Basic Thermo properties

2. - An indicator of the average thermal energy of the atoms or molecules
in the substance.
THERMAL ENERGY – the random kinetic energy of the atoms & molecules
in the substance.
Consider, a water molecule:
a.
TRANSLATIONAL
- they move or translate from one
position to another.

3. b.
VIBRATIONAL
- they vibrate in and out and
bend back and forth.
c.
ROTATIONAL
- they also spin and rotate as
they move through space.
- because all types of motion occur
simultaneously, there is much variation in the motion of
different molecules at any time resulting to thermal
energy.

4. Thermal energy is related
closely to the idea of
temperature:
If an object becomes
warmer, the motion of the
molecules increases as does the
thermal energy.
If an object becomes
cooler, the motion of the
molecules decreases as the
thermal energy.
This does not mean that all the particles in an object
have exactly the same amount of energy; they have
a wide range of energies.

5. - Also, it measures how hot or cold a body is with respect to a
standard object.
To discuss temperature changes, two basic concepts are
important: thermal contact and thermal equilibrium.
THERMAL CONTACT
- Two objects are in thermal contact if
they can affect each other’s temperature.
THERMAL EQUILIBRIUM
- exists when two objects in thermal
contact no longer affect each other’s
temperature.

6. 1.) ARBITRARY SCALE – based on the freezing & boiling point
of water at atmospheric pressure.
Note:
Celsius scale is name after Swedish astronomer Anders Celsius.
Fahrenheit scale is named after German Gabriel Fahrenheit.
On the Celsius scale, the ice
point is 0, and the steam point is
100. The interval between these
temperatures is divided into 100
equal parts called degrees.
On the Fahrenheit scale, the ice
point is 32 degrees, and the
steam point is 212 degrees. The
interval between these
temperatures is divided into 180
equal parts.

7. 1.) ARBITRARY SCALE – based on the freezing & boiling point
of water at atmospheric pressure.
212
o
F
TF
32
o
F
100
o
C
TC
0
o
C
TF – 32
212 – 32
TC – 0
100 – 0
Tc = 5/9 (TF – 32)
TF = 9/5Tc + 32
Celsius Scale
(°C)
Fahrenheit
Scale (°F)

8. 2.) ABSOLUTE TEMPERATURE SCALE
- in this scale, the lowest possible temperature is called
absolute zero & the temperature cannot be lowered further by
decreasing the motion of the molecules.
ABSOLUTE TEMPERATURE OR ABSOLUTE ZERO – the temperature at which all
molecular motion ceases.
- The zero in Celsius scale is shifted to the triple point of
water.
TRIPLE POINT OF WATER – exists when water within a closed vessel is in
equilibrium in all three states: ice, water, and vapor.
This point is defined as 273.16 Kelvin and equals .01 degrees Celsius;
therefore, to convert Celsius to Kelvin, simply add 273.15.
Note:
Kelvin scale is named after English Physicist William Thomson (Lord Kelvin).
Rankine scale is named after William John M. Rankine.

9. 2.) ABSOLUTE TEMPERATURE SCALE
The Rankine degree is exactly 5⁄9 as large as the Kelvin
degree. Conversely, the Kelvin degree is exactly 9⁄5, or 1.8
times, the size of the Rankine degree.
TK = TC + 273
Kelvin
Scale (K)
TR = TF + 460
Rankine
Scale
(°R)
Note:
Kelvin scale is named after English Physicist William Thomson (Lord Kelvin).
Rankine scale is named after William John M. Rankine.

10. EXAMPLES:
1.) A pan of water is heated from 25°C to
80°C. What is the change in its
temperature on the Kelvin scale and on
the Fahrenheit scale?
Ans: a. 55K
b. 99°F

11. 2.) Suppose that a person’s body
temperature, initially 37oC is
increased by 1% on the Kelvin
Scale. What new temperature
is on the Celsius & Fahrenheit
scale?
EXAMPLES:
Ans: a. Tc = 40.1°C
b. TF = 104.18°F

12. 3.) A Fahrenheit & a Celsius
thermometer are both immersed
in a fluid and indicate that the
Fahrenheit reading is numerically
twice that of the Celsius reading.
What is the temperature of the
fluid in oR & K?
EXAMPLES:
Ans: 433K or 780°R

13. 4.) At which temperature do the
Fahrenheit & Celsius scales
coincide?
EXAMPLES:
Ans: -40°

14. - The energy transferred between a
system and its environment
because of the difference in their
temperatures.
+Q, if a system gains energy
-Q, if a system loses energy

15. - When two objects with different
temperatures are in contact with
each other, heat flows from the
hotter system to the colder one.
An increase or decrease in mechanical energy in a
system always accompanies an equal decrease or
increase of heat, respectively.

