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Calorimetry

Physics 12 . Calorimetry. Objectives. Define specific heat capacity. Solve problems involving specific heat capacities. Explain the difference between solid, liquid, and gaseous phases. Explain in terms of molecular behavior why temperature does not change during a phase change.

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Calorimetry

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  1. Physics 12 Calorimetry

  2. Objectives • Define specific heat capacity. • Solve problems involving specific heat capacities. • Explain the difference between solid, liquid, and gaseous phases. • Explain in terms of molecular behavior why temperature does not change during a phase change. • Define latent heat. • Solve problems involving latent heats.

  3. Activities • Worksheet • Lab demonstration: Determining the specific heat capacity of a substance • Lab demonstration: Determining the latent heat of a substance • Design lab: Design a homemade calorimeter and test it

  4. Specific heat capacity • If heat flows into an object, its temperature rises. • What factors might affect the magnitude of the temperature change?

  5. Specific heat capacity • The amount of energy Q required to change the temperature of a given material is proportional to the mass m of the material and to the temperature change ΔT, shown by the simplistic expression: Q = mc ΔT where c is the characteristic of the material called its specific heat capacity

  6. Specific heat capacity • A high specific heat capacity means that more energy is required to achieve the same temperature change, i.e. it is more “difficult” to raise the temperature of that material. • If a material is a good heat conductor (e.g. metals) would you expect it to have a high or low specific heat capacity?

  7. Specific heat capacity Specific heat capacities of specific substances

  8. Specific heat capacity

  9. Specific heat capacity • How much heat input is needed to raise the temperature of an empty 20-kg vat made of iron from 10C to 90C? • What if the vat is filled with 20 kg of water?

  10. Specific heat capacity • 720 kJ • 7400 kJ

  11. Specific heat capacity You accidentally let an empty iron frying pan get very hot on the stove (approx. 200C). What happens when you dunk it into a few inches of cool water in the bottom of the sink? Will the final temperature be midway between the initial temperatures of the water and pan? Will the water start boiling? (Assume the mass of the water is roughly the same as the mass of the pan.)

  12. Calorimetry • In discussing heat and thermodynamics, we shall often refer to systems. • What is the difference between open, closed, and isolated systems?

  13. Calorimetry • Open system Mass and energy may leave and enter • Closed system Energy may leave and enter but mass may not • Isolated system Neither mass nor energy may leave or enter

  14. Calorimetry • We will often make the assumption that the systems we are dealing with are isolated. Why is this necessary?

  15. Calorimetry • In an isolated system, heat lost by one part of the system is equal to the heat gained by another part: heat lost = heat gained or energy out of one part = energy into another part

  16. Calorimetry If 200 cm3 of tea at 95C is poured into a 150-g glass cup initially at 25C, what will be the common final temperature T of the tea and cup when thermal equilibrium is reached, assuming no heat flows to the surroundings?

  17. Calorimetry T = 86C Would this be the case in the “real world”?

  18. Calorimetry • The exchange of energy (as shown in the previous example) is the basis for the technique known as calorimetry. • Calorimetry is the quantitative measurement of heat exchange. • A calorimeter is used.

  19. Calorimetry A simple water calorimeter

  20. Calorimetry An engineer wishes to determine the specific heat of a new metal alloy. A 0.150-kg sample of the alloy is heated to 540C. It is then quickly placed in 400 g of water at 10.0C, which is contained in a 200-g aluminum calorimeter cup. The final temperature of the system is 30.5C. Calculate the specific heat of the alloy.

  21. Calorimetry c = 500 Jkg-1C-1

  22. Calorimetry • In order to determine the specific heat of a particular substance, the following expression is used: Qlost = Qgained m1c1ΔT1 = m2c2 ΔT2 where the two substances share a final temperature • Thus, all other quantities except one must be measured or known.

  23. Phase change • Recall that matter most commonly exists in three states: solid, liquid, and gas. • What are the differences between these three states (or phases) in terms of molecular structure and motion?

  24. Phase change Comparison of the three common phases of matter (on Earth)

  25. Phase change Comparison of the three common phases of matter (on Earth)

  26. Phase change • When a material changes phase from solid to liquid or liquid to gas, a certain amount of energy is involved in this change of phase.

  27. Phase change Temperature as a function of heat added to10.0 g of ice

  28. Phase change • The heat required to change a substance from solid to liquid is called latent heat of fusion. • The heat required to change a substance from liquid to gas is called the latent heat of vaporization. • Values for latent heats will vary depending on the substance.

  29. Phase change Latent heats

  30. Phase change • What factors might affect the amount of energy needed to change the phase of a substance?

  31. Phase change • The heat involved in a change of phase Q depends not only on the latent heat but also on the total mass of the substance, i.e. Q = mL where m is the mass of the substance and L is the latent heat

  32. Phase change How much energy does a freezer have to remove from 1.5 kg of water at 20C to make ice at –12C?

  33. Phase change 6.6 x 105 J

  34. Phase change At a reception, a 0.50-kg chunk of ice at –10C is placed in 3.0 kg of tea at 20C. At what temperature and in what phase will the final mixture be? The tea can be considered as water. Ignore any heat flow to the surroundings, including the container.

  35. Phase change T = 5C

  36. Phase change The specific heat of liquid mercury is 140 Jkg-1C-1. When 1.0 kg of solid mercury at its melting point of –39C is placed in a 0.50-kg aluminum calorimeter filled with 1.2 kg of water at 20.0C, the final temperature of the combination is found to be 16.5C. What is the heat of fusion of mercury in Jkg-1?

  37. Phase change L = 11 x 103 Jkg-1

  38. Objectives • Define specific heat capacity. • Solve problems involving specific heat capacities. • Explain the difference between solid, liquid, and gaseous phases. • Explain in terms of molecular behavior why temperature does not change during a phase change. • Define latent heat. • Solve problems involving latent heats.

  39. Measuring specific heat Data collection

  40. Measuring specific heat Data processing Include propagation of uncertainty

  41. Measuring specific heat Data processing Include propagation of uncertainty Q = mcΔT Q (lost) = Q (gained)

  42. Measuring specific heat • Homework Write a conclusion and evaluation of the lab activity Design a homemade calorimeter using the materials available. Try to minimize the amount of energy lost to the surroundings. You will use this calorimeter for the next lab activity. The calorimeter with the smallest percent discrepancy gets bonus points.

  43. Measuring latent heat of fusion Data collection

  44. Measuring latent heat of fusion Data processing

  45. Measuring latent heat of fusion • Homework Complete your data processing. Make sure to include a sample calculation that shows the propagation of uncertainty. Write a conclusion for your data, is the known value for the latent heat of fusion of water within your range of uncertainty?

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