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Chapter 12 AIR The Sea We Live I

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• pg 1
CH104
Chapter 2: Energy & Matter
Temperature
Energy in Reactions
Specific Heat
Energy from Food
States of Matter
Heating & Cooling Curves
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Chapter 2 Energy and Matter
2.2
Temperature

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2
Units of Measurement

Metric        Common
SI          Conversions
Length        meter (m)      1 m = 1.09 yd
Volume        liter (L)      1 L = 1.06 qt
Mass          gram (g)       1 kg = 2.2 lb
Temperature   Celsius (oC)   C = (F-32)/1.8
Kelvin (K)      K = C + 273

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Temperature

o            o
180           100

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Learning Check
A. What is the temperature of freezing water?
1) 0 °F        2) 0 °C     3) 0 K

B. What is the temperature of boiling water?
1) 100 °F     2) 32 °F           3) 373 K

C. How many Celsius units are between the boiling
and freezing points of water?
1) 100          2) 180      3) 273
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5
Solution
A. What is the temperature of freezing water?
1) 0 °F        2) 0 °C     3) 0 K

B. What is the temperature of boiling water?
1) 100 °F     2) 32 °F           3) 373 K

C. How many Celsius units are between the boiling
and freezing points of water?
1) 100          2) 180      3) 273
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6
Trivial Pursuit Question
• At what temperature does
oF = oC?

- 40 oF = - 40 oC

- 40 o

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Temperature conversion
• Common scales used
• Fahrenheit, Celsius and Kelvin.
oF   = 1.8 oC + 32
oC   = (oF - 32)
1.8
SI unit       K    = oC + 273
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Solving a Temperature Problem
A person with hypothermia has a
body temperature of 34.8 °C. What
Is that temperature in °F?

F = 1.8(C) + 32 °

F = (1.8)(34.8 °C) + 32 °
exact tenth’s   exact

= 62.6 ° + 32 °
= 94.6 °F
tenth’s
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9
Practice: Temp Conversion

• What is 75.0 º F in ºC?
• ºC = (75.0 º F -32 º) = 23.9 ºC
1.8

• What is -12 º F in ºC?
• º F = 1.8 (-12) + 32 º F = 10 º C

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Learning Check
The normal temperature of a chickadee is 105.8
°F. What is that temperature on the Celsius scale?
1) 73.8 °C     2) 58.8 °C        3) 41.0 °C

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11
Solution
The normal temperature of a chickadee is 105.8
°F. What is that temperature on the Celsius scale?
1) 73.8 °C     2) 58.8 °C        3) 41.0 °C

TC    =     TF – 32 °
1.8
=     (105.8 – 32 °)
1.8
=     73.8 °F = 41.0 °C
1.8 °        tenth’s place
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12
Learning Check
A pepperoni pizza is baked at 455 °F. What
temperature is needed on the Celsius scale?
1) 423 °C             2) 235 °C         3) 221 °C

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13
Solution
A pepperoni pizza is baked at 455 °F. What
temperature is needed on the Celsius scale?
1) 423 °C             2) 235 °C         3) 221 °C

TF – 32 ° = TC
1.8

(455 – 32 °) = 235 °C
1.8       one’s place
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14
Learning Check
On a cold winter day, the temperature is –15 °C.
What is that temperature in °F?
1) 19 °F          2) 59 °F         3) 5 °F

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15
Solution
On a cold winter day, the temperature is –15 °C.
What is that temperature in °F?
1) 19 °F          2) 59 °F         3) 5 °F

TF   = 1.8TC + 32 °        Note: Be sure to use the
change sign key on your
calculator to enter the minus
TF = 1.8(–15 °C) + 32      (–) sign.
1.8 x 15 +/ – = –27
°
= – 27 + 32 °
= 5 °F
one’s place
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16
Temperatures

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17
Learning Check
What is normal body temperature of 37 °C in
kelvins?
1) 236 K      2) 310 K    3) 342 K

