Download 2 SHS Chem Ch 2 Lecture shs_chem_ch_2_lecture

10/27/2011
ENERGY
Energy - the capacity to do work
Reminder: Changes in matter Can Be Physical
or Chemical
Change When energy is involved when there is
a change in matter.
Every Change in matter involves a change in
energy
All physical and chemical changes involve a
change in energy.
Sometimes energy is released when a change in
matter occurs. Sometimes absorbed.
Process - Any change in matter in which energy
is changed.
THE SCIENTIFIC UNIVERSE
• The system is the part of the universe we
wish to focus our attention on. In the world of
chemistry, the system is the chemical reaction.
For example:
• The system consists of those molecules which
are reacting.
• The surroundings are everything else; the
rest of the universe. For example, say the
above reaction is happening in gas phase; then
the walls of the container are part of the
surroundings.
ENERGY TRANSFER
• Endothermic (+) and Exothermic Processes (-)
• Exo/Endo difference is direction of energy flow
SURROUNDINGS
EXO
ENDO
THE SCIENTIFIC UNIVERSE
• You are the system (that is, the chemical
reaction) and as you look outward, everything
else are surroundings.
• FOR THE SYSTEM: We will signify an
increase in energy with a positive sign (+)
and a loss of energy with a negative sign. (-)
ENERGY TRANSFER
• Energy is transferred back and forth between a
system and its surroundings.
• Endothermic Surroundings sends energy to
system
• The boiling of water & melting of ice physical
endothermic.
• Can also be chemical endothermic (Ice Packs).
• Endothermic processes, (energy is absorbed,
never destroyed.)
SURROUNDINGS
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ENERGY TRANSFER
• MOSH PIT ANALOGY
• Exothermic - system releases heat to
Surroundings
– Burning paper (chemical exothermic changes) /
freezing water of (exothermic physical
changes). Can also have chemical exothermic.
– Exothermic processes, energy is released, may
make it seem as if energy is being created.
THIS RELATIONSHIP BETWEEN ENERGY CHANGE AND CHANGE OF STATE IS
VERY IMPORTANT TO OUR MILK AND SCIENCE
http://sfgate.com/cgi-bin/object/article?f=/kron/archive/2001/05/03/frozen.DTL&o=0
Solid Particles
• Fixed very close together orderly, fixed arrangement.
• They are held in place by the attractive forces that are
between all particles.
• Because solid particles can vibrate only in place and
do not break away from their fixed positions, solids
have fixed volumes and shapes.
States of Matter
The physical properties of each state come from the arrangement of particles and bonds holding.
Liquid Particles
• Move easily past each other
• Some liquids can flow very readily, some not (viscosity).
• Held closely by attractive forces (fixed volume). Forces
between particles, which gives liquids special properties:
– Surface Wetting - Liquid particles have cohesion, attraction for
each other. They can also have adhesion, attraction for particles of solid
surfaces. The balance of these forces determines whether a liquid will
wet a solid surface
• Solid takes up the same amount of space. Solids
usually exist in crystalline
Gas
1. Particles are independent.
2. Gas particles are much farther apart
than the solids and liquids.
– Capillary Action - The adhesion of the water molecules pulls water
molecules up the sides of the tube. Critical to Biology.
– Surface Tension - Substances are liquids instead of gases because the
cohesive forces.
• Liquids tend to form spherical shapes, because a sphere
has the smallest surface area for a given volume. (rain
droplets are spherical)
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Changing States
• Melting (endothermic) - Solid becomes a liQuid
• Evaporation (endothermic) - liQuid becomes a gas
takes a lot of energy. The liQuid particles gain energy
when colliding. Sweating, low dew pt. steals energy.
• Sublimation (endothermic)- solids to a gas, Quick
change (mothballs)
• Condensation (exothermic)- gas becomes a liQuid
• Water on a pan lid – lid steals energy
• Deposition (exothermic)- gas become solids. Energy
is released (Night Frost/Iodine)
• Freezing (exothermic) - liQuid becomes a solid as
energy is stolen, particles slow down.
Changing States
• All state changes are physical changes.
• During the change of phase, energy is
used to realign.
• Realignment is seen by presence of both
states.
• Chemical properties are the same in
different states (but reaction rates will be
affected)
ENERGY VS TEMPERATURE
1) Temp increases as particle motion
increases.
2) The higher the temperature the more the
particle movement.
a) solids particles vibrate more rapidly.
b) liquid particles move more Quickly.
c) gas particles collide more often.
Once realignment is complete, temperature changes.
