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Part OneHeat and
Temperature
Part OneHeat and
Temperature
What is temperature?
What does it mean to have a temperature of 0 C?
Is temperature the same
thing as heat?
Temperature is a measure of how “hot” or “cold” something is.
Temperature is measured in arbitrary units, like Fahrenheit or Celsius.
Temperature is proportional to the average kinetic energy of the molecules of the substance.
Heat is the thermal energy transferred from a hot object to a cold object.
Heat is measured in energy units -- Joules or calories.
Heat has the symbol q and is calculated using …
q = mcT
q = mcT
Quantity of heat
mass
specific heat capacity
temperature change
q = mcT
Quantity of heat
specific heat capacity
The specific heat capacity of water
is 4.18 J/gC
q = m c T
How much heat is needed to raise the temperature of 25.6 grams of water from 20.0 C to 50.0 C?
q = (25.6g)(4.18J/gC)(30.0C)
q = 3210 J
q = m c T
What is the final temperature of 27.0 grams of liquid water, initially at 0C, after it absorbs 700.0 J of energy?
Hint: start by solving for T.
=qT m c
Answer: 6.20 C
Part TwoPhase Changes
Part TwoPhase Changes
We now know that heat is either absorbed or released during a phase change.
A process that gives off heat is called
exothermic.
A process that absorbs heat is called endothermic.
Exothermic:
Endothermic:
FreezingCondensationDeposition
Melting (fusion)
VaporizationSublimationHeat is absorbed.
Heat is released.
Ice And melts.
Heat is absorbed by the ice.
Heat is absorbed by the ice.
… making liquid water
One gram of ice at 0C absorbs 334 J as it melts to form water at 0C.
water
Heat is released by the water as it freezes.
334 joules is released when one gram of water freezes at 0C.
Ice
Ice
Ice absorbs 334 J per gram as it melts at 0C
Water releases 334 J per gram as it freezes at 0C
Hotplate
Water absorbs
2260 J/g as it boils at 100 C
Steam releases 2260 J/g as it condenses at 100 C
The heat gained or lost in phase changes can be calculated using …
q = mHf q = mHv
Heat of fusion
(melting)
Heat of vaporization
The values for water are …
Hf = 334 J/g
Hv=2260 J/g
Heat of fusion
(melting)
Heat of vaporization
How much heat is absorbed by 150.0 g of ice as it melts at 0C?q = m Hf q = (150.0 g)(334 J/g)
q = 50,100 J or 50.1 kJ
How much heat is released by 20.0 grams of steam as it condenses at 100C?q = m Hv q = (20.0 g)(2260 J/g)
q = 45,200 J or 45.2 kJ
Part ThreePhase Diagrams
Part ThreePhase Diagrams
Temperature
Pre
ssur
e
The phase diagram has three distinct regions.
2
31
Hint: What happens to ice as temperature increases?
Which phase is in each region?
The point where all three phases exist in equilibrium is called the
Temperature
S L
G
triple point. triple point.
Pre
ssur
e
At a pressure of 1 atm, most substances go through all three phases, as the temperature increases,
Temp.
S L
G1 atm
Solids melt to form liquids, which vaporize to form gases.
Temp.
S L
G1 atm
MP BP
Notice the melting point and boiling point.
But the phase diagram for CO2 is a little different.
Temperature
S L
G1 atm
Notice that the triple point is above 1 atm.
5 atm
At 1 atm CO2 goes directly from solid to vapor as the temperature increases.
Temperature
S L
G1 atm
The sublimation point is –78.5 C
What phase change is occurring?
Melting (fusion)
Temperature
Pre
ssur
e
S L
G
What phase change is occurring?
Temperature
Pre
ssur
e
VaporizationS L
G
What phase change is occurring?
Temperature
Pre
ssur
e
CondensationS L
G
What phase change is occurring?
Temperature
Pre
ssur
e
SublimationS L
G
What phase change is occurring?
Temperature
Pre
ssur
e
Liquefying a gas by increasing the pressure.
S L
G
Part FourHeating and Cooling
Curves
Part FourHeating and Cooling
Curves
Look at the different regions of the heating curve for water.
0
100
Time
Temp
Ice
Ice andwater
Water
Steam
Phase changes?
Water andsteam
Part FiveCalorimetry and
Specific Heat Capacity
Part FiveCalorimetry and
Specific Heat Capacity
Calorimetry is a collection of laboratory procedures used to investigate the transfer of heat.
In calorimetry experiments, one might be looking for a final temperature or a specific heat capacity.
What is the law of conservation of energy?
Energy is neither created nor destroyed, only changed in form.
The law of conservation of energy suggests that the heat lost by the hot object as it cools is equal to the heat gained by the cool water as it warms up.
Investigate:
To put it mathematically:
qlost = -qgained
And since q = mcT then
mocTo = -mwcTw
Heat lost by the hot object =
Heat gained by the cold water
Investigate:The convention for T is final temperature minus initial temperature or Tfinal – Tinitial
moc(Tf -Ti) = -mwc(Tf -Ti)Use your algebra skills, to solve for Tf , the final temperature.
mhcTo = -mccTw
becomes
Specific heat capacity …
• …varies from one substance to another.
• …a measure of how much heat something can “hold”.
• …the amount of heat needed to raise one gram of a substance by one Celsius degree.
Specific heat capacity lab suggestions:
1. Heat a metal to a known temp. 2. Transfer the metal to a known
quantity of water at a known temperature.
3. Measure the equilibrium temperature.
4. Use qlost = -qgained to compute the specific heat of the metal.
hotplate
Get the initial temperature of the metal.
The temperature of boiling water.
metal
Get initial temp of water in calorimeter cup.
Transfer the metal to the calorimeter.
Continue stirring.
Data: Mass of metalInitial temp of metal
Mass of waterInitial temp of water
Final temp of water and metal
qlost = -qgained
mmcmTm = -mwcwTw
-mwcwTw
mmTm
cm =
Mass of metal 40.0 gInitial T of metal 98.0 CMass of water in calorimeter 60.0 gInitial T of water 20.0 CFinal T of water and metal 22.9 C
Calculate the specific heat capacity of the
metal.
Aluminum 0.900Bismuth 0.123Copper 0.386Brass 0.380Gold 0.126Lead 0.128Silver 0.233Tin 0.225Zinc 0.387Mercury 0.140Ethanol 2.400Water 4.186Ice 2.050
Substancec in
J/g KTable of selected specific heats.
What is the unknown
metal?