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18
Solution
What is normal body temperature of 37 °C in
kelvins?
1) 236 K      2) 310 K    3) 342 K

TK   =     TC + 273
=     37 °C + 273
=     310. K
one’s place

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19
Chapter 2 Energy and Matter
2.1
Energy

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20
Energy
Energy = The capacity to cause change

Heat
Light
Wind

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Potential Energy = stored Energy
(Has potential for motion)

X

Kinetic Energy = Energy in motion
(Fulfilling its potential)

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Learning Check
Identify each of the following as potential
energy or kinetic energy.
A. roller blading
B. a peanut butter and jelly sandwich
C. mowing the lawn
D. gasoline in the gas tank

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23
Solution
Identify each of the following as potential
energy or kinetic energy.
A. roller blading      kinetic
B. a peanut butter and jelly sandwich potential
C. mowing the lawn kinetic
D. gasoline in the gas tank       potential

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24
Kinetic Energy
KE = 1 mv2
2           Which has more E?
•Truck moving at 5 mph

•Bicycle moving at 5 mph

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Units of Measurement
Metric            Common
SI              Conversions
Length         meter (m)       1 m = 1.09 yd
Volume         liter (L)       1 L = 1.06 qt

Mass           gram (g)        1 kg = 2.2 lb

Temperature    Celsius (oC)    C = (F-32)/1.8
Kelvin (K)      K = C + 273

Energy        calorie (cal)   1Kcal = 1000 cal = 1Cal
Joule (J)       1 cal = 4.18 J
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Examples of Energy Values in Joules

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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?

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28
Solution
How many calories are obtained from a pat of
butter if it provides 150 J of energy when
metabolized?
Given                  Need
150 J   1 cal      =   36     cal
4.184 J

Conversion Factors:
1 cal      and 4.184 J
4.184 J         1 cal
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Chapter 2 Energy and Matter
2.3
Specific Heat

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30
Energy
Units of Energy =
calorie       cal
kilocalorie   kcal   1000 cal = 1 kcal

Calorie      Cal    Cal = kcal

joule        J      4.18 J = 1 cal

British
BTU
Thermal Unit

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Energy
calorie:
•E to raise 1 g H20 by 1 oC
Specific Heat
H 2O               1 cal
1oC
1g 1oC

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Specific Heat
E to raise Temp of 1g substance by 1 oC
cal
g oC

H2O Al Sand Fe       Cu    Ag    Au
1.00 0.22 0.19 0.11 0.093 0.057 0.031
0oC = start                 Add 1 cal
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Specific Heat
E to raise Temp of 1g substance by 1 oC
30o                                               Au
cal                              0.031
20o                                        Ag
g oC
Cu 0.057
10o                            Fe
0.093
Al Sand    0.11
1o                0.22 0.19
H 2O
1.00
0oC = start                     Add 1 cal
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Specific Heat
30o                Low SpHt; Heats quickly      Au
Ag    0.031
20o
Cu 0.057
10o                          Fe
0.093
Al Sand    0.11
1o              0.22 0.19
H 2O   High SpHt; Resists change
1.00

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Specific Heat
Dehydrated person
Body temp rises
quickly

Hydrated person
Resists change in
body temp

Heats quickly
Sand   Gets hot
H2O Resists change 0.19
1.00 Stays cold
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Learning Check
A. For the same amount of heat added, a
substance with a large specific heat
1) has a smaller increase in temperature
2) has a greater increase in temperature
B. When ocean water cools, the surrounding air
1) cools 2) warms 3) stays the same
C. Sand in the desert is hot in the day and cool
at night. Sand must have a
1) high specific heat 2) low specific heat
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Solution
A. For the same amount of heat added, a
substance with a large specific heat
1) has a smaller increase in temperature
2) has a greater increase in temperature
B. When ocean water cools, the surrounding air
1) cools 2) warms 3) stays the same
C. Sand in the desert is hot in the day and cool
at night. Sand must have a
1) high specific heat 2) low specific heat
CH104
Sample Problem:                Specific Heat
What is the specific heat of a metal if 24.8 g absorbs
65.7 cal of energy and the temp rises from 20.2 C
to 24.5 C?
cal     = 65.7 cal       = 0.62 cal
SpHt =
g oC        24.8g 4.3oC       1g 1oC

