about temperature,heat and thermodynamics

about temperature,heat and thermodynamics

1. Chapter 17. Quantity of HeatChapter 17. Quantity of Heat
A PowerPoint Presentation byA PowerPoint Presentation by
Paul E. Tippens, Professor of PhysicsPaul E. Tippens, Professor of Physics
Southern Polytechnic State Southern Polytechnic State
UniversityUniversity
© 2007

2. FOUNDRY: It requires about 289 Joules of heat to
melt one gram of steel. In this chapter, we will
define the quantity of heat to raise the temperature
and to change the phase of a substance.
Photo © Vol. 05
Photodisk/Getty

3. Objectives: After finishing this Objectives: After finishing this
unit, you should be able to:unit, you should be able to:
•Define the quantity of heat in terms of the calorie,
the kilocalorie, the joule, and the Btu.
•Write and apply formulas for specific
heat capacity and solve for gains and
losses of heat.
•Write and apply formulas for
calculating the latent heats of fusion
and vaporization of various materials.

4. Heat Defined as EnergyHeat Defined as Energy
Heat is not something an object has, but rather
energy that it absorbs or gives up. The heat lost by
the hot coals is equal to that gained by the water.
Hot coals
Cool
water
Thermal Equilibrium

5. Units of HeatUnits of Heat
One calorie (1 cal) is the quantity of heat required
to raise the temperature of 1 g of water by 1 C
0
.
10 calories of heat will
raise the temperature of
10 g of water by 10 C
0
.
ExamplExampl
ee

6. Units of Heat (Cont.)Units of Heat (Cont.)
10 kilocalories of heat will
raise the temperature of 10
kg of water by 10 C
0
.
ExamplExampl
ee
One kilocalorie (1 kcal) is the quantity
of heat required to raise the temperature
of 1 kg of water by 1 C
0
.

7. Units of Heat (Cont.)Units of Heat (Cont.)
10 Btu of heat will raise
the temperature of 10 lb of
water by 10 F
0
.
ExamplExampl
ee
One British Thermal Unit (1 Btu) is the
quantity of heat required to raise the
temperature of 1 lb of water by 1 F
0
.

8. The Btu is an Outdated UnitThe Btu is an Outdated Unit
The British Thermal Unit (1 Btu) is discouraged, but
unfortunately remains in wide-spread use today. If it
is to be used, we must recognize that the pound unit
is actually a unit of mass, not weight.
1 lb (1/32) slug
When working with the BtuBtu, we must
recall that the pound-masspound-mass is not a
variable quantity that depends on
gravity --
one reason that the use of one reason that the use of
the Btu is discouraged!the Btu is discouraged!
1 lb

9. The SI Unit of HeatThe SI Unit of Heat
Since heat is energy, the joule is the preferred
unit. Then, mechanical energy and heat are
measured in the same fundamental unit.
1 cal = 4.186 J
Comparisons of Heat Units:Comparisons of Heat Units:
1 kcal = 4186 J
1 Btu = 778 ft lb
1 Btu = 252 cal
1 Btu = 1055 J

10. Temperature and Quantity of Temperature and Quantity of
HeatHeat
200 g
600 g
20
0
C
20
0
C
22
0
C
30
0
C
The effect of heat on temp-
erature depends on the
quantity of matter heated.
The same quantity of heat
is applied to each mass of
water in the figure.
The larger mass
experiences a smaller
increase in temperature.

11. Heat CapacityHeat Capacity
The heat capacity of a substance is the heat
required to raise the temperature a unit degree.
LeadGlassAlCopperIron
Heat capacities based on time to heat from zero
to 100
0
C. Which has the greatest heat capacity?
37 s52 s60 s83 s90 s
100
0
C100
0
C100
0
C 100
0
C100
0
C

12. Heat Capacity (Continued)Heat Capacity (Continued)
LeadGlassAlCopperIron
Iron and copper balls melt all the way through;
others have lesser heat capacities.
All at 100
0
C placed on Paraffin Slab
LeadGlassAlCopperIron

13. Specific Heat CapacitySpecific Heat Capacity
The specific heat capacity of a material is the
quantity of heat needed to raise the
temperature of a unit mass through a unit
degree.
;
Q
c Q mc t
m t
  

