAISE THE TEMP. OF AN OBECT 1 DEGREE. HEAT CAPACITY SPECIFIC HEAT CAPACITY 4. RAISES 1 g OF ANY SUBSTANCES 1 DEGREE. 5. EXOTHERMIC

  19. 295 CALORIE vs JOULE 1 CALORIE or 4.18 JOULESIS THE AMOUNT OF HEAT NEEDED TO RAISE THE TEMPERATURE OF 1 g ( 1 mL) OF WATER 1 DEGREE. UNITS FOR HEAT 1 GRAM OF WATER 56 DEGREES TO 57 DEGREES 4,18 JOULES 4 SMALL MATCHES 1 CALORIE LARGE MATCH EQUALS

  20. 295: SPECIFIC HEAT IRON ALCOHOL WATER IRON IS VERY EASY TO HEAT 1 CALORIE O.58 CALORIES 0.11 CALORIES 1 GRAM HEATED 1 DEGREE HOTTER. SPECIFIC HEAT IS THE AMOUNT OF HEAT ( CAL OR J) NEEDED TO RAISE 1 g OF ANY SUBSTANCE 1 DEGREE

  21. 297UNITS FOR SPECIFIC HEAT • HEAT CAN BE EXPRESSED AS: • CALORIES • JOULES HEAT ( C or J) SPECIFIC HEAT = GRAMS x TEMP. CHANGE JOULES GRAMS x -C

  22. 296 HEAT CAPACITY • THE HEAT CAPACITY IS THE AMOUNT OF HEAT NEEDED TO INCREASE THE TEMP. OF AN OBJECT ( OF ANY SIZE ) ONE DEGREE. 34.6 g 1 g SPECIFIC HEAT USED FOR A MASS OF 1 GRAM HEAT CAPACITY USED FOR ANY MASS.

  23. CALORIES vs kCAL vs JOULES 1 CALORIE = 4.18 J 1KCAL = 1000 CALORIES

  24. 299 SP 11-1 SPECIFIC HEAT THE TEMPERATURE OF A PIECE OF Cu WITH A MASS OF 40.0 g INCREASES FROM 25 -C TO 50 -C WHEN THE METAL ABSORBS 400 J OF HEAT. WHAT IS THE SPECIFIC HEAT ? FORMULA HEAT in J or C SPECIFIC HEAT = x MASS TEMP CHANGE 400 J 400 .40 J 1g x -C 1000 gxC 40 g 25 - C x

  25. 299 PP 1 FIND S.H. HEAT (J) MASS x TEMP WHEN 435 J OF HEAT IS ADDED TO 3.4 g OF OLIVE OIL AT 21 -C, THE TEMPERATURE INCREASES TO 85 -C. WHAT IS THE SPECIFIC HEAT? SPECIFIC HEAT = 435 J 435 J 2.0 J / g x -C = = 3.4 g x 64 -C 218

  26. 299 PP 2 FIND S.H. HEAT (J) MASS x TEMP A PIECE OF STEEL WEIGHING 1.55 g ABSORBS 141 J OF HEAT WHEN ITS TEMP. INCREASES 178 -C. FIND THE S.H.IN J. ? SPECIFIC HEAT = 141 J 141 J .51 J / g x -C = = 1.55 g x 178 -C 276

  27. START 11.2

  28. 300 CALORIMETRYUSING A SODA CAN • CALORIMETRY IS THE MEASUREMENT OF HEAT CHANGE IN CHEMICAL REACTIONS • CALORIMETER IS DEVICE USED TO MEASURE HEAT CHANGE ( SODA CAN) HOW MUCH HEAT, IS NEEDED TO RAISE 100g OF WATER 10 DEGREES ? ( CAL) CALORIES = MASS x SH x TEMP. CHANGE g CAL -C = KILOCALORIES = JOULES =

  29. 301 HEAT SIGNS • EXOTHERMIC REACTION • H IS NEGATIVE • THE WATER GETS HOTTER • ENDOTHERMIC REACTION • H IS POSITIVE • THE WATER GETS COLDER

  30. 312 MOLAR HEAT OF SOLUTION IS THE AMOUNT OF HEAT CREATED BY DISSOLVING 1 MOLE OF A SUBSTANCE IN WATER. 1000g 52º 22º HEAT OF SOLUTION= g OF WATER x SH x TEMP. CHANGE 1000g 4.18 30º = 125. 4 kj

  31. LIKE SP 11-6 HEAT OF SOLUTION HOW MUCH HEAT ( IN kJ) IS RELEASED WHEN 0.5 MOLES OF NaOH IS DISSOLVED IN WATER ?

  32. LIKE SP 11-6 HEAT OF SOLUTION HOW MUCH HEAT ( IN kJ) IS RELEASED WHEN 0.5 MOLES OF NaOH IS DISSOLVED IN WATER ? HEAT OF SOLUTION IS - 445.1 kj / MOLE HSoln = - 445 .1 kj / MOLE NaOH (s) ----> Na+1 (aq) + OH - (aq) WHEN 1 MOLE IS DISSOLVED IN WATER, 445 kj IS RELEASED 0.5 MOLES 1 - 445 kj 1 MOLE - 222.5 kj =

