Announcements & Agenda (01/15/07)
Last time: Making Measurements
Office Hours (SC 3063)
Permanent Change: Thurs 99-11am replaces Mon 33-5pm
This Week: must cancel Wed 33-5pm OH (will leave @ 2pm)
Quiz 1 (the CD quizzes) will be accepted until Wed
First inin-class quiz (Quiz 2) will be Wed. (Ch 1 & 2)
See overhead for your clicker number
You should currently be reading Ch 3 ☺
Today:
Heat and temperature (2.3, 2.4)
Phase changes (2.5, 2.6)
. l2. . . . l . . . . l3 . . . . l . . . . l4. .
cm
• The markings on the meter stick at the end of the
•
•
orange line are read as
the first digit
2
plus the second digit
2.7
The last digit is obtained by estimating.
estimating.
The end of the line might be estimated between 2.7–
2.7–
2.8 as halfhalf-way (0.5) or a little more (0.6), which gives
a reported length of 2.75
2.75 cm or 2.76
2.76 cm.
Measured #s always need units!
1
2
LAST TIME: Energy
Last Time:
Conversion Factors in Problem Solving
Energy – the ability to do work
•
•
•
Write the given and needed units.
Write a unit plan to convert the given unit to the needed unit.
Write equalities and conversion factors that connect the
units.
• Use conversion factors to cancel the given unit and provide
the needed unit.
1. Kinetic Energy: energy of motion
2. Potential Energy: stored energy
IMPORTANT EXAMPLES:
– ENERGY OF BONDS
– ENERGY OF FOOD MOLECULES
3. Thermal Energy
4. Radiant Energy
Unit 1
x
Given
unit
x
Unit 2
Unit 1
Conversion
factor
= Unit 2
= Needed
unit
PRACTICE!
3
4
1
Heat & Temperature
Three Important Temperature Scales
Fahrenheit
Particles are always moving.
When you heat water, the water molecules move faster.
When molecules move faster, the substance gets hotter.
Water boils
When a substance gets hotter, its temperature and total
energy content increases.
measure temperature changes with a thermometer
Celsius
Kelvin
100°C
373 K
212°F
180°
100°C
Water freezes 32°F
0°C
100K
273 K
5
6
Fahrenheit Formula
Celsius Formula
• On the Fahrenheit scale,
scale, there are 180°
180°F between the
freezing and boiling points and on the Celsius scale,
scale,
there are 100°
100°C.
180°F =
9°F =
1.8°F
100°C
5°C
1°C
• In the formula for the Fahrenheit temperature,
temperature, adding 32
adjusts the zero point of water from 0°
0°C to 32°
32°F.
TF
= 9/5 TC + 32°
32°
TF
= 1.8 TC + 32°
32°
• TC is obtained by rearranging the equation for TF.
TF =
1.8TC + 32
• Subtract 32 from both sides.
TF - 32
=
1.8TC ( +32 - 32)
TF - 32
=
1.8TC
• Divide by 1.8 = °F - 32
1.8
TF - 32
=
TC
1.8
or
7
= 1.8 TC
1.8
8
2
Solving A Temperature Problem
Specific Heat (2.4)
A person with hypothermia has a
body temperature of 34.8°C.
What is that temperature in °F?
TF
Different substances have different capacities for
storing energy
It may take 20 minutes to heat water to 75°C.
However, the same mass of aluminum might require
5 minutes and the same amount of copper may take
only 2 minutes to reach the same temperature.
= 1.8 TC + 32°
32°
TF = 1.8 (34.8°C)
exact
tenth's
+ 32°
exact
THINK ABOUT BOB MAKING MAC & CHEESE
= 62.6 + 32°
= 94.6°F
tenth’s
3 sig. figs. based on
measured numbers!
Copyright © 2005 by Pearson Education, Inc.
Publishing as Benjamin Cummings
9
Examples of Specific Heats
10
Specific Heat: Mathematical Description
Specific heat (SH)
TABLE 2.6
cal/g°C
0.214
0.0920
0.0308
0.108
0.0562
0.125
• is different for different substances.
• is the amount of heat that raises the temperature of 1 g
of a substance by 1°C
• Heat of a process = (Specific Heat) x (mass) x (∆T)
• in the SI system has units of J/g°
J/g°C.
0.488
0.588
0.207
0.100
• in the metric system has units of cal/g°
cal/g°C.
