Conservation of Energy

The most common energy transfer is in
the form of heat


Transferred heat is measured as a
temperature change
Temperature is the measurement of the
average kinetic energy of the particles in
the substance
Measuring Heat Energy

Joules


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4.184 Joules = 1 calorie
Calorie is the amount of heat that it takes to raise one
gram of water by one degree Celsius
Specific heat


The amount of heat necessary to raise one gram of a
substance by one degree Celsius
Unique for each substance


The specific heat of water is 1 calorie/gºC
Also 4.184 J/ gºC
Calculating Heat

Heat is calculated using the heat
equation

Q = (m) (∆T) (cp)

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
Q = heat
M = mass
∆T = change in temperature


Tf – Ti (final temp – initial temp)
Cp = specific heat
1
Example Problem

How much heat is lost when a solid
aluminum ingot with a mass of 4110 g
cools from 660.0 C to 25.0 C?


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


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q=(m) (T) (Cp)
Mass is 4110g
Initial temp is 660.0 C
Final temp is 25.0 C
Specific heat of aluminum is 0.903 J/gC
Q = (4110g) ( 25.0C – 660.0 C ) ( 0.903 J/gC)
Q = (4110g) ( -635C) ( 0.903 J/gC)
Q =- 2.35 x 106joules
Kinetic Theory

As energy in the particles of a
substance increases



Particles move faster
Particles spread out
As energy in the particles of a
substance decreases


Particles move more slowly
Particles move closer together
Energy and Change of State


As energy is absorbed or lost by a
liquid, the temperature will increase or
decrease byy a corresponding
p
g degree
g
until the boiling (heat increase) or
freezing (heat loss) point is reached
At the freezing or boiling point two
phases of matter can exist at the same
temperature
2



To make the change from one phase to
another more energy will be absorbed
(boiling or melting) or lost (condensing
or freezing) without a change in
temperature
This is because this energy is used
merely to overcome the bonds of one
state and move to the new state
Heat of vaporization Hv


Heat necessary to change from liquid to
gas or from gas
g
g to liquid
q
at the
boiling/condensation point
Heat of fusion Hf

Heat necessary to change from liquid to
q
at the
solid or from solid to liquid
freezing/melting point
3
Phase change graph


Slanted sections of the graph represent
change in kinetic energy M ∆TCp
Flat portions represent the change in
potential
t ti l energy necessary to
t change
h
from one phase or state of matter to
another – Hf or Hv; look at it again
Phase change graph
4

If one were to begin with a cold solid, and
add heat, each step as heat is added must be
calculated

The solid warms up to the melting point

At the melting point, some liquid will begin to form



the heat necessary to vaporize it
completely is calculated by mass of the
solid times the heat of vaporization
Vapor may continue to increase in
temperature


calculated by M ∆TCp (using the specific
heat for the liquid)
At the boiling point gas will begin to
form


the heat necessary to melt it completely is calculated by
mass of the solid times the heat of fusion
The liquid increases temperature until
reaching the boiling point


M ∆TCp (using the specific heat for the solid)
calculated by M ∆TCp (using the specific
heat for the gas)
The sum of all of these heats (from
initial temperature to final temperature)
is used to calculate the heat gained or
lost by the substance
5
If a 26.5g block of ice at –15.0 ºC is
heated to water at 20.0 ºC, how
much heat is used?

Ist heat the ice to its melting point
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Now melt the ice
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M(Hf)
26.5g
g (80
( cal/g)
g)
2120 cal
Now warm the water




M ∆TCp
p
(26.5g)(0.0C – (-15.0C))(0.53cal/gºC)
210.7 cal
M ∆TCp
26.5g (20 ºC - 0 ºC)(1.00cal/g ºC)
530 cal
Now add up the calories

210.7 cal + 2120 cal + 530 cal
2860 7 cal
2860.7
2860 cal (3 sig fig.)

Now try one yourself
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
6
How much heat is needed to heat
117g of water at 42 ºC to steam at
136 ºC

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117g (100 ºC - 42 ºC)(1.00 cal/g)
6786 cal
117g (540 cal/g)
63180 cal
117g(136 ºC - 100 ºC)(0.480 cal/g)
2021.76 cal
6786 cal + 63180 cal + 2021.76 cal
71988 cal
7.2 x 104 cal
How much heat does it take
to raise the temp of 112g of
ice at –4.0 ºC to 120 ºC?


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
112g (0 – ((-4
4.0))
0)) 0.53
0 53 cal/g ºC
C
237.44 cal (993.4j)
112g (80 cal/g)
8960 cal (37500 J)
7

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112 ( 100 ºC – 0 ºC)1.00 cal/g ºC
11200 cal
112 ( 540 cal/g)
60480 cal
112 ( 120 ºC - 100 ºC).48 cal/g ºC)
1075.2 cal
237.44 + 8960 + 11200 + 60480 +
1075.2
81952 64 cal
81952.64
82000 cal
8