Water: c = 1.0 cal/g C
0
or 1 Btu/lb F
0
or 4186 J/kg K
Copper: c = 0.094 cal/g C
0
or 390 J/kg K

14. Comparison of Heat Units:Comparison of Heat Units: How much heat is How much heat is
needed to raise 1-kg of water from 0needed to raise 1-kg of water from 0
00
to to
100100
00
C?C?
The mass of one kg of water is:
1 kg = 1000 g = 0.454 lb
m
1 kg
Q mc t 
For water: c = 1.0 cal/g C
0

or 1 Btu/lb F
0
or 4186 J/kg K
1 lb
m
= 454 g
The heat required to do this job The heat required to do this job
is:is:
10,000 cal 10 kcal
39.7 Btu 41, 860 J

15. Problem Solving ProcedureProblem Solving Procedure
;
Q
c Q mc t
m t
  

Water: c = 1.0 cal/g C
0
or 1 Btu/lb F
0
or 4186 J/kg K
1. Read problem carefully and draw a rough sketch.
2. Make a list of all given quantities
3. Determine what is to be found.
4. Recall applicable law or formula and constants.
5. Determine what was to be found.

16. Example 1:Example 1: AA 500-g 500-g copper coffee copper coffee
mug is filled with mug is filled with 200-g200-g of coffee. of coffee.
How much heat was required to heat How much heat was required to heat
cup and coffee from cup and coffee from 2020 to to 9696
00
CC??
1. Draw sketch of problem1. Draw sketch of problem.
2. List given information.2. List given information.
Mug massMug mass mm
mm = = 0.500 kg0.500 kg
Coffee massCoffee mass mm
cc = = 0.200 kg0.200 kg
Initial temperature of coffee and mug:Initial temperature of coffee and mug: tt
00 = 20 = 20
00
CC
Final temperature of coffee and mug:Final temperature of coffee and mug: tt
ff = 96 = 96
00
CC
Total heat to raise temp-
erature of coffee (water) and mug to 96
0
C.
3. List what is to be found: 3. List what is to be found:

17. Example 1(Cont.):Example 1(Cont.): How much heat needed to How much heat needed to
heat cup and coffee from heat cup and coffee from 2020 to to 9696
00
CC??
mm
mm = = 0.2 kg0.2 kg; m; m
w w = = 0.5 kg0.5 kg..
4. Recall applicable formula or law:4. Recall applicable formula or law:
Q = mc tHeat Gain or Loss:
5. Decide that TOTAL heat is that 5. Decide that TOTAL heat is that
required to raise temperature of mug required to raise temperature of mug
and water (coffee). Write equation.and water (coffee). Write equation.
QQ
TT = = mm
mmcc
mm t + mt + m
wwcc
w w tt
6. Look up specific 6. Look up specific
heats in tables:heats in tables:
Copper: cCopper: c
mm = 390 J/kg C = 390 J/kg C
00

Coffee (water): cCoffee (water): c
ww = 4186 J/kg C = 4186 J/kg C
00

18. t = 96
0
C - 20
0
C
= 76 C
0
Water: (0.20 kg)(4186 J/kgC
0
)(76 C
0
)
Cup: (0.50 kg)(390 J/kgC
0
)(76 C
0
)
Q
T = 63,600 J + 14,800 JQ
T
= 78.4 kJ
7. Substitute info and solve problem:7. Substitute info and solve problem:
Q
T
= m
m
c
m
t + m
w
c
w
t
Copper: cCopper: c
mm = 390 J/kg C = 390 J/kg C
00

Coffee (water): cCoffee (water): c
ww = 4186 J/kg C = 4186 J/kg C
00

Example 1(Cont.):Example 1(Cont.): How much heat needed How much heat needed
to heat cup and coffee from to heat cup and coffee from 2020 to to 9696
00
CC??
mm
cc = = 0.2 kg0.2 kg; m; m
w w = = 0.5 kg0.5 kg..