Heat As Energy Transfer
We often speak of heat as though it were a material
that flows from one object to another; it is not.
Rather, it is a form of energy.
3) After a certain point, adding more energy will
cause a substance to experience a change of
state instead of a temperature increase.
Heat As Energy Transfer
Definition of heat:
Heat is energy transferred from one object to
another because of a difference in temperature.
Unit of heat: calorie (cal)
1 cal is the amount of heat necessary to raise the
temperature of 1 g of water by 1 Celsius degree.
• Remember that the temperature of a gas is
a measure of the kinetic energy of its
molecules.
Don’t be fooled – the calories on our food labels are
really kilocalories (kcal or Calories), the heat
necessary to raise 1 kg of water by 1 Celsius
degree.
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Heat As Energy Transfer
If heat is a form of energy, it ought to be possible to
equate it to other forms. The experiment below found
the mechanical equivalent of heat by using the falling
weight to heat the water:
ENERGY
• Energy Types: Kinetic (motion) and
Potential (stored)
• Energy Forms: chemical, mechanical,
light, thermal, electrical, and sound.
http://www.youtube.com/watch?v=Ko5VpvE2btY&safety_mode=true&persist_safety_mode=1&safe=active
ENERGY
CONSERVATION OF ENERGY
Energy cannot be destroyed or created.
Potential
Kinetic
Chemical
Mechanical
Heater/Furnace
Chemical
Thermal
Hydroelectric
Gravitational
Electrical
Solar
Optical
Electrical
Nuclear
Nuclear
Battery
Chemical
Electrical
Food
Chemical
Heat, Kinetic
Photosynthesis
Optical
Chemical
Automobile Engine
Heat, Kinetic, Optical
During any physical or chemical change, the
total Quantity of energy remains constant.
Energy is never created or destroyed
Energy Can Be Absorbed/Released
Thermally As Heat
HEAT VS. TEMPERATURE
TEMPERATURE
• Heat - Thermal energy transferred between objects that are
at different temperatures.
• 1. Measurement of the average kinetic
energy of the random motion of particles in
a substance.
• 2. A measure of how hot (or cold)
• 3. As kinetic energy increases so does
temp.
• Thermal energy is always transferred from a warmer object
(higher energy) to a cooler (lower energy) object until
thermal equilibrium is reached.
• Heat Is Different from Temperature. Energy transfer can be
measured by temperature.
• The transfer of thermal energy (heat) does not always
result in a change of temperature.
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TEMPERATURE
Fahrenheito = (Co)(9/5)+32
NMC “Celsius has to work 9-5 for over a month to equal Fahrenheit.”
TEMPERATURE
• SI unit for temperature is the Kelvin, K.
• Why do we have use Kelvin? The scale is a
temperature scale where absolute zero—the
coldest possible temperature where there is no
heat energy—is defined as zero kelvin (0 K).
• Usefull for scientific calculations, since it begins
at absolute zero, meaning it has no negative
numbers.
• (The word "degrees" is NOT used with Kelvin
– it is not in degree but is an absolute number
from a fixed zero.)
ABSOLUTE ZERO
TEMPERATURE
Absolute zero is the point on the thermodynamic
(absolute) temperature scale where the heat
energy is minimum, that is, no more heat can
be removed from the system, corresponds to
zero kinetic energy.
0° Celsius scale is designated as the freezing
point of water.
0 Kelvin scale is designated as absolute zero
We will use both the Celsius and Kelvin scales.
Temperature change is the same in
Kelvin/Celsius.
T(K) = T(°C) + 273 °C
T(°C) = T(K) – 273 K
NMC “Why can’t I see that Celsius will be smaller by 273.”