11
12
3
Heat Equation
Learning Check
Rearranging the specific heat expression gives the
heat equation.
equation.
Heat = g x °C x
A. When ocean water cools, the surrounding air
1) cools.
2) warms.
3) stays the same.
J =J
g°C
B. 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.
The amount of heat lost or gained by a substance is
calculated from the
• mass of substance (g).
• temperature change (∆
(∆T).
• specific heat of the substance (J/g°
(J/g°C).
C).
Key Point: If one substance “heats up” by a certain
amount, another substance must exactly lose that same
amount of heat!
13
How many kilojoules are needed to raise the
temperature of 325 g of water from 15.0°C
to 77.0°C (SH of water = 4.184 J/(g
J/(goC)?
1) 20.4 kJ
2) 77.7 kJ
3) 84.3 kJ
0%
0%
0%
1
2
3
4
5
14
Solution
How many kilojoules are needed to raise the
temperature of 325 g of water from 15.5°C to
77.5°C?
3) 84.3 kJ
77.0°C – 15.0°C = 62.0°C
325 g x 62.0°C x 4.184 J x 1 kJ
g °C
1000 J
= 84.3 kJ
15
16
4
Another Example
Solution
What is the specific heat of a metal if 24.8 g absorbs
275 J of energy and the temperature rises from 20.2°
20.2°C to
24.5°
24.5°C?
What is the specific heat of a metal if 24.8 g absorbs
275 J of energy and the temperature rises from 20.2°
20.2°C to
24.5°
24.5°C?
Given: 24.8 g metal, 275 J of energy, 20.2°
20.2°C to 24.5°
24.5°C
Need: J/g°
J/g°C
Plan:
SH = Heat/g°
Heat/g°C
∆T = 24.5°
24.5°C – 20.2°
20.2°C = 4.3 °C
Set Up:
275 J
(24.8 g)(4.3°
g)(4.3°C)
= 2.6 J/g°
J/g°C
17
Why are we so interested in heat?
Chemical reactions
that produce heat
“Exothermic”
Chemical reactions
that absorb heat
“Endothermic”
18
States of Matter
Solid
Liquid
Gas
Heat is related to whether a chemical or
biological process will happen!
19
20
5
Phases Changes
States of Matter
Solid
Liquid
Gas
deposition
melting
sublimation
(e.g. freezefreeze-drying)
freezing
21
The Heat of Fusion
condensation
vaporization
22
Calculations Using 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°
0°C) is
80. cal
1 g water
23
The heat needed to freeze (or melt) a specific mass of
water (or ice) is calculated using the heat of fusion.
Heat = g water x 80. cal
1 g water
Example:
Example: How much heat in cal is needed to melt 15.0
g of water?
15.0 g water x
80. cal
1 g water
= 1200 cal
24
6
How many kilocalories (kcal) are released
when 50.0 g of steam from a volcano
condenses at 100°
100°C? (Heat of Vaporization
= 540 cal/g)
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.
1)
2)
3)
4)
0%
0%
0%
Boiling Point of Water = 100°C
0%
27 kcal
540 kcal
54 kcal
2700 kcal
Heat of Vaporization (water)
=
540 cal
1 g water
25
Summary of Phase
Changes: Heating Curves
1
2
3
4
5
26
Cooling Curve
A heating curve
• illustrates the
changes of state as
a solid is heated.
• uses sloped lines to
show an increase in
temperature.
• uses plateaus (flat
lines) to indicate a
change of state.
Using the heating curve of water as a guide,
draw a cooling curve for water beginning
with steam at 110°C and ending at -20°C.
27
28
7
Combined Heat Calculations
(continued.)
Combined Heat Calculations
To reduce a fever, an infant is packed in 250. g of ice.
If the ice (at 0°C) melts and warms to body
temperature (37.0°C), how many calories are
removed from the body?
Step 2: Calculate the heat to melt ice (fusion)
250. g ice x 80. cal
= 2.000 × 104 cal
1 g ice
Step 1: Diagram the changes
Step 3: Calculate the heat to warm the water from
0°C to 37.0°C (SH of water = 1 cal/g)
250. g x 37.0°C x 1.00 cal = 9 250 cal
g °C
∆T = 37.0°
37.0°C - 0°C = 37.0°
37.0°C
37°C
temperature increase
0°C
solid
liquid
melting
Total: Step 2 + Step 3
29
= 29 200 cal
(rounded to 3 SF)
30
8