Lets start with a new symbol

1. Lets start with a new symbol • Q. Q is for “Heat” • Measured in Joules. • Heat is a form of energy.

2. James Prescott Joule • Discovered that heat was form of energy. • Found that when water is stirred, temperature increases.

3. James Joule • Experimented with falling masses turning paddles in water. • Figured out how much mechanical energy is needed to change temperature. • Got a unit named after him.

4. Units of Heat • Joule is most common heat unit. • Also, there is the calorie. • Calorie: is the amount of energy necessary to raise the temperature of 1 g of water by 1° C . • Units to measure heat. • calorie: Energy needed to raise 1 g of water by 1C. • 1 cal = 4.186 Joules.

5. Calories and calories. • Calories and calories are not the same thing. • If C is uppercase, it is a “dietary calorie” (on your food label.) • One Calorie = 1000 calories One Calorie = 1 kilocalorie One Calorie = 4186 Joules

6. Units of Heat, cont. • Also – BTU • BTU stands for British Thermal Unit • A BTU is the amount of energy necessary to raise the temperature of 1 lb of water 1° F

7. New Symbol • Specific heat = c • Amount of heat needed to raise 1 kg of a material 1 C. • Units J / kg °C • Depends on the substance. Look it up. • Specific heat questions are on the JCCC final.

8. Specific Heat equation • Q = mcΔT • Heat = mass*specific heat*Temp. Change • If ΔT is positive, material gains heat. • If ΔT is negative, material gives up heat.

9. Problem • The Specific Heat of water is 4186 J/kgC. • How much heat is needed to raise 6.2 kg of water from 10 to 85 Celsius? • Do in PP notes. • Specific heat questions like this are on the JCCC final.

10. Problem • The Specific Heat of water is 4186 J/KgC. • How much heat is needed to raise 6.2 kg of water from 10 to 85 Celsius? • Q = mcΔT • Q = 6.2kg*4186 J/KgC*75C

11. Calorimetry • Measurement of heat flow from a hot material to a cold material (usually Water). • Heat flow into water is calculated based upon the temp. change of the water. • Conservation of energy applies to the isolated system. Don’t let any heat escape so that all goes into the water. (Insulated system) • The energy that leaves the warmer substance equals the energy that enters the water. • ΔQhot = - ΔQcold • Know Q = mcΔT for final.

12. Easy problem • Know for JCCC Final: • ΔQhot = - ΔQcold • mcΔThot = mcΔTcold

13. Easy problem • 1 kg of water at 70 C is mixed with 1 kg of water at 60 C. What is the final temp?

14. Easy problem • 1 kg of water at 70 C is mixed with 1 kg of water at 60 C. What is the final temp? • mcΔThot = mcΔTcold • 1 kg x 4186 J/kgK x (70C – Tf) • = 1 kg 4186 J/kgK x (Tf – 60C)

15. Easy problem • 1 kg of water at 70 C is mixed with 1 kg of water at 60 C. What is the final temp? • 2 kg of water at 70 C is mixed with 1 kg of water at 60 C. What is the final temp?

16. Problem • For breakfast you consume 320 dietary Calories. • You decide to work this off by lifting a 25 kg barbell .4 m. • How many times will you have to repeat this motion to burn off breakfast?

17. Problem • For breakfast you consume 320 dietary Calories. • Instead, you (60 kg) decide to spring from rest to 5 m/s. • How many repetitions of this will you need to do?

18. Still talking about breakfast • If instead of eating that food, you set it on fire, how much water could it bring from 0C to 100 C? • Q = mcΔT • m = Q /cΔT

19. What else does heat do? • If heat is added to a substance, what else could happen besides ΔT?

20. Phase Change • Temperature remains constant during phase change. • E.g. Ice will continue to absorb heat at 0 C without changing temp. The temp will not be 1 C until all of the ice has melted. • See graph next slide.

