changes in state
DESCRIPTION
Changes in State. warming the solid. warming the gas. warming the liquid. Temperature ->. Time ->. Warming Curve. Consider what happens to an ice cube which is heated. boiling. melting. warm gas. Temperature ->. warm liquid. warm solid. Time ->. Within One State. - PowerPoint PPT PresentationTRANSCRIPT
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Changes in
State
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Warming Curve
warming the solid
melting
boiling
warming the liquid
warming the gas
Time ->
Tem
pera
ture
->
Consider what happens to an ice cube which is heated.
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Within One State
warm solid
warm liquid
warm gas
Time ->
Te
mp
era
ture
->
This is measured by the specific heat (heat capacity): the calories needed to raise the temperature of 1 gram of a substance by 1°C.
Rising temperature indicates a change in kinetic energy: molecules are moving faster.
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Changing States
melt
boil
Time ->
Te
mp
era
ture
->
This is measured by the heat of vaporization (gas/liquid) or heat of fusion (solid/liquid): the calories needed to change the state of 1 gram.
A temperature plateau indicates a change in potential energy: bonds between molecules are changing, and energy can be stored in them.
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SUMMARY
warm so
lidmelt
boil
war
m li
quid
warm gas
Time ->
Tem
pera
ture
->
sp. heat solid
sp. heat liquid
sp. heat gas
heat of fusion
heat of vaporatization
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Energy
Amount
cal or kcal
Joules or kJ
grams or kg
moles
You will see a variety of units for both energy and amount of material in these problems.
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W happening at each stage of this graph?
Time ->
Te
mp
era
ture
->
1 2
34
5
When is kinetic energy affected? When is potential energy affected? Answers are in the notes.
What states are present at each stage?
What property describes each stage?
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Calculations with Specific Heat
1. How many calories are needed to warm up 3.45 g of water from 18.5°C to 60.0°C?
The specific heat of water is 1 cal/g°C.
Heat = (1 cal/g°C)(3.45 g)(41.5°C) = 143 cal
2. How many calories are removed (released) when 1.29 kg of H2O are cooled from 98.5°C to 20.0°C?Heat = (1 cal/g°C)(1,290 g)(78.5°C) = 101,000 cal
Technically, this is negative because heat is removed. We will not be using this convention, leaving all energy without signs and using works to decribe addition/loss of energy.
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More Calculations with Specific Heat
3. What is the final temperature when 85.0 J of heat energy are added to 23.5 g of iron (specific heat 0.125 J/g°C) at 19°C?
(0.125 J/g°C)* (23.5 g) * (T) = 85.0 J
T = 28.9 °C
Heat is being added, so Tfinal = 19°C + 28.9°C
Tfinal = 47.9 °C
Note: If the problem stated that heat was being removed the iron, then T would still be 28.9 °C but the final temperature would be -9.9 °C. Do you see why?
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Heats of Vaporization & Fusion1. How much heat must be released to freeze 25.0 g of
water? The heat of fusion of water is 1.14 kcal/mole.
(1.14 kcal/1 mole) x [25.0 g x (1 mole/18.0 g)]
= 1.77 kcal
2. How much heat is needed to boil 125 g of alcohol? The heat of vaporization of alcohol is 102.4 J/g.
(102.4J/1 g) x 125 g = 12,800 J
1.14 kcal = x kcal alternative proportional solution
18.0 g 25.0 g x = 1.77 kcal
102.4 J = x J alternative proportional solution
1 g 125 g x = 12,800 J
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How much heat energy must be removed from 35.0 g of water at 80.°C to turn it into ice at -25 °C?
Heat of fusion H2O = 1.14 kcal/mol
Sp. Heat of Ice = 0.50 cal/g°C
FIRST: Sketch a heating or cooling curve and mark the plateaus and your starting/ending points.
This tells you there are 3 stages to consider: (a) cooling the liquid, (b) freezing, (c) cooling the solid.
Time ->
Tem
per
atur
e ->
starts at 80 °C; liquid
ends at -25 °C; solid
100 °C
0 °C
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How much heat energy must be removed from 35.0 g of water at 80.°C to turn it into ice at -25 °C?
Heat of fusion H2O = 1.14 kcal/mol
Sp. Heat of Ice = 0.50 cal/g°C
Second: Find the properties that control each stage of the problem.(a) cooling the liquid = specific heat of liquid: 1 cal/g°C(b) freezing = heat of fusion: 1.14 kcal/mol(c) cooling the solid = specific heat of solid: 0.50 cal/g°C
The units of each property show you how to do the math!
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How much heat energy must be removed from 35.0 g of water at 80.°C to turn it into ice at -25 °C?
1. Cool the water to its freezing point. (1 cal/g°C)(35.0 g)(80.-0.0°C) = 2800 cal
Heat of fusion H2O = 1.14 kcal/mol
Sp. Heat of Ice = 0.50 cal/g°C
2. Freeze the water into ice. (1.14 kcal/1 mole) (35.0 g)(1 mole/18.0 g) = 2.22 kcal
3. Cool the ice to its final temperature. (0.50 cal/g°C)(35.0 g)(0-[-25°C]) = 440 cal
Last, add all the heats together. Make sure units agree!2800 cal + 2220 cal + 440 cal = 5460 cal
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crystalline solid
highly ordered
minimum entropy
liquid
some order
some entropy
gas
very random
maximum entropy
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can affect
is seen as a
has units of
can affect
is seen as a
includes
is characterized
by
has units of
includes
ischaracterized
by
has units of
is characterized by
includes
ischaracterized
by
includes
is characterized by
Heating & Cooling
kinetic energy
change in temperature
specific heat
cal/g°C
potential energy
change of state
freezing
heat of fusion
cal/g ORkcal/mole
boiling
heat of vaporization
condensingmelting
cal/g°CJ/g°C
cal/g, cal/molekcal/g, kcal/mole
J/g, J/mole