CHAPTER 10 ENERGY
NATURE OF ENERGY
Energy – the ability to do work
 May be potential or kinetic
 Potential energy is due to position or
composition
 Kinetic energy is due to motion depends on
mass and velocity KE=1/2 mv2
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TEMPERATURE AND HEAT
Temperature is a measure of the average
kinetic energy of the random motions (thermal
energy) of the components of a substance.
 Heat is the flow of energy due to temperature
differences – flows from high to low
 Exothermic – energy released
 Endothermic – energy absorbed
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THERMODYNAMICS
The study of energy
 Law of conservation of energy is often
referred to as the First Law of
Thermodynamics
 Thermodynamic quantities have 3 parts: a
number indicating magnitude of change, a
sign ( + or -) indicating the direction of
change, and a unit (J, kJ, cal, kcal)
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ENERGY CONVERSIONS
Energy (heat) may be expressed in joules,
kilojoules, calories or kilocalories.
 1 calorie (cal) = 4.184 joules (J)
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How many joules in 60.1 calories?
HOW MANY CALORIES IN 28.4 J?
SPECIFIC HEAT CAPACITY
Defined as the amount of energy required
to change the temperature of 1 gram of a
substance by 1oC
 Different substances have different
specific heat capacities. Metals tend to
have a low specific heat capacity compared
to water – they tend to heat quickly and
cool quickly, while water does not.
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CALCULATING ENERGY CHANGES
How much energy is required to heat 25 g of
liquid water from 25oC to 100.oC? Specific heat
capacity of liquid water is 4.18 J/goC.
 Q = mcΔT
 Q (energy) = m(mass) x c(specific heat) x ΔT
(change in temperature = Tf – Ti)
 Q = (25 g)(4.18 J/goC)(100.o – 25oC)
 If ΔT is positive, Q will be positive. If ΔT is
negative, Q will be negative.
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
How many joules of heat are given off when 5.0
g of water cools from 75oC to 25oc? (Specific
heat of water = 4.18 J/goC)
HEATING CURVE OF WATER
ENERGY DURING PHASE CHANGE
Energy is increasing/decreasing during phase
changes even though temp. remains constant.
 Q = mass(m) x heat of fusion (Hf) for melting
and freezing
 Q = mass(m) x heat of vaporization (Hv) for
boiling and condensing
 Values will be positive for melting and boiling
 Values will be negative for freezing and
condensing


How many calories are given off when 85 g of
steam condense to liquid water? (Hv= 539.4
cal/g)

How many joules does it take to melt 35g of ice
at 0oC? (Hf = 333 J/g)

How many joules are required to convert 10.0g
of ice at -10.0oC to steam at 150. oC? (specific
heat of ice = 2.06 J/goC and specific heat of
steam = 2.03 J/goC) A five step problem!
YOUR ASSIGNMENT:
Practice problems are in the yellow boxes.
 Page 329 Practice Problem 10.2
 Page 330 Practice Problem 10.3
 Page 332 Practice Problem 10.4
 Page 333 Section 10.2 Review Questions
2-6
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