TEMPERATURE CONVERSIONS
1. 126 °F =
2. 827 °F =
3. 323 °F =
4. 414 °F =
5. 443 °F =
6. 322 °F =
7. 322 °C =
8. 462 °C =
9. 486 °C =
10.52 °C =
11.32 °C =
12.-62 °C =
_______ °C
_______ °C
_______ °C
_______ °C
_______ °C
_______ °C
_______ °F
_______ °F
_______ °F
_______ °F
_______ °F
_______ °F
1. 126 °F =
2. 827 °F =
3. 323 °F =
4. 414 °F =
5. 443 °F =
6. 322 °F =
7. 322 °C =
8. 462 °C =
9. 486 °C =
10.52 °C =
11.32 °C =
12.-62 °C =
__52__ °C
__442__ °C
__162__ °C
__212__ °C
__228__ °C
__161__ °C
__612__ °F
__864__ °F
__907__ °F
_126__ °F
__90__ °F
_-80__ °F
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TEMPERATURE CONVERSIONS
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
126 K =
827 K =
323 K =
414 K =
443 K =
322 K =
322 °C =
462 °C =
486 °C =
52 °C =
32 °C =
-62 °C =
_______ °C
_______ °C
_______ °C
_______ °C
_______ °C
_______ °C
_______ K
_______ K
_______ K
_______ K
_______ K
_______ K
Round Up
1. 126 K = __-147__ °C
2. 827 K = __554__ °C
3. 323 K = ___50__ °C
4. 414 K = ___141_°C
5. 443 K = __170__ °C
6. 322 K = ___49__ °C
7. 322 °C = __595__ K
8. 462 °C = __735__ K
9. 486 °C = __759__ K
10.52 °C = __325__ K
11.32 °C = __305__ K
12.-62 °C = __211__ K
TEMPERATURE CONVERSIONS QUIZ
1) 32 °C = _______ °F
2) -62 °C = _______ °F
3) 32 °C = _______ K
4) -62 °C = _______ K
5) 283K = _______ °C
1) 32 °C =
2) -62 °C =
1. 32 °C =
2. -62 °C =
1) 283K =
__90__ °F
_-80__ °F
__305__ K
__211__ K
_10__ °C
A practical refrigerator cycle uses a special liquid
which, when the pressure is reduced,
evaporates to become a gas. Such a cycle is
illustrated below.
Uncertainty
Accuracy is how close one measurement relates to
the True Value; always considered in experiments
Bulls Eye = high accuracy
Precision -How closely several measurements agree
in exactness to one measurement.
How closely several measurements of the same quantity
made in the same way agree with one another.
Precision differs from accuracy.
Dartboard
The bull’s-eye of the dartboard represents the true
value.
1) Darts in the bull’s-eye mean high accuracy and
high precision.
2) Darts clustered within a small area but far from the
bull’s-eye mean low accuracy and high precision.
3) Darts scattered around the target and far from the
bull’s-eye mean low accuracy and low precision
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Dartboard
Uncertainty
Divide 9 by 55
Uncertainty
Uncertainty
Calculating Errors
There are two common ways to calculate your error:
Percentage Error is the most common way of measuring an
error, and often the most easy to understand.
1. Absolute Error is when you subtract the accepted value
from your measured value…
Absolute Error = Accepted Value - Measured Value
Percentage Error = Absolute Error/Accepted Value X 100
So, if a student measured a pencil to be 102mm long, and an
independent lab with high tech equipment measured it as
104mm, the percentage error is…
A negative answer means you are over the accepted value.
A positive answer means you are under the accepted value
Percentage Error =
(102mm - 104mm) / (104mm) =-0.02
But take the absolute value of this number
So, if a student measured a pencil to be 102mm long, and an
independent lab with high tech equipment measured it as
104mm, the absolute error is…
Which means you have a -2% error. The minus sign just means
that you were under the accepted value.
104-102=2mm
In high school labs, don’t be surprised if you obtain errors of
25%. The important part is, can you explain your errors!
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Uncertainty
Anytime an experiment is conducted, a certain degree of
uncertainty must be expected. There are three reasons
you might have an error in a measurement.
1. Physical errors in the measuring device
Example 1: Your thermometer was dropped and has small
air bubbles in it.
2. Improper or sloppy use of measuring device
Example 2: When you used your thermometer, you
measured the values in Fahrenheit instead of Celsius or
don’t place thermometer in center of beaker/same spot.
3. Ambient conditions (temperature, pressure, etc.)
Example 3: Measuring the length of a piece of wood
outdoors in the winter using a metal ruler, you forget that
metal contracts in the cold making the ruler shorter.
• ACCURACY AND PRECISION LAB
Scientific Method
Scientific Method
A series of steps followed to:
Form a
Hypothesis
Construct a
Theroy
No
No
Ask
Questions
Make
Observations
Test the
Hypothesis
Analyze the
Results
Do they
support your
Hypothesis?
Draw
Conclusions
Yes
Can others
confirm yourYes
results?
Publish
Results
1.
2.
3.
4.
5.
solve problems
collecting data
formulating a hypothesis
testing the hypothesis
stating conclusions
WHY????
Experiments May Not Turn Out As Expected
LAW
?
Roger Bacon (1220-1292)
For hundreds of years alchemists tried to change metals such
as lead into gold. Roger Bacon, though didn't "make" gold,
made many worthwhile scientific discoveries like gunpowder
and optic devices and had many interesting ideas about
flying machines. He is regarded by many as the forerunner
of experimental science.
