1 chapter 2 energy and matter 2.1 energy. 2 energy makes objects move makes things stop is needed to...
TRANSCRIPT
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Chapter 2 Energy and Matter
2.1Energy
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Energy
Energy• makes objects move• makes things stop• is needed to “do work”
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Work
Work is done when • you climb• you lift a bag of groceries• you ride a bicycle• you breathe• your heart pumps blood• water goes over a dam
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Potential Energy
Potential energy is energystored for use at a later
time.
Examples:• water behind a dam• a compressed spring• chemical bonds in
gasoline, coal, or food
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Kinetic Energy
Kinetic energy is the energy of matter in motion.
Examples:• swimming• water flowing over a dam• working out• burning gasoline
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Chapter 2 Energy and Matter
2.2Temperature
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Temperature
Temperature • is a measure of how hot or cold
an object is compared to another object
• indicates that heat flows from the object with a higher temperature to the object with a lower temperature
• is measured using a thermometer
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Temperature Scales
Temperature scales • are Fahrenheit,
Celsius, and Kelvin
• have reference points for the boiling and freezing points of water
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Units for Measuring Energy or Heat
Heat is measured in joules or calories.• 4.184 joules (J) = 1 calorie (cal) (exact) • 1 kJ = 1000 J• 1 kilocalorie (kcal) = 1000 calories (cal)
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• On the Fahrenheit scale, there are 180 °F between the freezing and boiling points, and on the Celsius scale, there are 100 °C.
180 °F = 9 °F = 1.8 °F 100 °C 5 °C 1 °C
• In the formula for the Fahrenheit temperature, adding 32 adjusts the zero point of water from 0 °C to 32 °F.
TF = 9 (TC) + 32 ° 5 or TF = 1.8(TC) + 32 °
Fahrenheit Formula
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• TC is obtained by rearranging the equation for TF.
TF = 1.8(TC) + 32 °• Subtract 32 from both sides.
TF – 32 ° = 1.8(TC) + (32 ° – 32 °)
TF – 32 ° = 1.8(TC)
• Divide by 1.8. TF – 32 ° = 1.8 TC
1.8 1.8
TF – 32 ° = TC 1.8
Celsius Formula
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Solving a Temperature Problem
A person with hypothermia has abody temperature of 34.8 °C. WhatIs that temperature in °F?
TF = 1.8(TC) + 32 °
TF = (1.8)(34.8 °C) + 32 ° exact tenth’s exact
= 62.6 ° + 32 ° = 94.6 °F tenth’s
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The Kelvin temperature• is obtained by adding 273 to the Celsius temperature
TK = TC + 273
In the Kelvin temperature scale: • There are 100 units between the freezing and boiling
points of water.100 K = 100 °C or 1 K = 1 °C
• 0 K (absolute zero) is the lowest possible temperature. 0 K = –273 °C
Kelvin Temperature Scale
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Chapter 2 Energy and Matter
2.3Specific Heat
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Specific Heat
Specific heat • is different for different substances• is the amount of heat that raises the
temperature of 1 g of a substance by 1 °C• in the SI system has units of J/g C • in the metric system has units of cal/g C
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Example of Calculating Specific Heat
What is the specific heat of a metal if 24.8 g absorbs65.7 cal of energy and the temperature rises from20.2 C to 24.5 C? STEP 1 Given: 24.8 g, 65.7 cal, ΔT = 20.2 C to 24.5 C
Need: SHmetal = cal/g C
STEP 2 Plan: ΔT = 24.5 C – 20.2 C = 4.3 C SH = Heat(cal) g C
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Example of Calculating Specific Heat (continued)
STEP 3 Set up to calculate SH: 65.7 cal = 0.62 cal/g C (24.8 g)(4.3 C)
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Heat Equation
Rearranging the specific heat expression gives theheat equation.
Heat = g x T x cal (or J) = cal (or J) g °C
The amount of heat lost or gained by a substance iscalculated from the• Mass of substance (g)• Temperature change ( T)• Specific heat of the substance (cal/g °C) or (J/g °C)
Guide to Calculations Using Specific Heat
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A hot-water bottle contains 750 g of water at 65 °C. If thewater cools to body temperature (37 °C), how manycalories of heat could be transferred to sore muscles?
STEP 1 Given: 750 g of water cools from 65 °C to 37 °C SHwater = 1.00 cal/g °C
Need: calories of heat transferred
STEP 2 Calculate the temperature change T: 65 °C – 37 °C = 28 °C
Sample Calculation for Heat
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STEP 3 Write the heat equation: Heat (cal) = mass(g) x T x SH
STEP 4 Substitute given values and solve for heat:
750 g x 28 °C x 1.00 cal g °C
= 21 000 cal
Sample Calculation for Heat
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Learning Check
How many calories are obtained from a pat of butter
if it provides 150 J of energy when metabolized?
