Specific heat and Latent Heat

1. HEAT

2. HEAT CAPACITY
• or thermal capacity is a physical property of matter, defined as the amount
of heat to be supplied to a given mass of a material to produce a unit change in its
temperature.
• Material with high heat capacity heats up and cools down slowly because it has to
absorb or give off more heat.
• Material with low heat capacity heats up and cools down at a faster rate because it
has less capacity to absorb or release heat.
• Comparing heat capacities of different materials is needed for us to identify the heat
properties of substances.
• Specific heat of any substance is the ratio of the amoung of heat necessary to raise
the temperature of that substance on a certain number of degrees to the amount of
heat required to raise the temperature of an equal mass of water the same number
of degrees.

3. SPECIFIC HEAT
• Defined as the number of calories needed to raise the temperature of one gram of a
substance 1Co. Meaning, if the specific heat of iron is 0.11 cal/g-Co, one gram
requires 0.11 calorie to raise the temperature by one Celsius degree.
Table of Specific heat of Some substance
Substance
Specific Heat
Co)
Substance
Specific Heat
Co)
Alcohol, ethyl 0.58 Iron or steel 0.11
Aluminum 0.22 Lead 0.031
Brass 0.92 Mercury 0.033

4. • Cal/g-Co is the unit of specific heat or Btu/lb-Fo and the SI unit is kJ/kg-K.
Q = mc∆T
• Formula used in determining the amount of heat needed to raise the temperature
of a substance.
• Q – quantity of heat gained or given off by an object
• m – mass of the object
• C – specific heat o ∆T – change in temperature (Tf-Ti) or (T2-T1)
Copper 0.093 Silver 0.056
Glass 0.16 Steam 0.48
Gold 0.031 Water 1.00
Ice 0.5 Zinc 0.092

5. LATENT HEAT
• Matter exist in different states (solid, liquid, gas, and plasma) because of the amount
of energy they have.
• They depend on the temperature and pressure exerted on them.
• There is a changing in its states if there is absorption and release of a certain
amount of heat that affects the arrangement of molecules.
• For example:
applying heat to ice – changing solid state to liquid state to gas state
applying heat to solid state will increase the motion of the molecules which
then becomes liquid, further add sufficient heat to liquid will take to a form of steam
(gas), adding more heat molecules will break into ions and electrons that give you
plasma. In reverse (removing heat, slows down molecules)

6. Condensation
Fusion
Vaporization
Solidification
HEAT GAIN (+Q)
HEAT LOSS (-Q)

7. LATENT HEAT OF FUSION
• Amount of heat absorb during the process
• If object undergo fusion solid for instance, it will turn into a liquid.
• Also refers to the melting process

8. LATENT HEAT SOLIDIFICATION
• Amount of heat released in the process
• Removing heat from the liquid and turns it to solid
• Also refers to the freezing process

9. • LATENT HEAT OF FUSION = 80 calories per gram = 80 cal/g = 334.72
J/g
• Heat Q required to melt a solid object with mass m is:
Q = mLf
Lf = the latent heat of fusion

10. Question: The heat of fusion of water is 334 J/g. How much energy
is required to melt 50 grams of ice into liquid water?
Given: m = 50 grams; Lf = 334 J/g
Solution: Q = m · Lf
Q = (50 g) · (334 J/g)
Q = 16700 J
ANSWER: Q= 16.7 kJ
EXAMPLE #1

11. Question: If it takes 41000 joules of heat to melt 200 grams solid copper
to liquid copper, what is the heat of fusion of copper?
GIVEN: m = 200 g; Q = 41000 J; Lf =?
SOLUTION: Q = m·Lf
41000 J = (200 g) · Lf
Lf = 41000 J/200 g
Lf = 205 J/g
Answer: The heat of fusion of copper is 205 J/g.
EXAMPLE #2

12. LATENT HEAT OF VAPORIZATION
AND CONDENSATION
HEAT OF VAPORIZATION - Amount of heat needed to change a unit mass of
the substance from liquid to gas without a change in temperature.
HEAT OF CONDENSATION – amount of heat required to change the unit
mass of substance from gas to liquid with a change of temperature.

13. • Heat of vaporization and condensation = 540 calories per gram at 100oC =
2259.36 J/g
Q = mLv
Lv – latent heat of vaporization

14. QUESTION: Determine the amount of heat released by nitrogen to change
phase from vapor to liquid. Heat of vaporization for nitrogen = 200 x 103 J/kg
GIVEN : Mass (m) = 1 gram = 1 x 10-3 kg; (LV) = 200 x 103 J/kg
Solution :
Q = m LV
Q = (1 x 10-3 kg)(200 x 103 J/kg)
ANSWER: Q = 200 Joule
EXAMPLE #1

15. Question: The heat of vaporization of water is 2259 Joules/gram or 540
calories/gram. What energy in Joules is required to convert 50 grams of water into
steam? How much energy in calories?
GIVEN: m = 50 grams; Lv = 2257 J/gram
SOLUTION: Q = m · Lv
Q = (50 g) · (2257 J/g)
Q = 112850 J = 112.85 kJ
In calories:
Lv = 540 cal/g
Q = m · Lv
Q = (50 g) · (540 cal/g)
Q = 27000 cal = 27 kcal
Answer: It takes 112850 Joules or 27000 calories of heat to convert 50 grams of
liquid water into gaseous water or steam.
EXAMPLE #2