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Energy Part II: Working with C, s, q, w & ΔH Chapter 7 Sec 3, 4 & 5 of Jespersen 6 th ed)

Energy Part II: Working with C, s, q, w & ΔH Chapter 7 Sec 3, 4 & 5 of Jespersen 6 th ed). Dr. C. Yau Fall 2013. 1. 1. Quick Review of Heat Capacity vs. Specific Heat. The amount of heat (q) transferred is proportional to the change in temperature Δ T .

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Energy Part II: Working with C, s, q, w & ΔH Chapter 7 Sec 3, 4 & 5 of Jespersen 6 th ed)

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  1. Energy Part II: Working with C, s, q, w & ΔHChapter 7 Sec 3, 4 & 5of Jespersen 6th ed) Dr. C. Yau Fall 2013 1 1

  2. Quick Review of Heat Capacity vs. Specific Heat The amount of heat (q) transferred is proportional to the change in temperature ΔT. The proportionality constant for this relationship is C, the heat capacity: q = CΔT y = k x Heat capacity depends on: 1) mass of sample (m) 2) specific heat of the sample (s) q = s mΔT (Remember ? C = s x m) 2

  3. Deciding on Which Eqn to Use q = C ΔT C = heat capacity q = s m ΔT s = specific heat capacity or specific heat Specific heat (s) is for pure substances, such as Fe, Cu, water, NaOH, glucose, NH4Cl etc. We examined these specific heat values in Table 7.1 p.261 back in Energy Part I. You either look them up and utilize them to calculate q or ΔT, or determine the specific heat of an unknown substance and identify the substance by comparing its specific heat to known values. 3

  4. Unit is Also could be Specific heat is the amount of heat need to change the temperature of 1 gram of substance by 1 degree (either oC or K) For example it takes 4.184 J to raise 1 g of water from 25oC to 26oC. However it takes only 0.4498 J to raise 1 g of Fe from 25oC to 26oC. This is consistent with the fact that iron conducts heat well. Water does not heat up as easily.

  5. Deciding on Which Eqn to Use q = s m ΔT Specific heat (s) has units of J g-1oC-1 (joules per gram per oC of change in temp) To determine how much heat has been transferred (q)… 1) you look up s, 2) measure the mass of the sample (m) 3) measure the initial and final temperature of the sample in order to calculate the change in temperature of the sample (ΔT) 5

  6. q = s m ΔT For example, if we have 5.00 g of water and its temperature rose from 25.0oC to 38.0 oC, we know how much heat the water has absorbed by looking up s for water. Specific heat of water is 4.184 J g-1oC-1 ΔT = (38.0 – 25.0)oC = 13.0 o = 272 J Ans. The water has absorbed 272 J. Note: This is a positive number. 6

  7. The Sign of q If an object is releasing heat, its temperature must be going from a higher T to a lower T. What is the sign of ΔT? ΔT = Tfinal – T initial smaller larger q = s m ΔT (s and m are ALWAYS positive) If ΔT is negative, then q must be negative. WHEN HEAT IS RELEASED, q is NEGATIVE. The process is said to be EXOTHERMIC. (Δ T must be negative.) 7

  8. The Sign of q If an object is absorbing heat, its temperature must be going from a lower T to a higher T. What is the sign of ΔT? ΔT = Tfinal – T initial larger smaller q = s m ΔT If ΔT is positive, then q must be positive. WHEN HEAT IS ABSORBED, q is POSITIVE. The process is said to be ENDOTHERMIC. 8

  9. KNOW THIS WELL!! Exothermic = heat released, q is – Endothermic = heat absorbed, q is + If heat is being transferred from one system to another, the magnitude of q (absolute value of q) is the same, but the sign must be opposite: | q(released) | = | q(absorbed)| q (released) = – q (absorbed) q (absorbed) = – q (released) 9