16. 1.) SPECIFIC HEAT CAPACITY (C)
– the amount of heat energy necessary to
raise the temperature of an object by one
degree.
Q = mcΔT
Where:
m = mass of the substance
ΔT = change in temperature
= the final temperature of the substance minus its
initial temperature before the heat was added or
removed
Units:
J/KgC
o
, cal/gC
o
, Btu/slug F
o
….
Note:
1J = 2.39x10
-4
Kcal
1Btu = 252 cal = 1.055KJ = 778.2ftlb

17. The calorie is defined as the amount of
energy required to raise 1 gram of water 1
degree.
(This energy is slightly dependent upon the temperature of
the water, so the temperature change is usually defined from
14.5 degrees to 15.5 degrees Celsius.)
The U.S. engineering unit of heat is the British
thermal unit (BTU). It is related to the calorie and
the joule:
BTU = 252 calories = 1.055 kJ.
These reversible conversions of heat energy and
work are called the mechanical equivalent of
heat.

18. Note that the temperature interval for Kelvins is the same as that for
Celsius degrees. For this reason, you can calculate T in Kelvins or in
degrees Celsius.

19. 1.) In a half hour, a 65kg jogger can
generate 8x105 J of heat. This heat is
removed from the jogger’s body by a
variety of means. If the heat were not
removed, how much would the body
temperature increase? Cbody = 3500
J/kgCo
EXAMPLES:
Ans: 3.5C°

20. 2.) You wish to take a bath and will
need to warm 160Kg of water by
14oC. How much heat is
required?
CH20 = 4180 J/KgCo
EXAMPLES:
Ans: Q = 9.4x10
6
J

21. 3.) A 30Kg child has a temperature
of 39oC. How much heat must be
removed from the child’s body to
lower his temperature to 37oC?
EXAMPLES:
Ans: Q = -2.1x10
5
J

22. EXAMPLES:
4.) An insulated aluminum can whose mass
is 0.2Kg contains 0.3Kg of water at 25oC.
A 0.1Kg metal block whose
temperature is 80oC is lowered into the
water. The final temperature of the
water, can & block is 30oC. What is the
specific heat capacity of the metal
block?
Cal=900J/KgCo
CH20=4180J/KgCo
Ans: C = 1434 J/kgC°

23. EXAMPLES:
5.) A 150g insulated aluminum
calorimeter containing 250g of
water is initially at 20oC. A 200g
metal block at 60oC is added to
the water, resulting in the final
temperature of 22.8oC. Calculate
the specific heat capacity of the
block.
Ans: C = 444.1 J/kgC°

24. 2.) LATENT HEAT OF FUSION (Lf)
– the amount of heat needed to melt or freeze a
substance at its melting or freezing
temperature.
Q = mLf
To melt a solid:
To freeze a liquid:
Q = -mLf
Where:
Q = heat needed to melt or
freeze a substance
Lf = latent heat of fusion
m = mass of the substance
Units:
J/Kg , cal/g , Btu/slug ….
Note: a substance does not change temperature
while it is melting or freezing, thus the
melting or freezing temperature of a
substance are the same.

25. Heat Fusion of Some Substances:
Substances Lf (kJ/kg)
Water 335
Ethanol 1042
Hydrogen (H2) 58.6
Oxygen (O2) 13.8
Nitrogen (N2) 25.7
Aluminum 398.4
Copper 134
Iron 272.1

26. Phase changes of water as heat is added:
In regions I, III, and V, the
addition of heat energy
increases the temperature
of the sample.
However, in regions II and IV,
additional heat does not
cause a change in
temperature because heat is
required to change the state.