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Scientific Method
Scientific Method
Hypothesis -
Theory -
A testable explanation for observations.
short and specific statement
guide scientists in making their
observations about the natural world
Takes cause effect, if-then statement
Explanation for observations
Result of repeated experiments
(and refined hypotheses)
Can be discarded for better theories
Scientific Method
Scientific Method
Variable: anything in the natural world that has levels which are
measurable. It could affect the results of an experiment.
Law – behavior of natural world how things work
example:
1. law of conservation of matter - matter cannot be
created or destroyed (mostly)
2. law of conservation of energy - energy cannot be
created or destroyed
True until an observation/experiment proves false.
SCIENTIFIC NOTATION
Chemistry deals with very large and very small numbers. Consider
this calculation:
(0.000000000000000000000000000000663 x 30,000,000,000) ÷
0.00000009116
The mess above is easily rewritten to:
(6.63 x 10¯ 31 x 3.0 x 1010) ÷ 9.116 x 10¯ 8
SN makes it much more compact, better represents significant
figures, and easier to manipulate mathematically. The trade-off,
of course, is that you have to be able to read scientific notation.
What you should remember
how to write numbers in scientific notation
how to convert to-from scientific notation.
As you work, keep in mind that a number like 9.116 x 10¯ 8 is
ONE number (0.00000009116) represented as a number 9.116
and an exponent (10¯ 8).
A scientist changes variables one at a
time to see which affects
experimental outcome
Independent (manipulated) variable
Dependent (responding) variable
Hypothesis – If I heat a 2 Liter bottle in an oven
then it will increase in pressure.
SCIENTIFIC NOTATION
Format for Scientific Notation
1. Used to represent positive numbers only.
2. Every positive number X can be written as:
(1 < N < 10) x 10 some positive or negative
integer
Where N represents the numerals of X with the
decimal point after the first nonzero digit.
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SCIENTIFIC NOTATION
1. Write, 1,202.5 in scientific notation.
1,202.5 = 1.2025 X 103
SCIENTIFIC NOTATION
2. Write, 0.0005203 in scientific notation.
0.0005203 = 5.203X10-4
OPTION #1: Move the decimal left so that
# is between 1 and 10.
OPTION #2: Move the decimal to the right so
that that # is between 1 and 10.
The number of places the decimal moves
left is the positive exponent of 10.
The number of places the decimal moves
right is the negative exponent of 10.
WRITE IN SCI NOTATION
Write in Scientific Notation
0.00087
8.7 x 10-4
9.8
9.8 x 100 (100 is seldom written)
23,000,000
2.3 x 107
0.000000809
8.09 x 10-7
250,000,000,000
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
2.10 x 10-2
1.0254 x 108
3.00 x 10-8
2.0 x 100
2.309 x 103
1.03 x 102
1.3 x 102
1.0001 x 103
3.03 x 100
1.0 x 10-1
2.0 x 101
7.3 x 100
0.0210
102540000
0.0000000300
2.0
23.09 x 102
103
130
1000.1
2 + 1.03
4.00 – 3.9
12 x 1.65
12 / 1.65
2.50 x 1011
Convert to Scientific
Notation:
1) 2.8 x 10
1) 28,000,000
2) 305,000
3) 0.000000463 (6)
4) 0.000201
5) 3,010,000
6) 0.000000000000057 (13)
7) 20,100
8) 0.00025
9)65,000,000,000,000,000 (15)
10) 8.54 x 1012
11) 2101 x 10-16
12) 305.1 x 107
13) 0.0000594 x 10-16 (4)
14) 0.00000827 x 1019 (5)
15) 386 x 10-22
16) 2511 x 1012
17) 0.000482 x 10-12 (3)
18) 0.0000321 x 1012 (4)
19) 288 x 105
20) 4.05 x 1011
7
2) 3.05 x 105
3) 4.63 x 10-7
4) 2.01 x 10-4
5) 3.01 x 106
6) 5.7 x 10-14
7) 2.01 x 104
8) 2.5 x 10-4
9) 6.5 x 1016
10) In SN
11) 2.101 x 10-13
12) 3.051 x 109
13) 5.94 x 10-21
14) 8.27 x 1013
15) 3.86 x 10-20
16) 2.511 x 1015
17) 4.82 x 10-16
18) 3.21 x 107
19) 2.88 x 107
20) In SN
MOLE DAY
• SMORE LAB
• ABSOLUTE ZERO VIDEO START
http://www.youtube.com/watch?v=y2jSv8PDDwA&oref=http%3A%2F%2Fwww.youtube.com%2Fresults%3Fsearch_query%3Dabsolute%2Bz ero%2Bnova%26aq%3Df&ytsession=jUiZHm
YrNwZw8kvsUrZtCU8gqGz ZfizRmTTIsNVCF2ofZCvHCJPihkrd4Shswa4Cq8rFhbbEhxwdoGWz OdXGjJmsDIuEwSm5_W1WIYwMX3Yamk0fjfvE3nuXWCUuYQucAhkKggD27Xw9FFpHpk1z 2B_b599pbafTHO198w53j43JZ7Wf
2Fcgcw1oTT25AIxgjOz WxWtxV3H10XiCRaNjAs_ZYmQwXZ915YPz R5uz XV08wUqlz Zt-egJEa0wCODftSH4LgPpUjuhoy76f6z nr0D4wJmYZ4_PfYy3aI_dF64&has_verified=1
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Significant Figures
• DIVIDE 7 by 22
• DIVIDE 65 by 88
Calculators Do Not Identify Significant Figures
If I have something that weights .005g and with a volume of .001mL,
what is the density?
