Younes Sina's Lecture at Pellissippi State Community College

1. Chapter 12
Temperature and Heat
Younes Sina

2. Temperature
The temperature of an object is a measure of how cold or hot that
object is. More precisely, the temperature of an object is a measure of
the average kinetic energy of the atoms and molecules of that object.
A hotter object has faster molecular vibrations. Temperature is a scalar
quantity.
Temperature Scales
Fahrenheit and Rankin (in the American system of units)
Celsius and Kelvin (in the Metric System)
Fahrenheit and Celsius are regular scales.
Rankin and Kelvin are absolute scales.

7. Example : The Fahrenheit scale reads 77o F, what is the reading on
the Celsius scale?
Solution:
°C = (5/9)[ F - 32 ]
oC = (5/9) [ 77 - 32] = 25
Example : A temperature difference of ΔC = 24oC is measured
between two points on Celsius scale. How much is this difference in
Fahrenheit scale?
Solution: 100oC difference on the Celsius scale corresponds to 180oF
difference on the Fahrenheit scale. Using a proportion, the difference
in Fahrenheit is
ΔF / ΔC = 180o / 100o
ΔF / 24o = 9/5
ΔF = 24o (1.8)
ΔF = 43oF.
C1 = (5/9)[ F1 - 32 ]
C1 = (5/9)[ F1 - 32
C 2 =(5/9)[ F2 - 32 ]
(C1-C2)= (5/9)[(F1-32)-(F2-32)]
ΔC=(5/9)ΔF

8. Example : At what temperature both Fahrenheit and Celsius scales read
the same temperature?
Solution:
°C = (5/9)[ F - 32 ]
F = (5/9)[ F - 32 ]
F=(5/9)F-17.777
(4/9)F=-17.777
F=-40

9. Absolute Scales:
The basis for absolute scales (Kelvin and Rankin scales) is the
temperature at which molecular motion ceases and stops. This
temperature cannot actually be reached; however, with great cooling,
temperatures very close to it have been reached. Experiments have
shown that when a gas is cooled down, its volume decreases.
At constant pressure, the volume decrease for a gas, is proportional to
the temperature decrease. In other words, The ratio ΔV/ΔT remains
constant. That means that if V( the gas volume) is plotted versus T
(the gas temperature) while pressure is kept constant, the graph is a
straight line as shown below:

10. Heat
Heat is a form of energy that transfers due to a temperature difference.
Units of Heat
The familiar unit often heard is "calorie."
One calorie (1 cal) is the amount of heat energy that can raise the temperature
of 1 gram of pure water by 1 oC.
Parallel to this definition is that of kilocalorie (kcal).
1 kcal is the amount of heat energy that can raise the temperature of 1 kg of
pure water by 1 oC.
A non-Metric unit for heat energy is Btu (British thermal unit).
1 Btu is the amount of heat energy that can raise the temperature of 1 lbm of
pure water by 1 oF.

11. Specific Heat (c)
Different substances take different amounts of heat energy for one unit
of mass of them to warm up by one degree. For example, if you place
1 kg of copper and 1 kg of aluminum on a burner, after the same length
of time, the copper piece will be much hotter than the aluminum. The
reason is the difference in their specific heat. Copper takes much less
heat to warm up compared to aluminum.

12. The specific heat (c) of a substance is the amount of heat that 1kg of
substance takes to warm up by 1oC. On this basis, the specific heat of
water is 1 kcal /(kg oC). The specific heat of a few elements are given
below:
Cwater =1.000 kcal / ( kg oC) or, cwater = 1.000 cal /(g oC)
ciron = 0.108 kcal / ( kg oC) or, ciron = 0.108 cal /(g oC)
cAl = 0.215 kcal / ( kg oC) or, cAl = 0.215 cal /(g oC)
Cbrass = 0.0924 kcal / ( kg oC) or, cbrass= 0.0924 cal /(g oC)

13. Heat calculation:
When heat is given to a pure substance or taken from it, its
temperature starts changing if a phase change does not start. During
a phase change (solid to liquid, liquid to solid, liquid to vapor, or
vapor to liquid ),temperature remains constant. We will look at the
following two cases:
1) Heat calculation for temperature change
2) Heat calculation for phase change

14. Heat Calculation When Temperature changes
Q = McΔT
No Phase Change
Mass of the object
specific heat
temperature change

15. Example :
Calculate the amount heat that must be given to 2.14 kg of iron to
warm up from 24.0 oC to 88.0 oC.
Solution:
Q = McΔT
Q = ( 2.14kg )[0.108 kcal /(kg oC)] ( 88 - 24 )oC = 14.8 kcal
ΔE

16. Example :
37.0 cal of heat is given to 2.00 gram of water at 12.0 oC. Find its
final temperature.
Solution:
Q = McΔT
ΔT = Q / Mc
ΔT = (37cal) / [(2gr)(1cal/ gr oC)] = 18.5oC
Tf - Ti = 18.5 °C
Tf - 12.0 oC = 18.5 oC
Tf = 30.5 oC
Example :
Calculate the amount heat that must be taken from 5.00 kg of
Aluminum to cool it down from 230oC to 30oC.
Solution:
Q = McΔT
Q =(5.00kg )[0.215 kcal / ( kg oC )](30 - 230 )oC = - 215 kcal

17. Thermal Equilibrium
- [ heat loss by hotter objects ] = Heat gain by colder objects

18. Example :
A 65-gram piece of aluminum at 180 oC is removed from a stove and
placed in 45 grams of water initially at 22 oC. Find the equilibrium
temperature (Teq).

