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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
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Chapter 6 NotesHEAT Heat & Temperature Calculations
Temperature= a measure of the AVERAGE kinetic energy in the substance. • Celsius (°C) • Fahrenheit (°F) • Kelvin (°K)
H20 distilled water (pure water) • melting point = 0°C • boiling 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 Joules • 1 BTU = 252 calories • 1 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 • Melting • Freezing • Vaporization • Condensation • Sublimation • physical 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 substance • freezing 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 liquid • boiling point= the temperature at which boiling occurs • the 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)
Heating of water STEAM 100°C WATER (liquid) heat of vaporization 0°C ICE heat of fusion
Heat Transfer • Conduction • Convection • Radiation
Conduction • transfer of heat by direct contact (molecule to molecule) • metals are good conductors • poor 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 waves • some wavelengths of infrared & ultraviolet • only type of transfer that can occur through empty space • sun Earth
Specific Heat The amount of heat needed to raise the temperature of one gram of a substance one degree Celsius.
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 Factors in Specific Heat
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
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 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 = 4.2 J/g°C m = 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 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?
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 constant • The heat of fusion of water = 340 J/g • Hv = mass x Hv constant • The 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/g • m = 500 g • H = Chf x m • H = (340 J/g)(500 g) • H = 170,000 J