All non-zero digits are significant.
Zeros in between non-zero digits are significant.
Example 1: zeros in 4506002 are all significant
Leading zeros (before the first non-zero digit) are not sig.
Example 2: in 0.0012, the zeros are not significant (only 2
significant figures)
Trailing zeros are significant only if a decimal point is present.
Example 3: in 13900 the trailing zeros are not significant, but in
13900.0 all zeros are significant
6.295 g has four significant figures.
46.3 m has three significant figures.
40.7 L has three significant figures.
87009 km has five significant figures.
0.009587 m has four significant figures.
0.0009 kg has one significant figure.
Rule for Additions and Subtraction
Determine the Number
of SF:
1) 8
1) 28,000,000.
2) 305,000
3) 0.000000463 (6)
4) 0.000201 (3)
5) 3,010,000
6) 0.000000000000057 (13)
7) 20,100
8) 0.00025
9)65,000,000,000,000,000 (15)
10) 8.54 x 1012
11) 2101 x 10-16
12) 305.1 x 107
13) 0.0000594 x 10-16 (4)
14) 0.00000827 x 1019 (5)
15) 386 x 10-22
16) 2511 x 1012
17) 0.000482 x 10-12 (3)
18) 0.0000321 x 1012 (4)
19) 288 x 105
20) 4.05 x 1011
2) 3
3) 3
4) 3
5) 3
6) 2
7) 3
8)2
9) 2
10)3
11) 4
12) 4
13) 3
14) 3
15) 3
16) 4
17) 3
18) 3
19) 3
20) 3
SIG FIGS
The resulting number has the same number of decimal places
than the number with the least decimal places.
Example 4:
5.36
+ 99.124
104.48 (2 decimal places)
Rule for multiplications and divisions
The resulting number has the same
number of significant figures than the
number with the least significant
figures.
Example 5:
5.5
- 1.12
4.4 (2 decimal places)
Example 6: 15.322
x 3.12
47.8 (3 significant figures)
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SIG FIG ROUNDING
SIG FIG ROUNDING
Example 7:
• Rules for Rounding
• Keep the extra digits until the end of the
calculations then round the final digit to the
correct number of significant digit.
• Round the extra digit after the last significant
one.
• If the extra digit is smaller than 5 the last
significant digit remains.
• If the extra digit is greater or equal than 5, add
round up last significant digit.
SIG FIG ROUNDING
Example 8:
Round 12.576 to 3 significant digits
12.576 the extra digit is 7 therefore the last
significant digit is increased by 1.
The result is 12.6
COMPLEX SIG FIGS
Example 9:
34.53 + 2.0342 x 16.2
According to the usual rules of arithmetic, we do
multiplication before the addition.
PEMDAS (PLEASE EXSCUSE MY DEAR AUNT SALLY)
How many significant figures should be in the result
of this multiplication? 32.95404
2.0342 x 16.2 =32.954
Next, we do the addition. How many significant
figures should be in the result of this addition?
34.53 + 33.0=67.53
The correct answer should have: 67.5 (3 SF)
Round 12.536 to 3 significant digits
In this examples: Digits in green are significant, the
digit in blue is the "extra" digit and the digit in red
is ignored.
12.536
the extra digit is 3 therefore the last significant digit
remains.