  10. Meaning of Heat Capacity In contrast, heat capacity has the units of J/oC. We generally talk about the heat capacity of an object with a fixed mass and usually not of one single compound or element. e.g. In this course we will be talking about the heat capacity of a bomb calorimeter, which consists of… 1) a steel container 2) a insulated vat and lid 3) a thermometer 4) a stirrer We do not talk about the mass of the calorimeter. The calorimeter is obviously not one single compound or element. Each calorimeter has its own heat capacity. (p.271 Fig7.9) 10

  11. Meaning of Heat Capacity We will also talk about a simpler form of calorimeter, the coffee cup calorimeter, which consists of… 1) double-nested coffee cup 2) a lid 3) a thermometer 4) a stirrer Again, we do not measure the mass of the calorimeter as it is made of different compounds. p.273 Fig.7.10 11

  12. Heat Capacity (C) Example (not in book) Similar to Ex 7.1 p.262 Central processing chips in computers generate a tremendous amount of heat – enough to damage themselves permanently if the chip is not cooled somehow. Aluminum "heat sinks" are often attached to the chips to carry away excess heat. Suppose that a heat sink at 71.3oC is dropped into a Styrofoam cup containing 100.0 g of water at 25.0oC. The temp of the water rises to 27.4oC. What is the heat capacity of the heat sink, in J/oC? (continued next slide) 12 12

  13. Example (cont'd) Suppose that a heat sink at 71.3oC is dropped into a Styrofoam cup containing 100.0 g of water at 25.0oC. The temp of the water rises to 27.4oC. What is the heat capacity of the heat sink, in J/oC? Tips: Recognize that there are 2 systems. Think… q (released) = - q (absorbed) Where is the heat coming from? Where is the heat going to? The heat is coming from the hot heat-sink and is going into the cool water. 13 13

  14. Example (cont'd) Suppose that a heat sink at 71.3oC is dropped into a Styrofoam cup containing 100.0 g of water at 25.0oC. The temp of the water rises to 27.4oC. What is the heat capacity of the heat sink, in J/oC? q (released) = -q (absorbed) out of heat-sink into the water First you decipher what are the ΔT's for each, remembering that ΔT is "final minus initial." 14

  15. Example (not in book)(similar to Example 7.2 p.263) If a gold ring with a mass of 5.509 g changes in temp from 25.00 to 28.00 oC, how much heat has it absorbed? Tips: In contrast to the previous problem, there is only one q (absorbed). A mass is given, and the substance is a pure substance. This tells you s not C is involved. Look up s for gold in Table 7.1 on p.261) Specific heat of gold is 0.129 J g-1oC-1. 15

  16. Example 7.3 p.264 If a 25.2 g piece of silver absorbs 365 J of heat, what will the final temp of the silver be if the initial temp is 22.2oC? Preliminary analysis: Does it involve one or two systems? (qsystem = - qsurroundings?) Do we use q = CT or q = smT ? (Look up specific heat of silver in Table 7.1: 0.235 J g-1oC-1) Do Example 7.1, 7.2 and Pract Exer 1, 2 & 3 p.265

  17. Heat Transfer During Rxns Remember we said when an object releases heat, its temperature must decrease. Note that was about an OBJECT releasing heat. If a REACTION is releasing heat, the temp of the reactants is NOT decreasing. This is because the heat is coming from the chemical energy and not from the thermal energy of the reactants. However, the temp of the SURROUNDINGS will increase due to the heat transfer. 17

  18. Hot coffee poured into a cold cup: qcoffee = negative (losing heat) exothermic T = T final – T initial = negative (cooler) (hotter) Reaction is exothermic, but... Cup/beaker feels hotter. T = T final – T initial = positive (hotter) (cooler) WHY? Heat is not thermal E of reactants but the reaction. T is of the surroundings, which is endothermic. CaO(s) + H2O(l) ? Highly exothermic