27. 1. Suppose that a certain substance melts and
freezes at 400°C. Imagine a block of this
material whose mass is 1.535 kg, and it is
entirely solid at 400°C. It is subjected to heating,
and it melts. Suppose that it takes 142,761 cal
of energy to melt the substance entirely into
liquid at 400°C. What is the heat of fusion for
this material?
EXAMPLES:
Ans: 93 cal/g

28. 2. Suppose that you are camping in the
mountains. You need to melt 1.50 kg of snow at
0.0°C and heat it to 70.0°C to make hot cocoa.
How much heat will be needed?
Cwater = 4180J/kgCo , Lfwater = 334kJ/kg
EXAMPLES:
Ans: 940 kJ

29. 3. How much heat is absorbed by 100 g of ice at
-20.0°C to become water at 0.0°C?
Cice = 2060J/kgCo , Lfwater = 334kJ/kg
EXAMPLES:
Ans: 37.5 kJ

30. 4. Your 500mL soda is at 20°C so you add 100g of
ice from the -20°C freezer. What is the final
temperature?
Cice = 2090J/kgCo , Lfwater = 333kJ/kg
Cwater = 4190J/kgCo
EXAMPLES:
Ans: 1.76°C

31. EXAMPLES:
5.) The 1x107Kg of ice in a small pond
has an average temperature of -
5oC during the middle of winter.
How much heat must be added to
the ice to convert it to water at 27oC
for summer swimming?
Cice=2090J/kgCo
Ans: Qt = 46x10
11
J

32. 3.) LATENT HEAT OF VAPORIZATION (Lv)
– the amount of heat needed to vaporize a
substance at its boiling temperature.
Q = mLv
To vaporize a liquid: Where:
Q = heat needed to vaporize
a substance
Lv = latent heat of vaporization
m = mass of the substance
Units:
J/Kg , cal/g , Btu/slug ….

33. 4.) LATENT HEAT OF CONDENSATION (Lc)
– the amount of heat needed to condense a
substance at its condensation temperature.
Q = -mLc
To condense a gas:
Where:
Q = heat needed to condense
a substance
Lf = latent heat of condensation
m = mass of the substance
Units:
J/Kg , cal/g , Btu/slug ….
Note: the boiling temperature is the same as
the condensation temperature.

34. Heat of Vaporization or Condensation of Some
Substances
Substances Lv/Lc (kJ/kg)
Water 2256
Ethanol 854
Hydrogen (H2) 452
Oxygen (O2) 213
Nitrogen (N2) 201
Aluminum 10,500
Copper 3,100
Iron 6,364

35. 1. How much heat is required to change 200mL of
ice at -20.0°C into steam?
ρice = 920kg/m3 , Lfwater = 333kJ/kg,
Cice = 2090J/kgCo , Cwater = 4190kJ/kgC°
Lvwater = 2260kJ/kg
EXAMPLES:
Ans: 562 kJ

36. 2. How much heat is needed to change 300g of ice
at -30.0°C to steam at 130.0°C?
Cice = 2060J/kgCo , Lfwater = 334kJ/kg,
Cwater = 4180J/kgCo , Lvwater = 2260kJ/kg
Csteam = 2020J/kgCo
EXAMPLES:
Ans: 940 kJ

37. 3. How much heat is needed to change 20g of
mercury at 20.0°C into mercury vapor at the
boiling point (357°C)?
CHg = 140J/kgCo , LfHg = 11kJ/kg (-39°C),
LvHg = 296kJ/kg
EXAMPLES:
Ans: 6860 J

38. 4.) A nuclear power plant generates
1000Mw of waste heat. If the heat
is disposed in an evaporative
cooling tower, how much water
must be evaporated (a) per
second; (b) per day?
Ans: a. m = 443.3 kg/s b. 3.83x10
7
kg/day
EXAMPLES:

39. 5.) While jogging, a man must lose
320J/s of thermal energy by
transferring it from his body to
moisture on his skin, thus causing
the moisture to evaporate. What
mass of perspiration must
evaporate per minute?
Lv = 2.4x106J/kg
EXAMPLES:
Ans: m = 8x10
-3
kg/min