24.5oC
DT = 4.3 Co
m = 24.8g
20.2oC

CH104
E = m DT SpHt
Specific Heat
Sample Problem:
How much energy does is take to heat
50 g’s of water from 75oC to 87oC?

87oC
DT = 12 Co
m = 50g H2O

75oC            1 cal
SpHt =
1g 1oC

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E = m DT SpHt
Specific Heat
Sample Problem:
How much energy does is take to heat
50 g’s of water from 75oC to 87oC?
m          DT    SpHt
50g H2O   12 Co    1 cal   = 600 cal
1g 1oC to heat water

CH104
E = m DT SpHt
Specific Heat
Sample Problem:
A hot-water bottle contains 750 g of water at 65 °C.
If the water cools to body temp (37 °C), how many
calories of heat could be transferred to sore
muscles?

65oC
m = 750g H2O                   DT = 28 Co

37oC
SpHt = 1 cal
1g 1oC
CH104
E = m DT SpHt
Specific Heat
Sample Problem:
A hot-water bottle contains 750 g of water at 65 °C.
If the water cools to body temp (37 °C), how many
calories of heat could be transferred to sore
muscles?

m       DT        SpHt
750g H2O     28 Co     1 cal     = 21000     cal
1g 1oC     from cool water

CH104
Learning Check
How many kilojoules are needed to raise the
temperature of 325 g of water from 15.0 °C to
77.0 °C?

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E = m DT SpHt                    Solution
How many kilojoules are needed to raise the
temperature of 325 g of water from 15.0 °C to
77.0 °C?
m       DT       SpHt
325g H2O       62 Co   4.184 J   1 KJ
= 84.3 KJ

1g 1oC    1000 J

77oC
DT = 62 Co
15oC
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Chapter 2 Energy and Matter

2.4
Energy and Nutrition

1 Cal     =   1000 calories
1 Cal     =   1 kcal
1 Cal     =   4184 J
1 Cal     =   4.184 kJ

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47
Energy
Units of Energy =
1 Cal    1000 cal   = 1 kcal
1 cal      = 4.18 J

1 Cal   4184 J     = 4.184 kJ

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Calorimeters
A calorimeter
• A reaction chamber
& thermometer in
H2O used to measure
heat transfer
• indicates the heat lost
by a sample
• indicates the heat
gained by water

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Energy from Food
4 kcal    17 kJ
g         g
Carbohydrate

9 kcal    38 kcal
Fat
g         g

Protein   4 kcal    17 kcal
g         g

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Energy from Food
Sample Problem:
How much energy in kcal (= Cal) would be
obtained from 2 Tbl peanut butter
containing 6 g carb, 16 g fat, & 7 g protein?
6 g carb    4 kcal    = 24 kcal
1 g carb                = 196 kcal

16 g fat    9 kcal    = 144 kcal   = 196 Cal
1 g fat

4 kcal                  = 200 Cal
7 g protein             = 28 kcal
1 g protein
CH104
Learning Check
A cup of whole milk contains 12 g of carbohydrate,
9.0 g of fat, and 9.0 g of protein. How many kcal
(Cal) does a cup of milk contain? (Round to the
nearest 10 kcal.)

