Hot Packs and Cold Packs

1. Hot Packs and Cold Packs • Common Medical “Over the Counter” products • “Universals” use mechanical heat storage • Put in freezer or microwave, then to injury • Temporary use, but can be recycled • “Instant” packs involve chemical reactions • Exothermic and Endothermic chemistry • Usually single-use, but no pre-heating or cooling required • We will explore these chemical types in today’s experiment

2. Thermal Semantics • Temperature • A quantitative measure of “hot and cold” • Arbitrary scales; Fahrenheit, Centigrade, Kelvin • An indicator of kinetic energy content • Does not depend on amount of material • Ocean & tea cup can have same temperature • Heat • A quantitative measure of energy transfer • Measured in Joules or Calories • Energy flows from hot to cold spontaneously • Transfer by conduction or radiation • Depends on amount of material involved • More material involves more heat transfer • Ocean has more heat than a tea cup of water

3. Temperatures around the Worldleft picture refers to January temperaturesHeat energy is delivered by solar radiationTemperature difference is a result of unequal heat delivery

4. Global Temperature HistoryShort term “warming”, but long term trend is “cooling”

5. Global WarmingCurrent trendbegan >10k years ago, are we now “between glacial periods”?

6. Changes in Energy Heat Energy change requires definitions Viewpoint is perspective of system changing Negative Energy considered as LOSS Heat flowing OUT OF a fireplace or oven Oven loses heat when oven door opens Positive Energy viewed as net GAIN Heat flows INTO an ice cube to melt it Kitchen warms due to open oven door Energy change is algebraic difference Definition: E after – Ebefore= ΔE change Depends only on the initial and ending conditions NOT dependent on the path taken Ice sample could be melted 10 times and frozen 9 Same result as melted once 6

7. James Joule experimentdemonstrated equivalence of potential energy and heat

8. Energy (Enthalpy) of Thermal Change “Enthalpy” or ∆H indicates Thermal energy Thermal Changes due to Chemical Reactions Exothermic reaction = heat generated Enthalpy sign is NEGATIVE (heat flowing away from system) Thermite reaction, neutralization burning gasoline, fireplace, hot tub body heat from food, rubbing hands to keep warm Endothermic reaction = heat absorbed Enthalpy sign is POSITIVE (heat flows into the system) Melting Ice, frozen foods Evaporation of water & other liquids absorb heat Cold can of soda warming on counter Choice of direction was arbitrary (like electron charge) Might not be intuitive, but consistent with other definitions 8

9. ThermodynamicsLosing Energy EX OTHERMIC Reactions which generate and/or lose heat Energy is transferred to surroundings Burning leaves, coffee cooling, moving automobile ΔH or “Enthalpy” is term for heat transfer -ΔH or “Enthalpy” is negative for Exothermic (-) Enthalpy becomes part of chemical equation Enthalpy usually in kJ per Mole Total energy depends on total quantity 9

10. EXOthermic reaction (-ΔH) Producing heat or thermal energy by burning fuel, converting chemical (or nuclear) into kinetic or heat energy

11. ThermodynamicsGaining energy ENDOTHERMIC Reactions which extract and/or gain heat Energy is transferred into the object Melting ice, coffee being made (water heated) +ΔH or “Enthalpy” is term for heat input ΔH “Enthalpy” positive (+) for Endothermic (+) Enthalpy becomes part of chemical equation Enthalpy usually in kJ per Mole Total energy depends on total quantity 11

12. ENDOthermic reaction (+ΔH)Absorbing heat energy from environment

13. Air Conditioning = ENDOthermic cooling results from evaporation reaction absorbing heataccompanied by exothermic condensation at radiator

14. 14

15. Water EnergyMaking water from elements releases heat energysplitting water into elements requires electrical energy 15

16. Bond Energy Heat results from rearranging chemical bonds Reducing available energy (reactants-products) releases energy Burning wood, animal metabolism Increasing chemical energy (products-reactants) absorbs energy Photosynthesis, melting ice Impractical to measure ΔH for every known reaction Billions of chemical combinations But use of common bonds provides a practical answer … Can use common “features” to divide and conquer Bond breaking energy can be determined for reference cases Carbon-Carbon bonds (single C-C, double C=C, triple C≡C) Diatomic molecules (Cl2, H2, O2, etc.) Use known bond energies to estimate new combinations Algebraic sum of the bond energy components Must use balanced equations and appropriate multipliers 16

