Download Energy - Moore Public Schools

Chapter 15 Section 1
ENERGY
Copyright © McGraw-Hill Education
Energy
The Nature of Energy
•
Energy is the ability to do work or produce heat.
•
Energy exists in two basic forms: potential energy and kinetic energy.
•
Potential energy is energy
due to composition or
position.
•
Kinetic energy is energy
of motion.
Copyright © McGraw-Hill Education
Energy
The Nature of Energy
•
The law of conservation of energy states that in any chemical reaction or
physical process, energy can be converted from one form to another, but it
is neither created nor destroyed—also known as the first law of
thermodynamics.
•
Chemical potential energy is energy stored in a substance because of its
composition.
•
Chemical potential energy is important in chemical reactions.
•
Heat is energy that is in the process of flowing from a warmer object to a
cooler object.
•
q is used to symbolize heat.
Copyright © McGraw-Hill Education
Energy
Measuring Heat
•
A calorie is defined as the amount of
energy required to raise the temperature
of one gram of water one degree Celsius.
•
The energy content of food is measured in
Calories, or 1000 calories (kilocalorie).
•
A joule is the SI unit of heat and energy,
equivalent to 0.2390 calories.
Copyright © McGraw-Hill Education
Energy
CONVERT ENERGY UNITS
SOLVE FOR THE UNKNOWN
Convert nutritional Calories to calories.
Use with Example Problem 1.
Problem
A breakfast of cereal, orange juice, and
milk might contain 230 nutritional
Calories. Express this energy in joules.
Response
ANALYZE THE PROBLEM
You are given an amount of energy in
nutritional Calories. You must convert
nutritional Calories to calories and then
convert calories to joules.
KNOWN
amount of energy = 230 Calories
UNKNOWN
amount of energy = ? J
Copyright © McGraw-Hill Education
•
Apply the relationship 1 Calorie = 1000 cal.
1000 cal
230 Calories ×
= 2.3 × 105 cal
1 Calorie
Convert calories to joules.
•
Apply the relationship 1 cal = 4.184 J.
4.184 J
2.3 × 105 cal ×
= 9.6 × 105 J
1 cal
EVALUATE THE ANSWER
The minimum number of significant figures used
in the conversion is two, and the answer correctly
has two digits. A value of the order of 105 or 106
is expected because the given number of
kilocalories is of the order of 102 and it must be
multiplied by 103 to convert it to calories. Then,
the calories must be multiplied by a factor of
approximately 4. Therefore, the answer is
reasonable.
Energy
Specific Heat
•
The specific heat of any substance is the
amount of heat required to raise one gram
of that substance one degree Celsius.
•
Some objects require more heat than
others to raise their temperatures.
Copyright © McGraw-Hill Education
Energy
Specific Heat
•
Calculating heat absorbed and released
Copyright © McGraw-Hill Education
Energy
CALCULATE SPECIFIC HEAT
KNOWN
energy released = 114 J
Ti = 50.4°C
Use with Example Problem 2.
mass of iron = 10.0 g Fe
Tf = 25.0°C
Problem
UNKNOWN
In the construction of bridges and
skyscrapers, gaps must be left between
adjoining steel beams to allow for the
expansion and contraction of the metal due
to heating and cooling. The temperature of a
sample of iron with a mass of 10.0 g changed
from 50.4°C to 25.0°C with the release of
114J. What is the specific heat of iron?
specific heat of iron, c = ? J/(g•°C)
SOLVE FOR THE UNKNOWN
Calculate ΔT.
ΔT = 50.4°C – 25.0°C = 25.4°C
•
q = c × m × ΔT
Response
ANALYZE THE PROBLEM
You are given the mass of the sample, the
initial and final temperatures, and the
quantity of heat released. You can calculate
the specific heat of iron by rearranging the
equation that relates these variables to
solve for c.
Copyright © McGraw-Hill Education
State the equation for calculating heat.
•
Solve for c.
c × m × ΔT
q
=
m × ΔT
m × ΔT
q
c=
m × ΔT
Energy
CALCULATE SPECIFIC HEAT
EVALUATE THE ANSWER
SOLVE FOR THE UNKNOWN (continued)
q
m × ΔT
Substitute q = 114 J, m = 10.0 g, and
ΔT = 25.4°C.
114 J
c=
(10.0 g)(25.4°C)
c=
•
•
Multiply and divide numbers and units.
c = 0.449 J/(g•°C)
Copyright © McGraw-Hill Education
The values used in the calculation have
three significant figures, so the answer is
correctly stated with three digits. The
value of the denominator of the equation
is approximately two times the value of
the numerator, so the final result, which is
approximately 0.5, is reasonable. The
calculated value is the same as that
recorded for iron in Table 2 (from slide 8).
Energy
Review
Essential Questions
•
What is energy?
•
How do potential and kinetic energy differ?
•
How can chemical potential energy be related to the heat lost or gained in
chemical reactions?
•
How is the amount of heat absorbed or released by a substance
calculated as its temperature changes?
Vocabulary
• energy
• law of conservation of
energy
• chemical potential energy
Copyright © McGraw-Hill Education
•
•
•
•
heat
calorie
joule
specific heat
Energy