Energy Change (Chapter 5) (Chapter 5) The Nature of Energy Part 1.
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Transcript of Energy Change (Chapter 5) (Chapter 5) The Nature of Energy Part 1.
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Energy Change (Chapter 5)
The Nature of Energy Part 1
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Outline
Energy
Thermochemistry
Transfer of Energy
Systems and Surroundings
Heat (Q)
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Nature of Energy
Energy is all around you! You can hear energy as sound. You can see energy as light. And you can feel it as wind.
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Forms of Energy
The five main forms of energy are: Heat Chemical Electromagnetic Nuclear Mechanical
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Energy
Energy- the capacity to do work or produce heat
Change in matter can involve the release or energy or absorption of energy
Regardless of chemical or physical change, all changes in matter involve changes in energy
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Thermochemistry
Thermochemistry- the study of the energy involved in chemical and physical processes
Observing and measuring the amount of heat that is released in these processes
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Types of Energy
Potential energy- (PE) due to position or composition ex. attractive or repulsive forces
Kinetic energy- (KE) due to motion of the object KE = ½mv2 :depends on mass and
velocity
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Transfer of Energy
Temperature- measure of the average kinetic energy of the particles
Two Ways to Transfer Energy: Heat- (Q) transfer of energy between
two objects because of a temperature difference
Work- (w) force acting over a distance
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Pathway
The specific conditions of energy transfer
Energy change is independent of pathway because it is a state function
Work and heat depend on pathway so are not state functions
State function- depends only on current conditions, not past or future
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Transfer of Energy
system- part of the universe you are focused on
surroundings- everything else in the universe
usually system: what is inside the container surroundings: room, container, etc.
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Types of Systems
Open Closed Isolated
Exchange: Mass & Energy Energy None
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Transfer of Energy
exothermic- energy is produced in reaction flows out of system container feels hot to the touch
endothermic- energy is consumed by the reaction flows into the system container feels cold to the touch
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Transfer of Energy
Combustion of Methane Gas is exothermic
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Transfer of Energy
Reaction between nitrogen and oxygen is endothermic
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Transfer of Energy
the energy comes from the potential difference between the reactants and products
energy produced (or absorbed) by reaction must equal the energy absorbed (or produced) by surroundings
usually the molecules with higher potential energy have weaker bonds than molecules with lower potential energy
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Internal Energy
(E) sum of potential and kinetic energy in system
can be changed by work, heat, or both
E = PE + KE ∆E = q + w
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Heat (Q) absorbed or released
Q= amount of heat transferred (Joules, J)m= mass of a substance (g)c= specific heat capacity (J/g• 0C)ΔT = Tfinal – Tinitial (change in temperature) ( 0C)
1 Joule = the amount of heat (Q) required to raise the temperature of one gram of the substance by one degree Celsius.The specific heat capacity (c) of a substance is the amount of heat (Q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius.
Q=mcΔT
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Specific Heat Capacities
Refer to page 280 in your textbook for a comprehensive chart
Substance c (J/g0C)
Ammonia (l) 4.70
Ammonia (g) 2.06
Ethanol 2.44
Water (s) 2.00
Water (l) 4.19
Water (g) 2.02
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Heat Examples
1) When a 1.25kg sample of water was heated in a kettle, its temperature increased from 16.40C to 98.90C. How much heat did the water absorb?
Q= 1.25x103g x 4.19 J/g0C x (+82.50C ) = 432 093.75 J = 4.32x105 J
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2) How much heat must be added to 128.62g of steam at 126.00C to increase its temperature to 189.50C?
Q = 128.62 gx 2.02J/g0Cx (189.5-126.0) 0C = 1.65x104 J
3) A solid substance has a mass of 250.0g. It is cooled by 25.000C and loses 4.937kJ of heat. What is the specific heat capacity of the substance?
c= Q/m(∆T) c= -4.937x103J/(250.0 g (-25.000C))
c= 0.7899 J/g0C
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Additional Examples1) How much heat is released when the temperature of 789g of liquid ammonia decreases from 82.70C to 25.00C? Q= mc∆T = 789 g x 4.70 J/g0C x (25.0-82.7) = - 213968.91 J change to -2.14 x105 J
2) On a warm day, how much solar energy does a 3.982kg piece of concrete absorb as heat if its temperature increases from 13.600C to 14.500C? Q= 3982 g x 0.88 J/g0C x (14.50-13.60) = 3153.744 J change to 3.2 x103J
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Homework
Readings (pgs. 278-283)
Questions (pg. 281) #7, 8
Questions (pg. 291) #2, 6, 7