Chapter 6 Notes HEAT Heat & Temperature Calculations
Jan 11, 2016
Chapter 6 NotesHEAT
Heat & Temperature Calculations
Temperature = a measure of the AVERAGE kinetic energy in the substance.
Celsius (°C)Fahrenheit
(°F)Kelvin (°K)
NEED TO FIND FORMULA
°F 1.8°C + 32
°C °F – 32/1.8
°K °C + 273
°C °K – 273
°F °K °C °F
0°K = absolute zero = all molecular motion stops
H20 distilled water (pure water)
melting point = 0°Cboiling point = 100°C
Melting Points examples
Gallium a# 31 M.P. 86oF
Iron a# 26 M.P. 2800oF
Mercury a# 80 M.P. -38oF
Gold a# 79 M.P. 1947oF
Copper a# 29 M.P. 1984oF
Boiling Pointsexamples
Gallium a# 31 B.P. 3999oF
Iron a# 26 B.P. 5182oF
Mercury a# 80 B.P. 674oF
Gold a# 79 B.P. 5173oF
Copper a# 29 B.P. 4644oF
Energy (heat) measure in Joules, BTUs (British Thermal Units) calories and Calories.
1 calories = 4.186 Joules1 BTU = 252 calories1 Calorie = 1000 calories
States of Matter Also called Phases of Matter
Solids
Liquids
Vapors (gases)
Solids
Have a definite shape
Have a definite volume
Particles VIBRATE in place
Liquids
Have NO definite shape
Have definite volume
particles SLIDE freely
Gases (vapor)
Have NO definite shape
Have NO definite volume
particles fill the volume of the container
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.
Phase Changes
MeltingFreezingVaporizationCondensationSublimationphysical changes
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
Freezing
the process of changing from a liquid to a solid energy is being pulled out of the substancefreezing point = same temperature as the melting point (used mainly in weather)
Vaporization
the process of changing from a liquid to a gas
energy in being put into the substance
evaporation
boiling
Evaporation
vaporization that occurs at the surface of the liquid
Boiling
vaporization that occurs throughout the liquidboiling point = the temperature at which boiling occursthe boiling point of water is 100ºC
Condensation
the process of changing from a gas to a liquid
energy is being pulled out of the substance
Sublimation
the process of changing from a solid to a gas
energy is being put into the substance
ex: dry ice (CO2)
heat of fusion
Heating of water
0°C
100°C
heat of vaporization
ICE
WATER (liquid)
STEAM
Heat Transfer
Conduction
Convection
Radiation
Conduction transfer of heat by direct contact
(molecule to molecule)metals are good conductorspoor conductors = insulators
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
Radiation transfer of heat by electromagnetic
wavessome wavelengths of infrared &
ultravioletonly type of transfer that can occur
through empty spacesun Earth
Specific Heat
The amount of heat needed to raise the temperature of one gram of a substance one degree Celsius.
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
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
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°Cm = 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
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?
C = 4.2 J/g°Cm = 500 grams∆T = 10°C (30-
20)
∆Q = m x ∆T x C ∆Q = (500 g)(10°C)(4.2 J/g°C) ∆Q = 21,000
Joules
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.
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.
Heat & Phase Changes
Hf = mass x Hf constantThe heat of fusion of water = 340 J/gHv = mass x Hv constantThe heat of vaporization of water = 2300
J/g
EXAMPLE:How many joules of heat are necessary to melt 500 g of ice?
Chf = 340 J/gm = 500 gH = Chf x mH = (340 J/g)(500 g)H = 170,000 J