COMPLEX SIG FIGS
While operating within the same rules (i.e. X
and ÷ ) use all the numbers in your
calculator. At the end of these type of
operations determine your significance.
You must do this before your move into a
different type of operations ( + and - ).
COMPLEX SIG FIGS
Example 10:
(34.53 + 2.0342) x 16.2
The parentheses around the first two terms indicate
that the addition should be done first. How many
significant figures should be in the result of this
addition?
34.53 + 2.0342 =36.564
Next, we do the multiplication. How many significant
figures should be in the result of this
multiplication?
36.564 x 16.2 = 592.337
The correct answer should have: 3 significant
figures.
592
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COMPLEX SIG FIGS
1) 3.461728 + 14.91 + 0.980001 + 5.2631
24.61
2) 23.1 + 4.77 + 125.39 + 3.581
156.8
3) 22.101 - 0.9307
21.170
4) 0.04216 - 0.0004134
0.04175
5) 564,321 - 264,321
300,000. = 3.00000 x 105
COMPLEX SIG FIGS
10) 5.26 x 5.25 x 6.34 +0.11
175
(3 SF)
11) 5.25 x1.1 + 6.342
12.1 (3SF)
COMPLEX SIG FIGS
6) (3.4617 x 107) ÷ (5.61 x 10-4)
(3 SF)
6.17 x 1010
7) [(9.714 x 105) (2.1482 x 10 -9)] ÷ [(4.1212)
(3.7792 x 10-5)]
1
1.3398 x 10 = 13.40 (4SF)
8) (4.7620 x 10-15) ÷ [(3.8529 x 1012)(2.813 x 10-7)
(9.50)]
4.62 x 10-22
(3 SF)
÷[(3.8529 x 1012) (2.813 x 10-7)
9) (4.7620 x 10-15)÷
(9.50)]
2.30 x 103
(3 SF)
COMPLEX SIG FIGS
1) 3.461728 + 14.91 + 5.2631x 3.46 ÷ 6.81
18.37 + 18.2 ÷ 6.81 = 21.04
2) 23.1 + 4.77 + 125.39 + 3.581 x 9.7 x 2.142 ÷ 4.12
153.26 + 18.1= 171.4
3) 9.23 x 2.6 ÷ 4.999 + 22.101 - 0.9307
4.8 + 21.170 = 25.9
4) 0.04216 ÷ 8.945 + 22.101 - 0.0004134
26.274
5) 564,321 - 264,321 x 0.544 ÷ 8.654 + 82.4533 0.0004134 18817.5 = 18818
COMPLEX SIG FIGS
LAB PARTNER EXERCISE
6) (3.4617 x 107) ÷ (5.61 x 10-4)
6.17 x 1010
(3 SF)
7) [(9.714 x 105) (2.1482 x 10 -9)] ÷ [(4.1212)
(3.7792 x 10-5)]
1.3398 x 101 = 13.40 (4SF)
8) (4.7620 x 10-15) ÷ [(3.8529 x 1012)(2.813 x 10-7)
(9.50)]
-22
4.62 x 10
(3 SF)
11)Why are significant figures important when taking
data in the laboratory?
12)Why are significant figures NOT important when
solving problems in your math class?
13) Using two different instruments, I measured the
length of my foot to be 27 centimeters and 27.00
centimeters. Explain the difference between these
two measurements.
14) I can lift a 20 kilogram weight over my head ten
times before I get tired. Write this measurement to
the correct number of significant figures.
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10/27/2011
LAB PARTNER EXERCISE
11) Why are significant figures important when taking data
in the laboratory?
Significant figures indicate the precision of the
measured value to anybody who looks at the data. For
example, if a weight is measured as being “1100
grams”, this means that the mass has been rounded
to the nearest hundred grams. If a weight is measured
as being “1100.0 grams”, this means that the mass
has been rounded to the nearest tenth of a gram.
Though the numbers plug into the calculator in exactly
the same way, they mean very different things.
12) Why are significant figures NOT important when
solving problems in your math class?
Math classes don’t deal with measured values. As a
result, all of the numbers are considered to be
infinitely precise.
LAB PARTNER EXERCISE
.
13) Using two different instruments, I measured the length
of my foot to be 27 centimeters and 27.00 centimeters.
Explain the difference between these two measurements.
As in problem 11, the first measurement implies that
my foot is somewhere between 26.5 and 27.4 cm long.
The second measurement implies that my foot is
between 26.995 and 27.004 cm long. Again, though
the numbers plug into the calculator in the same way,
they imply different precisions.
