Ch 21 Temperature, Heat, and Expansion

1. Ch 21 Temperature, Heat, and Expansion

2. Temperature A measure of the average kinetic energy of the particles in a substance.

3. Imagine a pail of warm water and a cup of a hot water. • A 1 & 2 liter bottle of boiling water.

4. Temperature is NOT a measure of the total KE of molecules in the substance.

5. Temperature Scales 1. Fahrenheit (oF) 2. Celsius (oC) 3. Kelvin (K)

6. Boiling Point 1. Fahrenheit 212 oF 2. Celsius 100 oC 3. Kelvin 373 K

7. Freezing Point • 1. Fahrenheit 32 oF • 2. Celsius 0 oC • 3. Kelvin 273 K

9. Rankine Temperature Scale • Temperature scale having an absolute zero, below which temperatures do not exist, and using a degree of the same size as that used by the Fahrenheit temperature scale. • Absolute Zero corresponds to a temperature of −459.67°F;

10. Absolute Zero Point at which all molecular motion has stopped. We have never reached it, but are very close. Scale is used in engineering.

11. Temperature Difference (DT) Is the primemover or force-like quantity in a thermal system. DT – “Delta T”

12. Ex: 110 oF inside and 40 oF outside. What is the DT? • DT = 110 – 40 = 70 Fo

13. Thermometer • Instrument used to measure temperature. • Based upon liquid expansion in the tube with respect to temperature.

14. Usually mercury or an alcohol mixture.

15. Converting Temperatures Fahrenheit to Celsius TC = 5/9(TF – 32o)

16. Ex: Convert 50 oF to oC TC = 5/9(TF – 32o) TC = 5/9(50 – 32o) TC = 5/9(18o) TC = 10 oC

17. Celsius to Fahrenheit • TF = 9/5(TC)+ 32o

18. Ex: Convert 20 oC to oF TF = 9/5(TC)+ 32o TF = 9/5(20)+ 32o TF = 36 + 32o TF = 68 oF

19. Convert Celsius to Kelvin Tk = Tc + 273 Tc = Tk - 273

20. Ex: Convert 72 oF to K TC = 5/9(TF – 32o) TC = 5/9(72 – 32o) TC = 22.2 oC

21. Tk = Tc + 273 Tk = 22.2 + 273 Tk = 295.2 K

22. Heat • Energy transferred from one body to another due to a DT between them.

23. Once its absorbed by the 2nd body/material it becomes internal energy.

24. Heat is energy in transit. • Heat flows from high to low temperatures.

25. Heat will flow out of the body at a higher temperature and into a body at a cooler temperature.

26. When the heat flows, the objects are said to be in thermal contact.

27. Two things can happen: • The temperature rises. • The object changes state.

28. Thermal Equilibrium

29. The state in which 2 bodies in physical contact with each other have identical temperatures. • No heat flows between them

30. Internal Energy The energy of a substance due to the random motions of its component particles and equal to the total energy.

31. Quantity of Heat • When heat is absorbed it raises the temp. or when it’s lost it lowers the temperature.

32. Unit for heat is the calorie.

33. calorie • The amount of heat energy required to raise the temperature of 1 gram of water 1 oC.

34. 1 kilocalorie (1000 calories) is used in rating food. • Written as Calorie (capital C)

35. Both are units of energy. • 1 calorie = 4.187 J • BTU – British Thermal Unit (English Unit)

36. Fuels are rated by how much heat is given off when a certain amount is burnt.

37. Heat Transfer • Specific Heat (c) • amount of heat energy required to raise the temp. of 1 kg of material by 1 degree Kelvin or Celsius • units: J/(kg·K) J/(g·°C), or cal/(kg·K) cal/(g·°C)

38. 50 g Al 50 g Cu Heat Transfer • Which sample will take longer to heat to 100°C? • Al - It has a higher specific heat. • Al will also take longer to cool down.

39. – Q = heat loss + Q = heat gain T = Tf - Ti Heat Transfer Q = mcT Q: heat (J) m: mass (kg) T: change in temperature (K or °C) Cp: specific heat (J/kg·K or J/g.oC)

40. Q = quantity of heat • m = mass of the substance • c = specific heat of the substance • DT = Temperature Difference

41. Coffee cup Calorimeter Heat Transfer • Calorimeter • device used to measure changes in thermal energy • in an insulated system, heat gained = heat lost

42. Specific heat of water is 4190 J/kg-Co • On HPg. 220 is a table of Specific Heat for different substances.

43. Ex : A 0.5 kg cast iron skillet is heated from 20 Co to 55 Co. How much heat is was absorbed by the iron?

44. m = .5 kg c = 448 J/kg-Co DT = (55 – 20) = 35 Co Q = ?

45. Q = mcDT Q=(.5 kg)(448 J/kg-Co) (35Co) Q = 7840 J

46. Ex: A 1 kg of lead at 100 oC is dropped into a bucket containing 1 kg of water at 0 oC. What is the final temperature of lead and water when it reaches equilibrium?

47. We know the heat lost by the lead is gained by the water. Qlead = Qlost = ? m = 1 kg c = 128 J/kg- oC DT = (100 oC – TF)

48. Qlost = (1 kg) (128 J/kg- Co) (100 oC - TF) • Qlost = (12800 - 128TF) J

49. Qwater = Qgained = ? m = 1 kg c = 4190 J/kg- oC DT = (TF – 0 oC)

50. Qgained = (1 kg)(4190 J/kg-Co) (TF - 0 oC) Qgained = (4190 TF )J