  19. Negative sign means sys lost E doing work. How Pressure Relates to Heat Flow ΔE = q + w change in = heat + work internal energy transfer and the work done is called "PV work" or "expansion work." Imagine a piston being pushed up by gas expansion in a cylinder. w = – P ΔV where w = work P = atmospheric pressure ΔV = change in volume Work is done pushing the atmosphere back. 19

  20. ΔE = q + w If a system does 48 J of work and receives 15 J of heat, what is the value of ΔE for this change? Since the system is receiving heat, q must be positive (+15 J). Since the system is doing work, it is using up energy, losing energy, so w must be negative (-48 J). ΔE = +15 J– 48 J = - 33 J. The system is losing energy overall.

  21. The Calorimeter The calorimeter is an apparatus used to measure the change in temperature due to chemical reactions. Some are designed to measure the change under constant pressure, some, under constant volume. In the bomb calorimeter, it is a tightly sealed system, qV is determined (q at constant volume). In an open vessel, it is under constant pressure (the pressure of the atmosphere) and we have qP(q at constant pressure). Learn the definition of qv and qp! 21

  22. Heats of Combustion Heat of combustion = heat released by a combustion reaction Heat of combustion is determined with the use of a bomb calorimeter (constant volume). Constant volume means ΔV = 0, so w = 0. ΔE = q + w where w = 0. ΔE = q Heat of combustion is qV. 22

  23. Example 7.5 p.271 a) When 1.000 g of olive oil was completely burned in pure oxygen in a bomb calorimeter, the temp of the water bath increased from 22.000 oC to 26.049 oC. How many dietary Calories are in olive oil, per gram? The heat capacity of the calorimeter is 9.032 kJ/oC. Tips: Recognize there are two systems: calorimeter and water absorbing heat and reaction producing heat. 23

  24. Example 7.5 p.271 (continued) b) Olive oil is almost pure glyceryl trioleate, C57H104O6. The eqn for its combustion is C57H104O6 + 80 O2 57CO2 + 52H2O Ans from part (a): 8.740 Cal from 1 g What is the change in internal energy, E, for the combustion of one mole of glyceryl trioleate? Assume olive oil burned in part a was pure glyceryl trioleate. MM = 885.4 g/mol Do Pract Exer 4 & 5 p.272 24

  25. Heat of Reaction (ΔH) H = enthalpy (potential heat not measurable) ΔH = enthalpy of reaction or heat of reaction = Hproduct – Hreactant (calculated from the ΔT of surroundings) 25

  26. Heats of Rxns in Solution Heat of rxns in solution are determined in open vessels (qP). ΔE = qP + w Enthalpy is defined as H = E + PV ΔH = ΔE+ P ΔV but P ΔV =- w, and ΔE =qP + w so ΔH = (qP+w)- w ΔH = qP

  27. Summary ΔE = q + w Heat of Combustion = ΔHcombustion measured at constant V (in steel container) ΔE = qV (w = zero) Heat of Reaction = Enthalpy of Reaction = ΔH defined to be ΔE + P ΔV ΔH = qP Significance: Experimentally determined q in an open container gives us ΔH.

  28. Example 7.6 p.273 HCl(aq) + NaOH(aq) H2O(l) + NaCl(aq) A student placed 50.0 mL of 1.00 M HCl at 25.5oC in a coffee cup calorimeter. To this was added 50.0 mL of 1.00 M NaOH soln also at 25.5oC. The mixture was stirred and temp quickly increased to a max of 32.2oC. What is ΔH in kJ per mole of HCl? Because the solns are relatively dilute, we can assume that their specific heats are close to that of water, 4.18 J g-1oC-1. The density of 1.00 M HCl is 1.02 g/mL and that of 1.00 M NaOH is 1.04 g/mL. (We will neglect the heat lost to the Styrofoam itself, to the thermometer or to the surrounding air.) Do Pract Exer 6, 7 & 8 p.275 28

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