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Calorimetry Energetics

Calorimetry Energetics. 1. Heat versus Temperature. Heat is energy that is transferred from one object to another due to a difference in temperature from hot to cold Temperature is a measure of the average kinetic energy of a body. 2.

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Calorimetry Energetics

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  1. CalorimetryEnergetics 1

  2. Heat versus Temperature • Heatis energy that is transferred from one object to another due to a difference in temperature from hot to cold • Temperature is a measure of the average kinetic energy of a body 2

  3. What must you measure to calculate the amount of heat lost or gained by a substance?

  4. Factors Affecting Heat Quantities • The amount of heat contained by an object depends primarily on three factors: • The mass of material • The temperature • The kind of material and its ability to absorb or retain heat. 4

  5. Units of Heat • The heat required to raise the temperature of 1.00 g of water 1 oC is known as a calorie • The SI unit for heat is the joule. • 1.00 calorie = 4.184 Joules • 1000 Joules = 1 kJ 5

  6. Calorimetry • Calorimetry involves the measurement of heat changes that occur in chemical processes or reactions. • You need to measure: • The mass of the substance • The temperature change • The heat capacity of the material 6

  7. Heat Capacity and Specific Heat • The heat capacity is a measure of a substance’s ability to absorb or release heat. • The specific heat capacity of a material is the amount of heat required to raise the temperature of 1 gram of a substance 1 oC (or Kelvin) 7

  8. Specific Heat values for Some Common Substances 8

  9. Heat Calculations The heat equation may be stated as DQ = m c DT where: DQ = Change in heat m = mass in grams c = specific heat capacity in J g-1oC-1 DT = Temperature change 9

  10. Calorimeters

  11. Temperature Changes Measuring the temperature change in a calorimetry experiment can be difficult since the system is losing heat to the surroundings even as it is generating heat. By plotting a graph of time v temperature it is possible to extrapolate back to what the maximum temperature would have been had the system not been losing heat to the surroundings. temperature graph vs time 11

  12. Heat Transfer Problem 1 Calculate the heat that would be required an aluminum cooking pan whose mass is 400 grams, from 20oC to 200oC. The specific heat of aluminum is 0.902 J g-1oC-1. Solution DQ = mCDT = (400 g) (0.902 J g-1oC-1)(200oC – 20oC) = 64,944 J 12

  13. Heat Transfer Problem 2 What change in energy if 50.0 grams of water at 20.0oC is heated to 60.0oC? Assume that the loss of heat to the surroundings is negligible. The specific heat of water is 4.184 J g -1oC-1 Solution:DQ = mCDT (50.0 g) (4.184 J g-1oC-1)(60.0 oC - 20.0oC) Q = 8368 = 8370 Joules 13

  14. Heat Transfer Problem 3 What is the final temperature when 50 grams of water at 20oC is added to 80 grams water at 60oC? Assume that the loss of heat to the surroundings is negligible. The specific heat of water is 4.184 J g-1 oC-1 Solution:DQ (Cold) = DQ (hot) mCDT= mCDT Let T = final temperature (50 g) (4.184 J g-1oC-1)(T- 20oC) = (80 g) (4.184 J g-1oC-1)(60oC- T) (50 g)(T- 20oC) = (80 g)(60oC- T) 50T -1000 = 4800 – 80T 130T =5800 T = 44.6 oC 14

  15. NaOH + HCl H2O + NaCl If 20.0 g of solid NaOH are added to 1000 mL of a 0.500 M HCl solution, the temperature of the solution rises from 20.0 oC to 26.0 oC. The specific heat of the solution is 4.184 J g-1oC-1. Calculate the heat released by this reaction. Then calculate ΔHrxn (i.e., the heat released per mole of NaOH).

  16. Calculations • Find moles of HCl and NaOH: 0.20 grams/40.0 g/mole = 0.50 moles NaOH 0.5 M = x moles/1.0 liter  0.50 moles HCl Q = c m DT Q = 4.184 J g-1oC-1 (1000 grams) 6 oC Q = -25104 J (Energy is released)

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