  33. 304 ENDO VS EXO LAB NaOH NaNO3 1 MASS OF H2O ____ 2 START TEMP ____ 3 END TEMP ____ 4 CHANGE ____ 5 SH IN JOULES 4.18 6 WATER H or C 7 TYPE ENDO or EXO 8 JOULES _____ 1 MASS OF H2O 21° 2 START TEMP ____ 3 END TEMP ____ 4 CHANGE ____ 5 SH IN JOULES 4.18 6 WATER H or C 7 TYPE ENDO or EXO 8 JOULES _____

  34. 304 ENDO VS EXO LAB NaOH NaNO3 1 10 g 2 3 4 5 6 7 8 10 g 1 2 21º 3 17º 4 5 6 7 8

  35. 301 HEAT and ENTHALPY • ENTHALPY, IN THE CLASS WILL BE THE SAME AS HEAT CONTENT. HEAT = MASS SH X TEMP. CHANGE X MASS OF WATER SH OF WATER CHANGE IN TEMP

  36. 25 ml OF HCl IS MIXED WITH 25 OF NaOH IN A FOAM CUP. THE STARTING TEMP. IS 25 -C AND THE ENDING TEMP. IS 32 -C. HOW MUCH HEAT IS RELEASED IN JOULES ? 302 SP 11-2 HCl + NaOH BECAUSE THE DENSITY = 1 …. 25 ml 25 ml 50 GRAMS + HEAT = MASS x SH x TEMP. x 50 g 4.18 x HEAT = 7 1463 j OR 1.5 kj

  37. 50 mL OF HCl IS MIXED WITH 50 OF NaOH IN A FOAM CUP. THE STARTING TEMP. IS 22.5 -C AND THE ENDING TEMP. IS 26 -C. HOW MUCH HEAT IS RELEASED IN KJ ? 302 PP 11 HCl + NaOH HEAT = MASS x SH x TEMP. x 100 g 4.18 x 1463 j OR 1.5 kj HEAT = 3.5

  38. 302 SP 11-2 HCl + NaOH 25 mL of 0.025 M HCl IS ADDED TO 25 mL OF 0.25 M NaOH IS A FOAM CUP. AT THE START ALL SOLUTIONS ARE AT 25 -C. DURING THE REACTION THE HIGHEST TEMP. IS 32 . ASSUMING THE DENSITIES ARE 1.0 g ./ ml, HOW MUCH HEAT IS RELEASED IIN KJ . • LIKE TEST QUESTION 28 HEAT = MASS x SPECIFIC HEAT x TEMP. CHANGE x HEAT = 50 g 4.18 x 7 HEAT = 1463 J 1.5 KJ OR

  39. 303 THERMOCHEMICALEQUATIONS • THERMOCHEMICAL EQUATION INCLUDES THE HEAT CHANGE. • HEAT OF REACTION IS THE HEAT CHANGE FOR A REACTION EXACTLY AS IT IS WRITTEN A NEW PART TO A CHEMICAL EQUATION JOULES OR CALORIES CaO + H2O ----> Ca(OH)2 + 65.2 kj 1 MOLE OF CaO ADDED TO WATER PRODUCES 65.2 kj OF HEAT.

  40. LAB: COLDER WATER • THE FIRST REACTION WAS ENDOTHERMIC. • THE WATER GOT COLDER • HEAT OF REACTION IS POSTIVE • THE SYSTEM GAINED HEAT /SURROUNDING LOST HEAT • IN THE EQUATION • HEAT GOES ON THE LEFT BLUE SALT + 400 kj ----> COLDER WATER = + 400 kj H BECAUSE THE SYSTEM ( SALT) GAINED HEAT SYSTEM CALORIE SURROUNDING

  41. LAB: WARMER WATER • THE SECOND REACTION WAS EXOTHERMIC. • THE WATER GOT WARMER • HEAT OF REACTION IS NEGATIVE • THE SYSTEM GAINED HEAT /SURROUNDING LOST HEAT • IN THE EQUATION • HEAT GOES ON THE RIGHT BLUE SALT ----> WARMER WATER+ 400 kj = - 400 kj H BECAUSE THE SYSTEM ( SALT)LOST HEAT SYSTEM CALORIE SURROUNDING

  42. GIVEN THIS EQUATION, HOW MUCH HEAT IS PRODUCED (kJ) WHEN 320 g OF O2 REACTS ? 304 LIKE TEST QUESTION 20 C3H8 + 5 O2 --> 3 CO2 + 4 H20 + 2220 kj CHANGE g TO MOLES CHANGE g TO MOLES FIND HEAT IN kj 320 g 1 MOLE 320 g 1 MOLE = 10 MOLES = 1 10 M 2220 kj 32 g 1 4440 kj 32 g = 1 5 MOLES O 2 = 16g + 16g = 32 g O 2 = 16g + 16g = 32 g

  43. 304 SP 11-3 H of REACTION • USING THE EQUATION BELOW, CALCULATE THE KJ OF HEAT NEEDED TO DECOMPOSE 2.24 MOLES OF NaHCO3 (s) ? 2 NaHCO3 + 129 kJ ---> Na2CO3 + H2O + CO 2 2.24 MOLES 1 129 kJ 2 MOLE 144 kj =

  44. 304 SP 11-3 H of REACTION • USING THE EQUATION BELOW, CALCULATE THE KJ OF HEAT NEEDED TO DECOMPOSE 2.24 MOLES OF NaHCO3 (s) ? 2 NaHCO3 + 129 kJ ---> Na2CO3 + H2O + CO2

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