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Solution
A cup of whole milk contains 12 g of carbohydrate,
9.0 g of fat, and 5.0 g of protein. How many kcal
(Cal) does a cup of milk contain? (Round to the
nearest 10 kcal.)
12 g carb 4 kcal          = 48 kcal
1 g carb                  = 149 kcal

9 g fat     9 kcal     = 81 kcal     = 149 Cal
1 g fat

4 kcal                     = 150 Cal
5 g protein             = 20 kcal
1 g protein
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54
Chapter 2 Energy and Matter
2.5
Classification of Matter

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•                        Matter
The stuff things are made of.
Has Mass and takes up space.
(Air, water, rocks, etc..)

The amount of stuff (in g’s)
(Bowling Ball > Balloon)
Weight on earth.
Pull of Gravity on matter.
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Classification of matter
Matter

Pure Substance             Mixture

Element       Compound

Fe          FeS         Fe + S

Mg     MgO         Mg + O2

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Mixture          Mixtures
Homogeneous               Heterogeneous
(Solution)
Uniform composition     Non-uniform composition

Air
Pizza

Urine                  Fe + S
Gasoline                Tea w/ice
Sand
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Physical Separation of A Mixture
Mixtures can be separated
• involves only physical changes
– Like Filtering & distilling

Like when pasta and water are separated with a
strainer
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Learning Check
Identify each of the following as a pure
substance or a mixture.
A. pasta and tomato sauce
B. aluminum foil
C. helium
D. Air

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Solution
Identify each of the following as a pure
substance or a mixture.
A. pasta and tomato sauce         mixture
B. aluminum foil            pure substance
C. helium                   pure substance
D. Air                            mixture

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Learning Check
Identify each of the following as a
homogeneous or heterogeneous mixture.
A. hot fudge sundae
B. shampoo
C. sugar water
D. peach pie

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Solution
Identify each of the following as a
homogeneous or heterogeneous mixture.
A. hot fudge sundae         heterogeneous
B. Shampoo                  homogeneous
C. sugar water              homogeneous
D. peach pie                heterogeneous

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65
Chapter 2 Energy and Matter
2.6
States and Properties
of Matter

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Elemental states at 25oC
H                            Solid                                         He

Li Be
Liquid                 B    C    N   O   F    Ne
Gas
Na Mg                                               Al   Si   P   S   Cl   Ar

K       Ca Sc   Ti    V   Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr

Rb Sr      Y    Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te                I    Xe

Cs Ba Ls Hf Ta            W Re Os     Ir   Pt Au Hg Tl Pb Bi Po At Rn

Fr Ra Ac
Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu

Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr
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6-2
Properties of matter
Physical State
Property    Solid    Liquid      Gas
Density     High       High         Low
like solids
expands
Shape       Fixed    Shape of
container     to fill
container
Compressibility
Small     Small        Large
Thermal
expansion    Very                  Moderate
small
Small
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Three States of Water

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71
Hydrogen Bonding of Water

•

Frozen H2O:
Slow moving molecules
H-Bond in patterns
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Learning Check
Identify each as a S) solid, L) liquid, or G) gas.
__ A. It has a definite volume but takes the
shape of the container.
__ B. Its particles are moving very rapidly.
__ C. It fills the volume of a container.
__ D. It has particles in a fixed arrangement.
__ E. It has particles that are close together
and are mobile.

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Solution
Identify each as a S) solid, L) liquid, or G) gas.
L A. It has a definite volume but takes the
shape of the container.
G B. Its particles are moving very rapidly.
G C. It fills the volume of a container.
S D. It has particles in a fixed arrangement.
L E. It has particles that are close together
and are mobile.

