1. Heat & Temperature Calculations
Chapter 6 Notes HEAT
Heat & Temperature Calculations

2. Temperature = a measure of the AVERAGE kinetic energy in the substance.
Celsius (°C)
Fahrenheit (°F)
Kelvin (°K)

3. 0°K = absolute zero = all molecular motion stops
NEED TO FIND
FORMULA °F
1.8°C + 32 °C
°F – 32/1.8 °K
°C + 273
°K – 273
°K °C °F
0°K = absolute zero = all molecular motion stops

4. H20 distilled water (pure water)
melting point = 0°C
boiling point = 100°C

5. Melting Points examples
Gallium a# 31 M.P. 86oF
Iron a# M.P oF
Mercury a# 80 M.P. -38oF
Gold a# M.P oF
Copper a# 29 M.P oF

6. Boiling Points examples
Gallium a# B.P oF
Iron a# B.P oF
Mercury a# B.P oF
Gold a# B.P oF
Copper a# B.P oF

7. Energy (heat) measure in Joules, BTUs (British Thermal Units) calories and Calories.
1 calories = Joules
1 BTU = 252 calories
1 Calorie = 1000 calories

8. States of Matter Also called Phases of Matter Solids Liquids
Vapors (gases)

9. Solids Have a definite shape Have a definite volume
Particles VIBRATE in place

10. Liquids Have NO definite shape Have definite volume
particles SLIDE freely

11. Gases (vapor) Have NO definite shape Have NO definite volume
particles fill the volume of the container

12. Solids, Liquids & Gases
Solids = can form crystals = solid where the particle are arranged into repeating patterns.
Liquids = physical property of Viscosity = “thickness” – the resistance to flow.
Gases = volume of gases depend greatly on pressure and temperature.

13. Phase Changes Melting Freezing Vaporization Condensation Sublimation
physical changes

14. Melting the process of changing from a solid to a liquid
energy is being put into the substance
melting point = the temperature at which melting occurs – physical property
the melting point of water is 0ºC

15. Freezing the process of changing from a liquid to a solid
energy is being pulled out of the substance
freezing point = same temperature as the melting point
(used mainly in weather)

16. Vaporization the process of changing from a liquid to a gas
energy in being put into the substance
evaporation
boiling

17. Evaporation
vaporization that occurs at the surface of the liquid

18. Boiling vaporization that occurs throughout the liquid
boiling point = the temperature at which boiling occurs
the boiling point of water is 100ºC

19. Condensation the process of changing from a gas to a liquid
energy is being pulled out of the substance

20. Sublimation the process of changing from a solid to a gas
energy is being put into the substance
ex: dry ice (CO2)

21. Heating of water heat of vaporization heat of fusion STEAM 100°C
WATER (liquid)
heat of vaporization
0°C
ICE
heat of fusion

22. Heat Transfer
Conduction
Convection
Radiation

23. Conduction transfer of heat by direct contact (molecule to molecule)
metals are good conductors
poor conductors = insulators

24. Convection
transfer of heat by “convection currents” warm fluids are less dense than colder fluid thus warm fluids rise and cold fall.
not possible in solids
fluid = anything that flows (liquids & gases)
hot air balloons, “convection” ovens

25. Radiation transfer of heat by electromagnetic waves
some wavelengths of infrared & ultraviolet
only type of transfer that can occur through empty space
sun  Earth

26. Specific Heat
The amount of heat needed to raise the temperature of one gram of a substance one degree Celsius.

27. Factors in Specific Heat
types of substance (C)
mass of the substance (m)
how much of a temperature change (∆T)
C = specific heat constant
m = mass
∆T = difference in the temperature

28. Specific Heat Calculations
∆Q = amount of heat absorbed (difference in the heat or heat change)
∆Q = m x ∆T x C
The specific heat of water
= 1.0 cal/g°C or
= 4.2 joules/ g°C

29. EXAMPLE #1: How many calories are absorbed by a pot of water with a mass of 500 grams in order to raise the temperature from 20°C to 30°C?
C = 1.0 cal/g°C
m = 500 grams
∆T = 10°C (30-20)
∆Q = m x ∆T x C
∆Q = (500 g)(10°C)(1.0 cal/g°C)
∆Q = 5000 calories

30. EXAMPLE #2: How many joules are absorbed by a pot of water with a mass of 500 grams in order to raise the temperature from 20°C to 30°C?
∆Q = m x ∆T x C
∆Q = (500 g)(10°C)(4.2 J/g°C)
∆Q = 21,000 Joules
C = 4.2 J/g°C
m = 500 grams
∆T = 10°C (30-20)

31. Phase Changes Heat of fusion (Hf)
the heat energy needed to melt (or freeze) a substance.
All heat being put into the substance goes to the melting process
thus the temperature does not change while the substance is melting.

32. Phase Changes Heat of vaporization (Hv)
the heat energy needed to boil (or condense) a substance.
All heat being put into the substance goes to the boiling process
thus the temperature does not change while the substance is boiling.

33. Heat & Phase Changes Hf = mass x Hf constant
The heat of fusion of water = 340 J/g
Hv = mass x Hv constant
The heat of vaporization of water = 2300 J/g

34. EXAMPLE: How many joules of heat are necessary to melt 500 g of ice?
Chf = 340 J/g
m = 500 g
H = Chf x m
H = (340 J/g)(500 g)
H = 170,000 J