Download Chapter 3: Matter and Energy

CHEMISTRY 100
CHAPTER 3: MATTER & ENERGY
Homework problems: 3, 6, 7, 10, 14, 15, 20, 21, 31, 33, 35, 37, 41, 43, 49, 57, 62, 65, 67, 73, 77,
79, 81, 91, 99
Outline
I.
Matter
A. Has mass, occupies space
1. Made of atoms and/or molecules (more on these later)
B. States
1. Solid: definite shape, volume, incompressible
2. Liquid: definite volume, no shape, incompressible
3. Gas: no shape or volume, very compressible
C. Pure or Mixture
1. Pure
a. Pure: all one type of atom or all one type of molecule
2. Mixture
a. Homogeneous: the same throughout (think homogeneous milk- different
pronounciation, same spelling, same thought
b. Heterogeneous: not the same throughout; think homogenous milk that’s been
left out of the fridge for a week or two.
II. Changes
A. Physical and Chemical changes
1. physical changes
a. changes that don’t alter the nature of the molecule
i. changes in amount of stuff
ii. changes in physical state (gas to liquid, liquid to solid, , etc) but not to
the molecules of the stuff
ice = H2O (s), when melted becomes H2O (l) → physical state changes from s to l, but
molecule stays H2O. This may become more clear after we discuss what molecules are
2. chemical changes
a. change that does alter the molecule (paper burning, etc.)
B. Physical and Chemical Properties
1. extrinsic physical properties
a. depends on amount of stuff
b. mass, volume, etc.
2. intrinsic physical properties
a. independent of amount of stuff
b. color, density, temperature, specific heat capacity, etc.
3. chemical properties
a. does the molecular formula change under a given set of circumstances
b. does paper burn? does paper rust? etc.
C. Separations
1. Based on physical properties
a. distillation
b. filtration
III. Conservation of mass
A. Mass (and energy) neither created nor destroyed
IV. Energy
A. Units
1. Calories (1 Cal = 1000 cal or 1 kcal)
2. calories
3. Joules (4.184 J = 1 cal)
4. kW/hrs = 3.60 x 106 J
B. Energy in Physical and Chemical changes
1. Exothermic
2. endothermic
V. Temperature
A. Measure of heat energy in substance
B. Fahrenheit, Celsius (Centigrade), Kelvin
Water boils
212° F
=
100° C
Water freezes
32° F
=
0° C
=
=
373 K
273 K
We will keep sig figs in temp conversions simple. If a temp is given to the nearest degree in one
scale, report it to the nearest degree in all scales. This gives the same result as doing it the long
way almost all of the time, but we do not have to keep track of which of these numbers are counts
and which are measurements. So, for Temp sig figs, keep the placeholder of the uncertainty in all
temps the same.
The Celsius and Kelvin temperatures have the same scale: the difference between the boiling and
freezing pts in both systems is 100. The only difference is the starting point. There is NEVER a
negative Kelvin temperature, but you can have negative Celsius. So, +273 to C to get K and K-273
=C
There is no direct method to go from K to F, but you can easily go from C to F and F to C. Notice
that there are 180 Fahrenheit degrees between the boiling and freezing pts of water while there are
100 Celsius degrees. The Fahrenheit numbers are bigger by a scale of 180/100 or 9/5. To go from
F to C scale, multiply by 5/9 so number gets smaller and to go from C to F, multiply by 9/5 so that
number gets bigger (you may notice that the book says to multiply and divide by 1.8. That is
because 1.8 is 9/5). But multiplying by 9/5 or 5/9 just changes the scale. F and C have different
starting points. Multiplying gets us how many degrees away from a point on those scales. Let’s
choose the freezing point.
Now, if I want to find out what a reading of 27°C is in F, then I find out how many degrees from
the freezing pt of water that reading is: it is 27° above the freezing point. Now, multiply it by 9/5
(to get the larger number scale) and I get 49°F above freezing. Since freezing in F is 32, I add 49 to
it and get 71° F. In math terms:
F = (C x9/5) +32 = 71°C
For a hot day, 101°F, convert this to C: Well, 101°F is 69°F from the freezing point of water. This
will translate into 69 x 9/5 = 38°C from the freezing point in C which is 38°. In math terms:
C = (F-32) x 5/9 = 38°C
You may notice that in math, we did exactly the opposite in exactly the opposite order. In one case
we multiplied by 9/5 and added 32, in the other we divided by 9/5 and subtracted 32.
VI. Heat Capacity
A. Specific heat capacity (called “s” or “c” depending on author of book)
1. How much heat it takes to raise 1 g by 1°C
2. For liquid water, s = 1 cal/g°C
a. For liquid water, this is a definition, so it has infinite # of sig figs
b. For all other materials, this is a measurement and has set # of sig figs
c. i.e, iron = 0.11 cal/g°C (2 sig figs) and aluminum = 0.22 cal/g°C
B. Q = smΔT
1. Q = heat, s = specific heat, m = mass, ΔT = Tf - Ti (final – initial T)
How much heat involved in taking 200.0 g water from 25.0°C to 100.0° C?
Q = smΔT = 1 cal/g°C (from VI.A.2, above) x 200.0 g x (100.0-25.0°C) = 15.0 kcal (or
1.50 x 104cal)
How much heat involved in taking 25.0 g water from 98.0°C to 0.0° C?
Q = smΔT = 1 cal/g°C (from VI.A.2, above) x 100.0 g x (0.0-98.0°C) = -2450 cal (notice
the sign of the heat is negative!)
What is the final temperature if 444 cal is applied to 100.0 g of water starting at 25.0°C?
ΔT = Q/sm = 444 cal/ ((1 cal/g°C) x 100.0 g)) = 4.44°C
And ΔT = Tf – Ti ; so 4.44°C = Tf – 25.0°C and Tf = 29.4°C
What is the final temperature if 444 cal is applied to 100.0 g of iron starting at 25.0°C?
ΔT = Q/sm = 444 cal/ ((0.11cal/g°C) x 100.0 g)) = 4.0 x 102 °C
And ΔT = Tf – Ti ; so 4.0 x 102 °C = Tf – 25.0°C and Tf = 65°C
What is the final temperature if 444 cal is applied to 25.0 g of aluminum starting at 25.0°C?
ΔT = Q/sm = 444 cal/ ((0.11cal/g°C) x 100.0 g)) = 4.0 x 102 °C
And ΔT = Tf – Ti ; so 2.0 x 102 °C = Tf – 25.0°C and Tf = 45°C
Objectives
Knowledge:
Memorize the formula for specific heat problems (Q = smΔT)
Know that ΔT = Tfinal – Tinitial
Learn new terms heterogeneous (not same throughout) and homogeneous (same throughout)
Remember the three states (s, l, g) and their characteristics (shape, volume, compressibility)
Comprehension:
Distinguish between pure substances and mixtures
Distinguish between heterogeneous and homogeneous mixtures
Understand that the specific heat capacity of a material is the amount of heat it takes to raise 1 g of
a substance by 1 degree Celsius
Realize a chemical change changes what the stuff is (changes its molecules), but a physical change
does not alter the make up of the stuff
Understand that a change in the amount of stuff, or the state (solid to liquid) is a physical change
Application:
Convert a temperature from any one scale (F, C, K) to another
Convert between the different units of energy if given conversions (I give you the conversion
factor of 4.184 J = 1 cal and you convert a number from cal to J or vice-versa)
Analysis:
If given any three of Q, s, m, or ΔT, find the fourth
If given any four of Q, s, m, Tf or Ti, find the fifth