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
Changes in
State
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.
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.
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.
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
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.
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?
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.
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?
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
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
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!
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
crystalline solid
highly ordered
minimum entropy
liquid
some order
some entropy
gas
very random
maximum entropy
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
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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