19. A Word About UnitsA Word About Units
The substituted units must be consistent with those of The substituted units must be consistent with those of
the chosen value of specific heat capacity.the chosen value of specific heat capacity.
QQ
== mm
wwcc
w w tt
For example: Water c
w = 4186 J/kg C
0
or 1 cal/g C
0
The units for The units for QQ, , m, m, and and t t
must be consistent with must be consistent with
those based on the value of those based on the value of
the constant the constant c.c.
If you use 4186 J/kg C
0
for c,
then Q must be in joules, and
m must be in kilograms.
If you use 1 cal/g C
0
for c,
then Q must be in calories,
and m must be in grams.

20. Conservation of EnergyConservation of Energy
Whenever there is a transfer of heat within a
system, the heat lost by the warmer bodies must
equal the heat gained by the cooler bodies:
Hot
iron
Cool
water
Thermal Equilibrium
 (Heat Losses) =  (Heat Gained)

21. Example 2:Example 2: A handful of copper A handful of copper
shot is heated to shot is heated to 9090
00
CC and then and then
dropped into dropped into 80 g80 g of water in an of water in an
insulated cup at insulated cup at 1010
00
CC. If the . If the
equilibrium temperature is equilibrium temperature is 1818
00
CC, ,
what was the mass of the what was the mass of the
copper?copper?
90
0
shot
10
0

water
Insulator
t
e
= 18
0
C
c
w
= 4186 J/kg C
0
; c
s
= 390 J/kg C
0
m
w = 80 g; t
w= 10
0
C; t
s = 90
0
C
Heat lost by shot = heat gained by water
m
s
c
s
(90
0
C - 18
0
C) = m
w
c
w
(18
0
C - 10
0
C)
Note: Temperature differences are [High - Low] to
insure absolute values (+) lost and gained.

22. 2679 J
0.0954 kg
28,080 J/kg
sm  m
s
= 95.4 g
m
s
(390 J/kgC
0
)(72 C
0
) = (0.080 kg)(4186 J/kgC
0
)(8 C
0
)
m
s
c
s
(90
0
C - 18
0
C) = m
w
c
w
(18
0
C - 10
0
C)
90
0
shot
10
0

water
Insulator
18
0
C
Heat lost by shot = heat gained by water
Example 2: (Cont.)Example 2: (Cont.)
80 g of Water
m
s = ?

23. Change of PhaseChange of Phase
Solid Liquid
Gas
Q = mL
f
Q = mL
v
fusion
Vaporization
When a change of phase occurs, there is only a
change in potential energy of the molecules. The
temperature is constant during the change.
Terms: Fusion, vaporization, condensation, latent
heats, evaporation, freezing point, melting point.

24. Change of PhaseChange of Phase
The The latent heat of fusionlatent heat of fusion ( (LL
ff) of a substance is ) of a substance is
the heat per unit mass required to change the the heat per unit mass required to change the
substance from the solid to the liquid phase of substance from the solid to the liquid phase of
its melting temperature.its melting temperature.
The The latent heat of vaporizationlatent heat of vaporization ( (LL
vv)) of a of a
substance is the heat per unit mass required substance is the heat per unit mass required
to change the substance from a liquid to a to change the substance from a liquid to a
vapor at its boiling temperature.vapor at its boiling temperature.
For Water: L
f
= 80 cal/g = 333,000 J/kg
For Water: L
v = 540 cal/g = 2,256,000 J/kg
f
Q
L
m

v
Q
L
m


25. Melting a Cube of CopperMelting a Cube of Copper
The heat The heat QQ required to melt a required to melt a
substance at its melting temperature substance at its melting temperature
can be found if the can be found if the massmass and latent and latent
heat of fusionheat of fusion are known. are known.
Q = mL
v
2 kg
What Q
to melt
copper?
L
f = 134 kJ/kg
Example:Example: To completely melt To completely melt
2 kg of copper at 10402 kg of copper at 1040
00
C, we need:C, we need:
Q = mLQ = mL
ff = (2 kg)(134,000 J/kg)= (2 kg)(134,000 J/kg)Q = 268 kJ