17. Sample Bond Energy CalculationBurning of Hydrogen in Air, producing heatTables of data differ, but have similar values

18. A few common bond energies

19. Table of Bond Energies combustion heat output 19

20. A home furnace exampleWe can predict heat from burning methane via bond energies • Burning Methane CH4+ 2O2 CO2 + 2H2O • Reactants: • 4 * C-H bonds x 414kJ/mol * 1mol= 1656kJ • O=O bond = 498kJ/mol * 2mol = 996 • Total reactants bond energies = 2652kJ • Products: • 2 * C=O bonds x 803kJ/mol = 1606kJ • 2 * H-O bonds x 464kJ/mol * 2 mol = 1856kJ • Total products bond energies = 3462 kJ • Change = 2652 - 3462 = - 810 kJ • Literature value comparison = - 803 to - 889kJ/mole • Negative energy change means Exothermic • Products more tightly bonded than reactants • Takes more energy to pull products apart • Excess energy released as Heat

21. Standard State Must define “state” of material for reference Gas, Liquids, Solids have different energy content Evaporation of water cools (energy loss 44kJ/mole) Compression of refrigerant heats it (energy gain) “STP” is a definition for reference (standard) state Reference temperature (typically 0 or 25 degrees Celsius) 1 atmosphere of pressure Concentration of 1.00 Moles per Liter (usually) 21

22. Heats of reactionsyour home furnace in chemical termsone last step involves the water vapor Two reactions can be combined (both of these exothermic) Burning of Methane, and condensation of water vapor. A thermodynamic model of the furnace in your house. CH4(g) + 2O2(g) CO2(g) + 2 H2O(g) ΔH = - 810kJ/mol H2O(g)  H2O(aq) a change of stateΔH = - 44 kJ/mol Evaporation absorbs heat, so condensation yields heat Stoichiometry requires consistent number of moles 2H2O(g)  2H2O(aq)ΔH = - 88 kJoule 22

23. Heats of reactions Can add the equations, molecules AND reaction energy Net reaction, adding the two: CH4(g) + 2O2(g) CO2(g) + 2 H2O(g) ΔH = -810 kJoule 2H2O(g)  2H2O(aq)ΔH = - 88 kJoule ----------------------------------------------------------------------------- CH4(g) + 2O2(g) CO2(g) + 2 H2O(aq)ΔH = -898kJoule Magnitudes of heat energy combine same as for the molecules in a chemical reaction 23

24. Some mechanisims

25. Mechanism for heat of solution

26. Heat of solution (dissolving)

27. Calorimeter in a cup

28. We will use a simple calorimeterstyrofoam cups for insulationswirling better than stirring

29. What’s wrong with this picture?(recall James Joule’s experiment)

30. Energy Dimensions • Original definition is “calorie” (small c) • Energy to raise temp.1 gram (1 ml) water 1.0oC • Turned out to be inconveniently small • Usual quotation in kcal = “Calorie” (big C) • Energy to raise temp 1.00 liter water by 1.0oC • Calories are NOT in S.I. (MKS) dimensions • Commonly used for food products • SI or ISO unit of energy is “Joule” • 1 watt for one second = 1 Joule • Conversion is 4.184 Joule/calorie • Same thing is 4.184 kJ/kcal = 4.184 kJ/Calorie

31. Our Procedure • Perform an EXOTHERMIC reaction • CaCl2 dissolving in water produces heat • Make a plot to determine maximum temp. • Use Q=m*ΔT*c = calories • C is a constant for water = 1.00 • Calculate kcal per mole • Moles from mass of salt & formula weight • Compare to literature values, how close?

32. Calculations • Similar to burning of food experiment • Heat is delivered to measured mass of water • Calories into water + salt, Q = m*c*∆T • Q = heat in calories • M = actual mass of water + salt ( ≈ 120gram) • C = specific heat of water = 1 cal/(gm*∆T) • Q = 120gm*1cal/(gm*∆T)*∆T = calories • If Q positive, solution gets COLD • If Q negative, solution gets HOT

33. Procedure • Water • Weigh empty cup and with ≈100mL water • Obtain mass of water in grams • Salt • Weigh container without & with salt • Obtain mass of salt in grams • Temperature • take initial temperature of water • Mix salt and water • Take temp. every 10 seconds for first 3 minutes • Take temp. every 30 seconds for another 2 minutes • Swirl water in cup to mix between readings • Plot the data

