Last rev. 5/12/04

Last rev. 5/12/04

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Last rev. 5/12/04

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1. Heat, Temperature And Phase Changes Pisgah High School Chemistry Mr. Jones Last rev. 5/12/04

2. Part One Heat and Temperature

3. What temperature does the thermometer indicate?

4. What might be going on that would cause this temperature?

5. This is the view out the window, past the thermometer.

6. Yep. It is snowing.

7. Why would snow cause the air temperature to be at precisely 0C? What occurs at 0C? Waterfreezes and icemelts.

8. Ice, in the form of snow, falls through the slightly warmer air. The snow melts and absorbs heat from the air, causing the air to cool. Ice melts at 0 C, so the air cools to that temperature.

9. The temperature hovers at zero Celsius as the snow melts.

10. So why is there snow on the ground if it is melting? Yep. That’s what allows the snow to accumulate.

11. As the snow melts, it absorbs heat and cools the ground, the car, and the grill.

12. This allows more snow to lay. It doesn’t melt because the ground is now at 0C.

13. What does it mean to have a temperature of 0 C? What is temperature? Is temperature the same thing as heat?

14. Temperature is a measure of how “hot” or “cold” something is. Temperature is measured in arbitrary units, like Fahrenheit or Celsius.

15. Temperature is proportional to the average kinetic energy of the molecules of the substance. T µ KE KE =½ mv2

16. Velocity or speed Temperature is therefore proportional to the speed of the molecules of a substance. T µ KE KE =½ mv2 T µv

17. Velocity or speed The higher the temperature, the greater the average speed of the molecules. T µ KE KE =½ mv2 T µv

18. Heat is the thermal energy transferred from a hot object to a cold object. Heat is measured in energy units -- Joules or calories.

19. The heat transferred is proportional to the mass of the object, the specific heatcapacity of the object and the temperature change the object undergoes.

20. specific heat capacity Quantity of heat mass temperature change q = mcDT

21. specific heat capacity Quantity of heat q = mcDT The specific heat capacity of water is 4.18 J/gC

22. How much heat is needed to raise the temperature of 25.6 grams of water from 20.0 C to 50.0 C? q = m c DT q = (25.6g)(4.18J/gC)(30.0C) q = 3210 J

23. q DT = m c What is the final temperature of 27.0 grams of liquid water, initially at 0C, after it absorbs 700.0 J of energy? q = m c DT Hint: start by solving for DT. Answer: 6.20 C

24. Part Two Calorimetry and Specific Heat Capacity

25. Calorimetry is a collection of laboratory procedures used to investigate the transfer of heat. In calorimetry experiments, one might be looking for a final temperature or a specific heat capacity.

26. Investigate: Suppose two different masses of water at different temperatures are mixed. Can you predict the final temperature?

27. Investigate: Will the final temperature be cooler than the cool water, or will it be warmer than the warm water? Or will the final temperature be somewhere in between?

28. Investigate: Develop a procedure where you could mix a known mass of cool water with a different mass of water at an elevated temperature and measure the final (equilibrium) temperature. What equipment would you need?

29. Investigate: Develop a procedure where you could mix a known mass of cool water with a different mass of water at an elevated temperature and measure the final (equilibrium) temperature. You could use a balance, a thermometer, a coffee cup calorimeter, and a hot plate.

30. Investigate: Mass of calorimeter cup Mass of cool water Initial temperature of cool water Mass of warm water Initial temperature of hot water Final temperature after mixing What do you need in a data table? Feel free to make additions.

31. Investigate: Whenever we design an experiment we make some assumptions. Here are a couple, can you add any more? The calorimeter cup is a perfect insulator and no heat is exchanged with the surroundings. Note: Hot plates and boiling water can cause severe burns.

32. Investigate: You might need a hint about how to calculate the results. What is the law of conservation of energy? Energy is neither created nor destroyed, only changed in form.

33. Investigate: You might need a hint about how to calculate the results. The law of conservation of energy suggests that the heat lost by the hot water as it cools is equal to the heat gained by the cool water as it warms up.

34. Investigate: To put it mathematically: qlost = -qgained Heat lost by the hot water Heat gained by the cold water = And sinceq = mcDT then mhcDTh = -mccDTc

35. Investigate: Finally, since DT includes the final temperature (DT = Tfinal – Tinitial ) then replace DT: mhc(Tf -Th) = -mcc(Tf -Tc) Write an equation that solves for the final temperature.

36. Investigate: Use your equation to solve the following problem: Calculate the final temperature when 20.0 grams of water at 85 C is added to 35.0 grams of water at 10.0 C in an insulated container.

37. Investigate: The answer to the following problem is 37.3 C. Calculate the final temperature when 20.0 grams of water at 85 C is added to 35.0 grams of water at 10.0 C in an insulated container.

38. In the next investigation you will … develop a method to find the specific heat capacity of a metal.

39. Specific heat capacity … • …varies from one substance to another • …a measure of how much heat something can “hold” • …the amount of heat needed to raise one gram of a substance by one Celsius degree

40. Specific heat capacity … • …varies from one substance to another • …a measure of how much heat something can “hold” • …the amount of heat needed to raise one gram of a substance by one Celsius degree

41. Specific heat capacity lab suggestions: • Heat a metal to a known temp • Transfer the metal to a known quantity of water at a known temperature • Measure the equilibrium temperature • Use qlost = qgained to compute the specific heat of the metal.

42. Get the initial temperature of the metal. The temperature of boiling water. metal hotplate

43. Get initial temp of water in calorimeter cup.

44. Transfer the metal to the calorimeter.

45. Continue stirring until thermal equilibrium is reached.

46. Data: Mass of metal Initial temp of metal Mass of water Initial temp of water Final temp of water and metal

47. qlost = -qgained mmcmDTm = -(mwcwDTw) -(mwcwDTw) cm = mmDTm

48. Data: Suppose 104.58 g of a metal is heated to 100. C. It is placed into 54.21 g of water at 17.5 C. What would be the specific heat capacity of the metal? The final temp of water and metal is 22.0 C.

49. Data: Suppose 104.58 g of a metal is heated to 100. C. It is placed into 54.21 g of water at 17.5 C. What could be the identity of the metal? The final temp of water and metal is 22.0 C.