26. Example 3:Example 3: How much heat is needed to How much heat is needed to
convert convert 10 g10 g of ice at of ice at -20-20
00
CC to steam at to steam at
100100
00
CC??
First, let’s review the process graphically as shown:
temperature
t
Qice
steam
only
-20
0
C
0
0
C
100
0
C
steam
and
water
540 cal/g
ice and
water
80 cal/g
water
only
1 cal/gC
0
icesteam
c
ice
= 0.5 cal/gC
0

27. Example 3 (Cont.):Example 3 (Cont.): Step one is Q Step one is Q
11 to convert to convert
10 g of ice at 10 g of ice at -20-20
00
CC to ice at to ice at 00
00
CC (no water (no water
yet).yet).
t
Qice
-20
0
C
0
0
C
100
0
C
c
ice
= 0.5 cal/gC
0
Q
1
= (10 g)(0.5 cal/gC
0
)[0 - (-20
0
C)]
Q
1
= (10 g)(0.5 cal/gC
0
)(20 C
0
)
Q
1 = 100 cal
-20
0
C0
0
C
Q
1 to raise ice to 0
0
C: Q
1 =
mct

28. t
Q
-20
0
C
0
0
C
100
0
C
Example 3 (Cont.):Example 3 (Cont.): Step two is Q Step two is Q
22 to convert to convert
10 g of ice at 10 g of ice at 00
00
CC to water at to water at 00
00
CC..
Melting
Q
2 to melt 10 g of ice at 0
0
C: Q
2 = mL
f
80 cal/g
ice and
water
Q
2
= (10 g)(80 cal/g) = 800 cal
Q
2
= 800 cal
Add this to Q
1
= 100 cal:
900 cal used to this point.

29. t
Q
-20
0
C
0
0
C
100
0
C
water
only
1 cal/gC
0
Example 3 (Cont.):Example 3 (Cont.): Step three is Q Step three is Q
33 to change to change
10 g10 g of water at of water at 00
00
CC to water at to water at 100100
00
CC..
0
0
C to 100
0
C
Q
3 to raise water at 0
0
C to 100
0
C.
Q
3 = mct ; c
w= 1 cal/gC
0

Q
3
= (10 g)(1 cal/gC
0
)(100
0
C - 0
0
C)
Q
3
= 1000 cal
Total = Q
1
+ Q
2
+ Q
3
= 100 +900 + 1000
= 1900 cal

30. Example 3 (Cont.):Example 3 (Cont.): Step four is Q Step four is Q
44 to convert to convert
10 g of water to steam at 10 g of water to steam at 100100
00
CC? (? (QQ
44 = mL = mL
vv))
Q
-20
0
C
0
0
C
100
0
C
vaporization
Q
4 to convert all water at 100
0
C
to steam at 100
0
C. (Q = mL
v
)
Q
4
= (10 g)(540 cal/g) = 5400 cal
100 cal
ice
water
only
ice and
water
800 cal
1000
calsteam
and
water
5400 calTotal Heat:
7300 cal

31. Example 4:Example 4: How many grams of ice at How many grams of ice at 00
00
CC
must be mixed with four grams of steam must be mixed with four grams of steam
in order to produce water at in order to produce water at 6060
00
CC??
Ice must Ice must meltmelt and then and then riserise to 60 to 60
00
C. C.
Steam must Steam must condensecondense and and dropdrop to 60 to 60
00
C.C.
Total Heat Gained = Total Heat Lost
m
i
L
f
+ m
i
c
w
t = m
s
L
v
+ m
s
c
w
t
Note: All losses and gains are absolute values (positive).Note: All losses and gains are absolute values (positive).
Total Gained: m
i
(80 cal/g) + m
i
(1 cal/gC
0
)(60 C
0
- 0
0
C

)
Lost: (4 g)(540 cal/g) + (4 g)(1 cal/gC
0
)(100 C
0
- 60
0
C

)
Total Gained: m
i
(80 cal/g) + m
i
(1 cal/gC
0
)(60 C
0
)
Total Lost: (4 g)(540 cal/g) + (4 g)(1 cal/gC
0
)(40 C
0
)
m
i
= ?
4 g
t
e
= 60
0
C
ice
steam