34. HEATING REACTION

35. Calculation Procedure • Use graph to determine maximum temperature reached • Calculate calories produced • (water grams + salt grams) * ΔT = calories • Calculate energy per mole derived from salt • Calories / moles = energy per mole • Convert to kJ/mole for literature comparison • Calories / 4.182 = Joules • Joules / 1000 = kJoules • Compare to literature values • - 82.0 kJ/mole for CaCl2 (exothermic) is customary value • How close did you get ? • Calculate error = (literature-experimental) / literature • Answer *100 = percent error

36. Sample Calculation

37. 2nd half of experiment • Repeat for ENDOTHERMIC reaction • NH4Cl in water absorbs heat • Measure masses, initial temperature • Mix and measure temperature changes • Plot data • Calc energy absorbed per mole of salt • Compare to literature value

38. COOLING REACTION

39. Endothermic data example

40. Now you try it • Report due next week

41. Los Alamos National Laboratory's Periodic Table

42. Ice meltingclassic example of entropy increase described in 1862 by Rudolf Clausiusas an increase in the disagregation of the molecules of the body of ice. 42

43. Calories in the food • Calories delivered into water, Q = m*c*∆T • Q = heat in calories • M = actual mass of water heated ( ≈ 100gram) • C = specific heat of water = 1 cal/(gm-∆T) • Q = 100gm*1cal/(gm*∆T)*∆T = calories • Calories into water came from food • Calories transferred / mass of food = cal/gram • If 0.5 gram food (preburn-postburn) yields 2 kcal • 2 kcal / 0.5 gram = 4 kcal/gram for the food • 1.0 pound (454 gm) of this food yields ≈ 1800 kcal

44. Carbon Fuels Heat output of fuel results from breaking bonds, releasing energy “Heat of Combustion” for carbon fuels (e.g. gasoline, jet fuel) Called ΔH of combustion, or Combustion Enthalpy Source material always contains C and H Numbers of C & H varies tremendously Natural products full of variants: linear + branched + ring structures Combustion products always contain H2O and CO2 Sometimes also CO and NOX (N2O, NO, NO2, NO3) Depends on amount of oxygen available and temperature Theoretically possible to calculate heat of combustion for any fuel Works for simple materials (hydrogen, methane, benzene) See table for typical values Not too practical for “real world” bulk materials Too many variations and uncertainties with natural products Dissolved dinosaurs and vegetation don’t yield pure chemical products 44

45. Carbon Fuels Fuels have 3 entangled physical properties Density (grams per cm^3) Molecular Weight (grams per mole) Combustion Energy (bond breaking) Application defines which is “best” Higher density (liquid) fuels good for Automobiles 5 to 11 carbons in gasoline (depends on season) More moles per gas tank, drive farther between fillings Diesel fuel more energy than Gasoline, 11-14 carbons Low density (gas) fuels good for domestic use Vapor state fuels (methane, propane) easy to handle Constant pressure, simple distribution using pipes Weight and size of delivery system not important 45

46. Common Fuels Burning Hydrogen (proposed by CA) 1 H-H bond = 436kJ/mol (22.4 Liters or 5.9 gal ) or 436kJ/gram of H2 Burning Methane (natural gas) 4 C-H bond = 1,656kJ/mol,(22.4Liters or 5.9 gal) or 1656kJ/mol/ 16gm/mol = 106kJ/gm CH4 Burning Gasoline (octane=C8H18) 18C-H & 7C-C bond = 7848+2429 =10,277 kJ/mol Or 10,277kJ/mjol/114g/mol= 90kJ/gram of octane density = 0.72 gm/mL, 114g/0.72g/mL= 0.16 Liter

47. ISO Energy Definition Units of Energy, definition of Joule Auto data SUV is 4000 lb= 1842kg Speed of 62mi/hr = 100km/hr= 27.7 m/sec kg*(m/s)^2= 1842*27.7*27.7 = 1.41E6 W-S 47

48. Gasoline efficiency • 18miles/gallon (my Ford Explorer) • 18mi/g/3.84L/g*1.6km/mile  7.4 km/Liter • At 700 gr/Liter, gasoline  10.6 meter/gm • Gasoline energy = 43.6 kJ/gram • Energy expended = 43.6/10.6 = 4.11 kJ/meter