14) I can lift a 20 kilogram weight over my head ten times
before I get tired. Write this measurement to the correct
number of significant figures.
The answer is written as 20.0, with a line drawn above
the zero in the tenths place. This is one of the few
cases where you can measure data with infinite
significant figures. After all, I can either lift it or I can’t
– there’s no “half-lift” that would result in a decimal.
LAB PARTNER EXERCISE
How many significant figures are in each of the following
numbers?
1)
2)
3)
4)
5)
6)
7)
8)
9)
10)
5.40
210
801.5
1,000
101.0100
1.2 x 103
0.00120
0.0102
9.010 x 10-6
2,370.0
3
2
4
1
7
2
3
3
4
5
Specific Heat (c)
The specific heat - energy transferred as heat
to a substance vs. that substance’s temperature
change
1. The Quantity of energy as heat that must be
transferred to raise the temperature of 1 g of a
substance (1°C).
2. Specific heat is expressed in joules (energy)
per gram Kelvin (J/gK).
3. Metals tend to have low specific heat
(less energy must be transferred as heat to raise
their temperatures. ) Water’s specific heat it is
the highest of most common substances.
Specific Heat (s)
• Transfer of heat affects dissimilar substances in
different ways.
• Why does a pan of water heat slower than an empty
pan?
• Why does a tongue stick on a cold metal flagpole vs. a
cold telephone pole?
• Why does a 400 mL heat slower than 40 mL?
Specific Heat (S)
Q=cm∆
∆ T NMC: Q-SMAT
Q
s m ∆T
The relationship between
heat and temperature
change is usually expressed
in the form shown above
where c is the specific heat.
The relationship does not
apply if a phase change is
encountered, because the
heat added or removed
during a phase change does
not change the temperature.
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10/27/2011
Specific Heat (c)
Specific Heat (c)
If the total energy required to increase the temperature of a
substance is much larger, its specific heat is much larger.
J/kg/o C
cal/g/o C
Water (0 oC to 100 oC)
4186
1.000
Methyl Alcohol
2549
0.609
2093
0.500
Steam (100 C)
2009
0.480
Wood (typical)
1674
0.400
Soil (typical)
1046
0.250
Air (50 oC)
1046
0.250
Aluminum
900
0.215
Marble
858
0.205
Glass (typical)
837
0.200
Iron/Steel
452
0.108
Copper
387
0.0924
Silver
236
0.0564
Mercury
138
0.0330
Gold
130
0.0310
Lead
128
0.0305
Substance
o
o
Ice (-10 C to 0 C)
o
Specific Heat (s) Q unknown
Step 4. Substitute your values into the formula
Q= sm(∆
∆ T)
Q = 23.984 g x 393.0 oC x 0.902 J/g x oC
Step 5. Cross out units where possible, and solve
for unknown.
Q= 23.984 g x 393.0 oC x 0.902 J/g x oC
Q= 8501.992224 J
Step 6. Round to the correct number of significant
digits and check answer makes sense.
Q = 8.50 x 103 J
Specific Heat (s) m unknown
Step 3: List the known and unknown
factors. Looking at the units in the word problem
will help you determine which is which.
Q = 24 500 J m = ? ∆ T = 69.5 oC s = 4.18 J/goC
Step 4. Substitute your values into the formula.
m = Q/(∆
∆ T)s
m = 24 500 J/69.5 oC x 4.18 J/goC
Step 5. Cross out units where possible, and solve for
unknown.
m = 24500 J/(69.5 oC x 4.18 J/goC)
m = 84.3344463184 g
Example #1 If Heat Transferred (Q) is the unknown:
Ex. Aluminum has a specific heat of 0.902 J/g x
oC. How much heat is lost when a piece of
aluminum with a mass of 23.984 g cools from a
temperature of 415.0 oC to a temperature of 22.0 oC?
Step 1: First read the Question and try to understand
what they are asking you.
Step 2: Write the original formula. Q= sm(∆
∆ T)
Step 3: List the known and unknown factors.
Q= ?
m=23.984g ∆ T=(415.0oC-22.0oC)=393.0oC
(remember, asked for the change in temperature)
s = 0.902 J/goC
Specific Heat (s) m unknown
Example #2 If mass (m) is the unknown:
The temperature of a sample of water increases by 69.5 o C
when 24,500 J are applied. The specific heat of liquid water
is 4.18 J/g x oC. What is the mass of the sample of water?
Step 2: Write the original formula, and then modify it isolate the
unknown.