32. Total Gained: m
i
(80 cal/g) + m
i
(1 cal/gC
0
)(60 C
0
)
Total Lost: (4 g)(540 cal/g) + (4 g)(1 cal/gC
0
)(40 C
0
)
m
i
= ?
4 g
t
e = 60
0
C
80m
i
+ 60m
i
= 2160 g +160 g
Total Heat Gained = Total Heat Lost
2320 g
140
i
m m
i
= 16.6 g
Example 4 (Continued)Example 4 (Continued)

33. Example 5:Example 5: Fifty gramsFifty grams of ice of ice
are mixed with are mixed with 200 g200 g of of
water initially at water initially at 7070
00
CC. Find . Find
the equilibrium temperature the equilibrium temperature
of the mixture.of the mixture.
Ice melts and rises to t
e
Water drops from 70 to t
e
.
Heat Gained: m
i
L
f
+ m
i
c
w
t ; t = t
e
- 0
0
C
Gain = 4000 cal + (50 cal/g)t
e
Gain = (50 g)(80 cal/g) + (50 g)(1 cal/gC
0
)(t
e - 0
0
C

)
0
0
C 70
0
C
t
e
= ?
50 g 200 g
icewater

34. Example 5 (Cont.):Example 5 (Cont.):
0
0
C 70
0
C
t
e
= ?
50 g 200 g
GainGain = 4000 cal + (50 cal/g)t
e
Lost = (200 g)(1 cal/gC
0
)(70
0
C- t
e
)
Heat Lost = m
w
c
w
t
Lost = 14,000 cal - (200 cal/C
0
) t
e
t = 70
0
C - t
e [high - low]
Heat Gained Must Equal the Heat Lost:
4000 cal + (50 cal/g)t
e
= 14,000 cal - (200 cal/C
0
) t
e

35. 0
0
C 70
0
C
t
e = ?
50 g 200 g
Simplifying, we have:(250 cal/C
0
) t
e = 10,000 cal
0
0
10,000 cal
40 C
250 cal/C
e
t 
t
e = 40
0
C
Heat Gained Must Equal the Heat Lost:
4000 cal + (50 cal/g)t
e = 14,000 cal - (200 cal/C
0
) t
e
Example 5 (Cont.):Example 5 (Cont.):

36. Summary of Heat UnitsSummary of Heat Units
One calorie (1 cal) is the quantity of heat
required to raise the temperature of 1 g of
water by 1 C
0
.
One kilocalorie (1 kcal) is the quantity of
heat required to raise the temperature of 1
kg of water by 1 C
0
.
One British thermal unit (Btu) is the
quantity of heat required to raise the
temperature of 1 lb of water by 1 F
0
.

37. Summary: Change of Phase Summary: Change of Phase
The The latent heat of fusionlatent heat of fusion ( (LL
ff) of a substance is ) of a substance is
the heat per unit mass required to change the the heat per unit mass required to change the
substance from the solid to the liquid phase of substance from the solid to the liquid phase of
its melting temperature.its melting temperature.
For Water: L
f
= 80 cal/g = 333,000 J/kg
f
Q
L
m

The The latent heat of vaporizationlatent heat of vaporization ( (LL
vv)) of a of a
substance is the heat per unit mass required substance is the heat per unit mass required
to change the substance from a liquid to a to change the substance from a liquid to a
vapor at its boiling temperature.vapor at its boiling temperature.
For Water: L
v = 540 cal/g = 2,256,000 J/kg
v
Q
L
m


38. Summary: Specific Heat Summary: Specific Heat
CapacityCapacity
The specific heat capacity of a
material is the quantity of heat to
raise the temperature of a unit
mass through a unit degree.
;
Q
c Q mc t
m t
  


39. Summary: Conservation of Summary: Conservation of
EnergyEnergy
Whenever there is a transfer of
heat within a system, the heat lost
by the warmer bodies must equal
the heat gained by the cooler
bodies:
 (Heat Losses) =  (Heat Gained)

40. Summary of Formulas:Summary of Formulas:
;
Q
c Q mc t
m t
  

 (Heat Losses) =  (Heat Gained)
;
v v
Q
L Q mL
m
 
;
f f
Q
L Q mL
m
 

41. CONCLUSION: Chapter 17CONCLUSION: Chapter 17
Quantity of HeatQuantity of Heat