19. Solution:
-[ heat loss by hotter objects ] = heat gain by colder objects
Q = McΔT
-Mc [Tf - Ti] Al = Mc[Tf - Ti] water
- (65)(0.215)(Teq - 180) = (45)(1.00)(Teq - 22)
-14(Teq - 180) = (45)(Teq - 22)
-14Teq + 2520 = 45Teq - 990
3510 = 59Teq
Teq = 59 oC

20. Example :
A 225-gram piece of hot iron is removed from an electric oven at an
unknown temperature. It is known that the iron piece has been in the
oven long enough so that its initial temperature can be thought as the
temperature of the oven. The iron piece is placed in 75.0 grams of water
that is held by a 45.0-gram aluminum container initially at an equilibrium
temperature of 25.0 oC. The final equilibrium temperature of iron,
aluminum, and water becomes 41.0 oC. Find the initial temperature
of iron (oven). Assume that the whole system is thermally isolated
from the surroundings.
hot iron
225 gram
75.0 gram
water
45.0 gram
aluminum
25.0 oC
41.0 oC

21. Solution:
-[ heat loss by hotter objects ] = heat gain by colder objects
- Mc [Tf - Ti]iron = Mc [Tf - Ti]Al + Mc [Tf - Ti]water
-(225)(0.108)(41-Ti ) = (45)(0.215)(41-25) + (75)(1)(41-25)
-24.3(41-Ti ) = 1354.8
-41-Ti = 55.8
Ti = 96.8 oC

22. Heat Calculation
Phase Change Only
During a phase change such as following, temperature remains constant:
solid to liquid
liquid to solid
liquid to vapor
vapor to liquid

23. Q = M Lf
latent heat of fusion
Lf is measured for different substances at their melting/freezing points or
temperatures. Typical values may be found in texts or handbooks.

24. Example : How much heat should be removed from 250 grams of water
already at its freezing point (0 oC) to convert it to ice at (0 oC)? The latent
heat of fusion (freezing) for water is Lf = 80 cal/gr.
Solution: In this problem, the heat calculation involves a phase change
only. There is no temperature change.
Q = M Lf
Q = (250gr)(80 cal/gr ) = 20,000 cal
0 oC 0 oC

25. Example :
How much heat should be given to 250 gr of ice at 0oC to convert it to water
at 40.oC ? The latent heat of freezing/melting for water is Lf = 80 cal /gr and
the specific heat of water is 1 cal/[gr oC].
Solution: In this problem, the heat calculation involves a phase change and
a temperature change.
Q = M Lf + Mc (Tf - Ti)
Q = (250)(80) + (250)(1)(40 - 0) = 30,000cal

26. -20 oC
0 oC
250 gr
100 oC
Example :
How much heat should be given to 250 gr of ice at -20 oC to bring it to boil (100 oC)?
0 oC

27. Solution:
This problem has 3 steps:
A temperature change ( ice from -20oC to 0oC )
A phase change ( ice at 0oC to water at 0°C )
A temperature change ( water from 0oC to water at 100oC)
cice = 0.48 cal / (gr oC)
Lf = 80 cal/gr
cwater = 1cal / (gr oC)
Q = Mcice (Tf - Ti) + M Lf + Mcwater (Tf - Ti)
Q = (250)(0.48)[ 0 - (-20) ] + (250)( 80) + (250)( 100 - 0) = 47,000 cal
2400cal 20,000cal 25,000cal
phase changetemperature change temperature change

28. Example :
Draw a diagram that shows temperature change vs. heat consumed for previous Example.
Solution:

29. Example :
2650 cal of heat is given to 225 grams of ice at -12.0 oC. Find the final
temperature of the result.
Solution: All of the ice may not melt. Let's first calculate the heat necessary
for the ice to warm up to 0 oC-ice.
Q = Mc ( Tf - Ti ) = (225gr)(0.480 cal/gr oC)[ 0 - (-12.0)]oC = 1300 cal
The remaining heat is 2650cal - 1300cal = 1350 cal
Now, let's see if this heat is enough to melt the ice
Q = MLf = (225gr)(80. cal /gr ) = 18000 cal
No, 1350 cal is not enough to melt all of the ice. We need to see how much of
the ice does melt.
Q = MLf
1350cal = M (80 cal / gr)
M = 17 grams.
The final result is a mixture of
17 grams of water and (225 - 17) grams
of ice, of course both at 0oC.

30. Example :
26500 cal of heat is given to 225 grams of ice at -12.0 oC. Find the final temperature.
Solution:
All of the ice may not melt. First calculate the needed heat for the ice to warm up
to 0 oC-ice.
Q = Mc ( Tf - Ti ) = (225gr)(0.480cal /(gr oC)[ 0.0 - (-12.0)]oC = 1300 cal
The remaining heat is 26500cal - 1300cal = 25200 cal
Now, let's see if this heat is enough to melt the ice.
Q = MLf = (225gr)(80cal /gr ) = 18000 cal
Yes, 25200 cal is enough to melt all of the ice.
The remaining heat is 25200cal - 18000cal = 7200 cal. This heat warms up water
from 0oC to Tf .
Q = Mc ( Tf - Ti )
7200cal = (225gr)(1 cal/gr oC)( Tf - 0oC )
Tf - 0.0oC = 7200 / 225 = 32oC
Tf = 32oC
The final result is 225 grams of water at 32oC.

31. Homework :
problems
chapter 12