Q= sm(∆
∆ T)
Q = m(∆T)c
(∆T)s (∆T)s
m = Q/(∆
∆ T)s
Specific Heat (s) m unknown
Step 6. Round to the correct number of
significant figs and check to see that you
answer makes sense.
m = 84.3 g
Our answer makes sense because grams
are the correct units for mass, and the
value should be positive.
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10/27/2011
Specific Heat (s) ∆ T unknown
850 joules of heat are applied to a 250 g sample of
water with an initial temperature of 13.0 oC. Find a)
the change in temperature and b) the final
temperature. (remember, the specific heat of liquid
water, in joules, is 4.186 j/g x oC.)
Q = m(∆ T)s
Q = m(∆T)s
ms
ms
∆ T = Q/ms
Specific Heat (s) ∆ T unknown
Step 5. Cross out units where possible, and
solve for unknown.
∆ T = 850 cal/250 g x 4.186/g x oC
Answer to step a) ∆ T = .81 o C
Answer to step b) Tf= 13.0 oC + .81 oC = 13.8 oC
Step 6. Round to the correct number of
significant digits and check to see that you
answer makes sense. 13.8 oC
Specific Heat (s) s is unknown
Step 3: List the known and unknown
factors. Looking at the units in the word
problem will help you determine which is
which
Q = 34700 J
m = 350 g s = ?
∆ T = (173.0oC - 22.0oC) = 151.0 oC
Step 4. Substitute your values into the
formula
s = Q/m(∆ T)
s = 34700 J/(350 g x 151.0 oC)
Specific Heat (s) ∆ T unknown
Step 2: Write the original formula, and then
modify it isolate the unknown.
Step 3: List the known and unknown
factors. Looking at the units in the word
problem will help you determine which is
which.
Q = 850 J m = 250 g s = 4.86/goC ∆ T = ?
Step 4. Substitute your values into the
∆ T = Q/mc
formula
∆ T = 850 cal/(250 g x 4.86/g oC)
Specific Heat (s) c is unknown
When 34, 700 J of heat are applied to a 350 g
sample of an unknown material the
temperature rises from 22.0 oC to 173.0
oC. What must be the specific heat of this
material?
Step 2: Write the original formula, and then
modify it to isolate the unknown.
s = Q/m(∆
∆ T)
Specific Heat (s) s is unknown
Step 5. Cross out units where possible, and
solve for unknown.
s = 34 700 J/350 g x 151.0 oC
s = 0.65657521286 J/g x oC
Step 6. Round to the correct number of
significant digits and check to see that you
answer makes sense.
s= 0.66 J/g x oC
16
10/27/2011
Specific Heat Problem 1
Specific Heat Problem 2
Calculate the amount of heat needed
to increase the temperature of 250g
of water from 20oC to 56oC.
Calculate the specific heat capacity of
copper given that 204.75 J of energy
raises the temperature of 15g of copper
from 25oc to 60oc
m = 204.75g
m = 250g s = 4.18 J/goC Tf = 56oC Ti = 20oC
Q = s x m x (Tf - Ti)
Q = 250g x 4.18 J/goC x (56 - 20)oC
Q = 250g x 4.18 J/goC x 36oC
Q = 37620 J = 38000J
Specific Heat Problem 3
216 J of energy is required to raise the
temperature of aluminum from 15oC to
35oC. Calculate the mass of aluminum.
(Specific Heat Capacity of aluminum is
0.90 J/goC.
Q = 216 J s = 0.90 J/goC Ti = 15oC Tf = 35oC
Q = m x x (Tf - Ti)
216J = m x 0.90 J/goC x (35 - 15) oC
216J = m x 0.90 J/goC x 20 oC 216
216J = m x 18g
m = 216J ÷ 18J/g= 12g
m = 15g
Tf = 60oC Ti = 25oC
Q = s x m x (Tf - Ti)oC
204.75J = 15g x s x (60 - 25)oC
204.75J = 15g x s x 35oC
204.75J = 525goC x s
s = 204.75J/goC ÷ 204.75 = 0.39 J/goC
Specific Heat Problem 4
The initial temperature of 150g of ethanol
was 22oC. What will be the final
temperature of the ethanol if 3240 J was
needed to raise the temperature of the
ethanol? Specific heat capacity of ethanol
is 2.44 J/goC).
Q = 3240 J m = 150g s = 2.44 J/goC Ti = 22oC
Q = m x s x (Tf - Ti)
3240J = 150 x 2.44 x (Tf - 22)
3240J = 366 (Tf - 22)
8.85 = Tf - 22
Tf = 30.9oC
17