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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
Physical Properties of Copper

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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

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76
Examples of Physical Change

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77
Chemical Properties
Chemical properties
• describe the ability of a substance to
change into a new substance

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Chemical Change
During a chemical
change,
• reacting substances
form new substances
with different
compositions and
properties
• a chemical reaction       Iron   Iron (III)
takes place                        oxide
Fe
Fe2O3
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Examples of Chemical Change

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80
Summary

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81
Learning Check
Classify each of the following as a 1)
physical change or 2) chemical change.
A. ____ burning a candle
B. ____ ice melting on the street
C. ____ toasting a marshmallow
D. ____ cutting a pizza
E. ____ polishing a silver bowl

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82
Solution
Classify each of the following as a
1) physical change or 2) chemical change.
A. ____ burning a candle         2) chemical
B. ____ ice melting on the street 1) physical
C. ____ toasting a marshmallow 2) chemical
D. ____ cutting a pizza         1) physical
E. ____ polishing a silver bowl 2) chemical

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Chapter 2 Energy and Matter
2.7
Changes of State

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84
Changes of State
Fast, far apart,
Random                Vapor
Vaporize          Boiling Pt       Condense
Moderate, close,
Liquid              Random
arrangement
Melting Pt =
Melt         Freezing Pt        Freeze
Slow, close,
Fixed                  Solid
arrangement
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Changes of State
Vapor

Deposit
Liquid            Frost
Sublime
Freeze Dry

Solid
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Heating Curve Specific Heat of Steam
0.48 cal to heat vapor
g oC

Heat of Vaporization
Specific Heat of H2O           540 cal to vaporize water
g
1.00 cal to heat water
g oC                            Heat of Fusion
80 cal to melt ice
g
Specific Heat of Ice
0.50 cal to heat ice
CH104         g oC
Heating Curve

0.48 cal to heat vapor
g oC

540 cal to vaporize water
g
1.00 cal to heat water
g oC
80 cal to melt ice
g
0.50 cal to heat ice
g oC

CH104
Example: Calculate the total amount of heat needed
to change 500. g of ice at –10 oC into 500. g of steam
at 120oC.                   o
500g 20 C 0.48 cal   = 4800 cal
g  oC
to heat vapor

500g 540 cal         = 270,000 cal
g           to vaporize water
500g 100 o C 1.00 cal = 50,000 cal
g o C to heat water

500g   80 cal        = 40,000 cal
g            to melt ice
500g 10 o C 0.50 cal = 2500 cal
g oC    to heat ice
367,300 cal = 3.67 x 105 cal
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Learning Check
A. A plateau (horizontal line) on a heating
curve represents
1) a temperature change
2) a constant temperature
3) a change of state

B. A sloped line on a heating curve represents
1) a temperature change
2) a constant temperature
3) a change of state
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90
Solution
A. A plateau (horizontal line) on a heating
curve represents
1) a temperature change
2) a constant temperature
3) a change of state

B. A sloped line on a heating curve represents
1) a temperature change
2) a constant temperature
3) a change of state
CH104
Learning Check
Use the cooling curve for water to answer each.
A. Water condenses at a temperature of
1) 0 °C      2) 50 °C 3) 100 °C
B. At a temperature of 0 °C, liquid water
1) freezes 2) melts 3) changes to a gas
C. At 40 °C, water is a
1) solid          2) liquid 3) gas
D. When water freezes, heat is
1) removed 2) added
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Solution
Use the cooling curve for water to answer each.
A. Water condenses at a temperature of
1) 0 °C      2) 50 °C 3) 100 °C
B. At a temperature of 0 °C, liquid water
1) freezes 2) melts 3) changes to a gas
C. At 40 °C, water is a
1) solid     2) liquid 3) gas
D. When water freezes, heat is
1) removed         2) added
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93
Learning Check
To reduce a fever, an infant is packed in 250 g of ice.
If the ice (at 0 °C) melts and warms to body temp
(37.0 °C), how many calories are removed?

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94
Solution
To reduce a fever, an infant is packed in 250 g of ice.
If the ice (at 0 °C) melts and warms to body temp
(37.0 °C), how many calories are removed?

37.0 °C           250g    37 o C 1.00 cal = 9,250 cal
g o C to heat water
STEP 2
250g    80 cal         = 20,000 cal
0.0 °C                        g             to melt ice
STEP 1
29,000 cal = 2.9 x 104 cal

CH104

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