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Lets start with a new symbol

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Title: Chapter 11 Author: Marilyn Akins Last modified by: tt Created Date: 8/11/2002 10:41:27 PM Document presentation format: On-screen Show (4:3) Company – PowerPoint PPT presentation

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Title: 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 mc?T
  • Heat massspecific heatTemp. Change
  • If ?T is positive, material gains heat.
  • If ?T 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 mc?T
  • Q 6.2kg4186 J/KgC75C

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. Dont 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.
  • ?Qhot - ?Qcold
  • Know Q mc?T for final.

12
Easy problem
  • Know for JCCC Final
  • ?Qhot - ?Qcold
  • mc?Thot mc?Tcold

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?
  • mc?Thot mc?Tcold
  • 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 mc?T
  • m Q /c?T

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

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 -gt Ice from -30 to 0
  • Temp. change, so use specific heat.
  • Q1 micecice?Tice
  • .001kg 2090 J/kgº C (0 - -30)

34
Melt the Ice
  • Qtotal 63 J Q2 Q3 Q4 Q5
  • Q2 -gt Ice from 0 to water _at_ 0
  • Phase change, so use latent heat.
  • Q2 micelfice
  • .001kg 3.33 X 105 J/kg

35
Warm the Water
  • Qtotal 63 J 333 J Q3 Q4 Q5
  • Q3 -gt Water from 0 to 100
  • Temp. change, so use specific heat.
  • Q3 mwatercwater?Twater
  • .001kg 4186 J/kgº C (100 - 0)

36
Boil the water
  • Qtotal 63 J 333 J 419 J Q4 Q5
  • Q4 -gt Ice from 0 to water _at_ 0
  • Phase change, so use latent heat.
  • Q4 mwaterlfwater
  • .001kg 2.26 x 106 J/kg

37
Warm the Steam
  • Qtotal 63 J 333 J 419 J 2260 J Q5
  • Q5 -gt Steam from 100 to 120
  • Temp. change, so use specific heat.
  • Q5 msteamcsteam?Tsteam
  • .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
  • S?Q 0
  • Or
  • ?Q1 ?Q2 ?Q3 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?
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