chapter 6 notes heat
DESCRIPTION
Chapter 6 Notes HEAT. Heat & Temperature Calculations. Temperature = a measure of the AVERAGE kinetic energy in the substance. Celsius (°C) Fahrenheit (°F) Kelvin (°K). 0°K = absolute zero = all molecular motion stops. H 2 0 distilled water (pure water). melting point = 0°C - PowerPoint PPT PresentationTRANSCRIPT
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Chapter 6 NotesHEAT
Heat & Temperature Calculations
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Temperature = a measure of the AVERAGE kinetic energy in the substance.
Celsius (°C)Fahrenheit
(°F)Kelvin (°K)
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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
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H20 distilled water (pure water)
melting point = 0°Cboiling point = 100°C
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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
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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
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Energy (heat) measure in Joules, BTUs (British Thermal Units) calories and Calories.
1 calories = 4.186 Joules1 BTU = 252 calories1 Calorie = 1000 calories
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States of Matter Also called Phases of Matter
Solids
Liquids
Vapors (gases)
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Solids
Have a definite shape
Have a definite volume
Particles VIBRATE in place
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Liquids
Have NO definite shape
Have definite volume
particles SLIDE freely
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Gases (vapor)
Have NO definite shape
Have NO definite volume
particles fill the volume of the container
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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.
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Phase Changes
MeltingFreezingVaporizationCondensationSublimationphysical changes
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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
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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)
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Vaporization
the process of changing from a liquid to a gas
energy in being put into the substance
evaporation
boiling
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Evaporation
vaporization that occurs at the surface of the liquid
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Boiling
vaporization that occurs throughout the liquidboiling point = the temperature at which boiling occursthe boiling point of water is 100ºC
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Condensation
the process of changing from a gas to a liquid
energy is being pulled out of the substance
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Sublimation
the process of changing from a solid to a gas
energy is being put into the substance
ex: dry ice (CO2)
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heat of fusion
Heating of water
0°C
100°C
heat of vaporization
ICE
WATER (liquid)
STEAM
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Heat Transfer
Conduction
Convection
Radiation
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Conduction transfer of heat by direct contact
(molecule to molecule)metals are good conductorspoor conductors = insulators
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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
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Radiation transfer of heat by electromagnetic
wavessome wavelengths of infrared &
ultravioletonly type of transfer that can occur
through empty spacesun Earth
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Specific Heat
The amount of heat needed to raise the temperature of one gram of a substance one degree Celsius.
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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
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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
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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
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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
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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.
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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.
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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
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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