1) 36 cal2) 150 cal 3) 630 cal
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Solution
STEP 1 Given: 150 J Need: caloriesSTEP 2 Plan: J calSTEP 3 Write equality and conversion factors: 1 cal = 4.184 J 1 cal and 4.184 J 4.184 J 1 calSTEP 4 Set up problem: 150 J x 1 cal = 36 cal (1)
4.184 J
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Learning Check
How many kilojoules are needed to raise thetemperature of 325 g of water from 15.0 °C to
77.0 °C?1) 20.2 kJ2) 84.3 kJ3) 105 kJ
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Solution
2) 84.3 kJ STEP 1 Given: 325 g of water warms from 15.0 °C
to 77.0 °C SHwater = 4.184 J/g °C
1 kJ = 1000 J Need: kilojoules of heat neededSTEP 2 Calculate the temperature change T:
77.0 °C – 15.0 °C = 62.0 °CSTEP 3 Write the heat equation:
Heat (joules) = mass (g) x T x SH
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Solution (continued)
STEP 4 Substitute given values and solve for heat: 325 g x 62.0 °C x 4.184 J x 1 kJ
g °C 1000 J = 84.3 kJ (2)
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Chapter 2 Energy and Matter
2.4Energy and Nutrition
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CalorimetersA calorimeter • is used to measure heat
transfer• contains a reaction
chamber and thermometer in water
• indicates the heat lost by a sample
• indicates the heat gained by water
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Chapter 2 Energy and Matter
2.5Classification of Matter
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MatterMatter • is the material that makes up a substance• makes up the things we see everyday, such as water, wood, cooking pans,
clothes, and shoes
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A pure substance is classified as • matter with a specific composition• an element when composed of one type of atom• a compound when composed of two or more
elements combined in a definite ratio
Pure Substances
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Mixtures
A mixture is matter that consists of• two or more substances that are physically
mixed, not chemically combined• two or more substances in different
proportions• substances that can be separated by
physical methods
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Homogeneous Mixtures
In a homogeneous mixture,
• the composition is uniform throughout
• the different parts of the mixture are not visible
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Heterogeneous Mixtures
In a heterogeneous mixture,
• the composition is not uniform; it varies from one part of the mixture to another
• the different parts of the mixture are visible
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Chapter 2 Energy and Matter
2.6States and Properties
of Matter
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Solids
Solids have • a definite shape• a definite volume• particles that are close
together in a fixed arrangement
• particles that move very slowly
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Liquids
Liquids have• an indefinite shape
but a definite volume• the same shape as
their container• particles that are close
together but mobile• particles that move
slowly
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Gases
Gases have • an indefinite shape• an indefinite volume• the same shape and
volume as their container• particles that are far apart• particles that move very
fast
Physical Properties
Physical properties• are observed or measured without
changing the identity of a substance• include shape and color• include melting point and boiling point
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Physical Change
In a physical change,• the identity and
composition of the substance do not change
• the state can change or the material can be torn into smaller pieces
Chemical PropertiesChemical properties• describe the ability of a substance to change into a new substance
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During a chemical change, reacting substances form new
substances with different compositions and properties
a chemical reaction takes place
IronFe
Iron (III) oxideFe2O3
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Chapter 2 Energy and Matter
2.7Changes of State
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Melting and Freezing
A substance • is melting when it changes from a solid to a
liquid• is freezing when it changes from a liquid to a
solid• such as water has a freezing (melting) point of
0 °C
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Heat of Fusion
The heat of fusion • is the amount of heat released when 1 gram
of liquid freezes (at its freezing point) • is the amount of heat needed to melt 1 gram
of a solid (at its melting point)• for water (at 0 °C) is
334 J or 80 cal 1 g of water
Guide to Calculations Using Heat of Fusion (or Vaporization)
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Calculations Using Heat of Fusion
How much heat in calories is needed to melt 15.0 g ofice at 0 °C ?STEP 1 Given: 15.0 g of water(s)
change of state: melting at 0 °CSTEP 2 Plan: g of water(s) g of water(l)
STEP 3 Write conversion factors:1 g of water = 80 cal
1 g of water and 80 cal 80 cal 1 g of water
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Calculations Using Heat of Fusion (continued)
STEP 4 Set up the problem to calculate calories:
15.0 g water x 80 cal = 1200 cal 1 g water
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SublimationSublimation• occurs when a solid changes directly to a gas• is typical of dry ice, which sublimes at 78 C• takes place in frost-free refrigerators• is used to prepare freeze-dried foods for long-
term storage
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Evaporation and CondensationWater• evaporates when molecules
on the surface gain sufficient energy to form a gas
• condenses when gas molecules lose energy and form a liquid
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Heat of Vaporization
The heat of vaporization is the amount of heat• absorbed to vaporize 1 g of a liquid to gas at the
boiling point• released when 1 g of a gas condenses to liquid at
the boiling point
Boiling Point of Water = 100 °C
Heat of Vaporization (water) = 2260 J or 540 cal 1 g of water
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Classify each of the following as a 1) physical change or 2) chemical change.A. ____ burning a candleB. ____ ice melting on the streetC. ____ toasting a marshmallowD. ____ cutting a pizzaE. ____ polishing a silver bowl
Learning Check
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Classify each of the following as a 1) physical change or 2) chemical change.A. 2 burning a candleB. 1 ice melting on the streetC. 2 toasting a marshmallowD. 1 cutting a pizzaE. 2 polishing a silver bowl
Solution
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Learning Check
How many joules are released when 25.0 g of water
at 0 °C freezes? 1) 334 J
2) 2000 J 3) 8350 J
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Solution
3) 8350 J
STEP 1 Given: 25.0 g of water(l) change of state: freezing at 0 °C
STEP 2 Plan: g of water g of water
STEP 3 Write conversion factors:1 g of water = 334 J 1 g of water and __334 J___ 334 J 1 g of water
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Solution (continued)
STEP 4 Set up the problem to calculate joules:
25.0 g water x 334 J = 8350 J (3) 1 g water
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Learning Check
How many kilocalories (kcal) are released when 50.0 g
of water(g) as steam from a volcano condenses at
100 °C?1) 27 kcal2) 540 kcal 3) 2700 kcal
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Solution
1) 27 kcalSTEP 1 50.0 g of water(g) = 50.0 g of water(l) Change of state: water condensing at 100 °CSTEP 2 Plan: g of water(g) g of water(l)STEP 3 Write conversion factors:
1 g of water = 540 cal 1 g of water and 540 cal
540 cal 1 g of water 1 kcal = 1000 cal
1 kcal and 1000 cal 1000 cal 1 kcal