21. Graph of Ice to Steam

22. Phase Change • 3 Basic types of Phase Change: • Melting/Freezing – Liquid/Solid • Boiling/Condensing – Gas/Liquid • Sublimation/Deposition – Gas/Solid

23. New Symbol • l – “l” is for latent heat. • Amount of heat absorbed, per kilogram, for a change from one phase to another. • Heat absorbed during a phase change does not change temperature, so it is “latent.” • Units: J/kg

24. Latent Heat, cont. • Depends on both substance and phase change. Just something to look up. • Latent heat of fusion, Lf, is used for melting or freezing • Latent heat of vaporization, Lv, is used for boiling or condensing • Latent heat of sublimation, Ls, is used for sublimation or deposition.

25. Equation • Q = m l

26. Problem • A gram of H20 is at -30 C. How much heat total is needed to raise it to H20 at 120 C?

27. Multistep processes. • A lot of problems will involve both a temperature change and a phase change. Evaluate the Q for each step separately, and then add them together.

28. Problem • How many steps?

29. Graph of Ice to Steam • How many steps? 5 • 1- Ice to 0 C • 2- Melting • 3- Water to 100 C • 4- Boiling • 5- Steam to 120 C • Qtotal = Q1 + Q2 + Q3 + Q4 + Q5

30. End here in 2012

31. Problem • How many steps? • 5 • 1- Ice to 0 C • 2- Melting • 3- Water to 100 C • 4- Boiling • 5- Steam to 120 C • Qtotal = Q1 + Q2 + Q3 + Q4 + Q5

32. Useful Information • For H20: • cice = 2090 J/kgº C • lf = 3.33 X 105 J/kg • cwater = 4186 J/kgº C • lv = 2.26 x 106 J/kg • csteam = 2010 J/kgº C

33. Warm the Ice • Qtotal = Q1 + Q2 + Q3 + Q4 + Q5 • Q1 -> Ice from -30 to 0 • Temp. change, so use specific heat. • Q1 =mice*cice*ΔTice • = .001kg* 2090 J/kgº C * (0 - -30)

34. Melt the Ice • Qtotal = 63 J + Q2 + Q3 + Q4 + Q5 • Q2 -> Ice from 0 to water @ 0 • Phase change, so use latent heat. • Q2 =mice*lfice • = .001kg* 3.33 X 105 J/kg

35. Warm the Water • Qtotal = 63 J + 333 J + Q3 + Q4 + Q5 • Q3 -> Water from 0 to 100 • Temp. change, so use specific heat. • Q3 =mwater*cwater*ΔTwater • = .001kg* 4186 J/kgº C * (100 - 0)

36. Boil the water • Qtotal = 63 J + 333 J + 419 J + Q4 + Q5 • Q4 -> Ice from 0 to water @ 0 • Phase change, so use latent heat. • Q4 =mwater*lfwater • = .001kg* 2.26 x 106 J/kg

37. Warm the Steam • Qtotal = 63 J + 333 J + 419 J + 2260 J + Q5 • Q5 -> Steam from 100 to 120 • Temp. change, so use specific heat. • Q5 =msteam*csteam*ΔTsteam • = .001kg* 2010 J/kgº C * (120 - 100)

38. Warm the Steam • Qtotal = 63 J + 333 J + 419 J + 2260 J + 40 J Qtotal = 3115 J

39. Graph of Ice to Steam

40. Problem • The latent heat of lead is 24500 J/kg, and the melting point is 327.5 C. How much heat is needed to bring 17 kg of solid lead at 327.5 C to 17 kg of liquid lead at 327.5 C?

41. Problem • Back in October, we learned that the most powerful person in the class was about 1050 Watts. • If we gave him an electrical generator bike, how long would it take him to boil a kg of water from room temp?

42. Problem • 400 g of Copper (c = 386 J/kgC) at 95 C is placed into a kilogram of water at 5 C. What will the final temperature of the mixture be?

43. Another problem. • 10 kilograms of Aluminum (c = 900 J/kgC) are at 130 C. A 50 g Ice cube at -5 C is dropped onto the aluminum and melts, then boils away. What is the final temperature of the Al?

44. More than 2 things • ΣΔQ = 0 • Or • ΔQ1 + ΔQ2 + ΔQ3 … = 0

45. Problem • 400 ml of water at 40 C is poured into a .3 kg glass beaker (cglass = 837 J/C) and a .5 kg block of Al at 37 C (cAl 900 J/C) is dropped in. • What will the equilibrium temp be?