WHAT ARE SOME OF THE WAYS THAT
HEAT ENERGY CAN BE TRANSFERRED?
1) CONDUCTION – IF TWO OBJECTS
ARE IN CONTACT, THERMAL ENERGY
CAN BE TRANSFERRED THROUGH
COLLISIONS OF THE
ATOMS/MOLECULES OF THE OBJECTS
AT THE INTERFACE BETWEEN THE
TWO OBJECTS.
CONDUCTION DOES NOT INVOLVE MATERIAL TRANSFER.
2) CONVECTION IS THE PROCESS OF HEAT
TRANSFER FROM ONE LOCATION TO ANOTHER
BY THE MOVEMENT OF FLUIDS.
AS A FLUID IS HEATED, IT BECOMES LESS
DENSE AND RISES. COOLER FLUID WOULD
TAKE ITS PLACE.
3) RADIATION IS THE TRANSFER OF HEAT
THROUGH ELECTROMAGNETIC RADIATION.
AS THE TEMPERATURE OF AN OBJECT
INCREASES, THE AMOUNT OF RADIATION
ALSO INCREASES.
RATE OF RADIATION = k x T4
THE AMOUNT OF RADIATION INCREASES
WITH INCREASE IN TEMPERATURE. THE
WAVELENGTHS OF EMISSION ALSO CHANGE
WITH THE WAVELENGTH OF MAXIMUM
EMISSION SHIFTING TO SHORTER
WAVELENGTH AS THE TEMPERATURE
INCREASES.
SUPPOSE THAT OBJECT A AND OBJECT B
HAVE REACHED A THERMAL EQUILIBRIUM. DO
THE PARTICLES OF THE TWO OBJECTS STILL
COLLIDE WITH EACH OTHER? IF SO, DO ANY
OF THE COLLISIONS RESULT IN THE
TRANSFER OF ENERGY BETWEEN THE TWO
OBJECTS? EXPLAIN.
OBJECTS DO NOT CONTAIN HEAT.
OBJECTS ARE MADE UP OF ATOMS AND
MOLECULES AND THEY CONTAIN ENERGY
(THERMAL ENERGY).
HEAT IS THE TRANSFER OF ENERGY FROM
AN OBJECT TO ANOTHER OBJECT OR TO
ITS SURROUNDINGS.
SO, WHAT DOES HEAT DO?
1) HEAT CHANGES THE TEMPERATURE OF
AN OBJECT.
2) HEAT CHANGES THE STATE OF AN
OBJECT.
3) HEAT CAN CAUSE WORK TO BE DONE.
UNITS FOR ENERGY ARE JOULES, ALTHOUGH
CALORIES ARE SOMETIMES USED.
1 CALORIE = 4.184 JOULES
IF WE HEAT DIFFERENT SUBSTANCES AT THE
SAME RATE, THE TEMPERATURE CHANGES WE
WILL SEE WILL BE DIFFERENT. DIFFERENT
SUBSTANCES HAVE DIFFERENT HEAT
CAPACITIES.
HEAT CAPACITY – THE HEAT REQUIRED TO
RAISE THE TEMPERATURE OF 1 GRAM OF A
SUBSTANCE BY 1 DEGREE CELCIUS.
SUBSTANCE
Water
Aluminum
HEAT CAPACITY,
J/g, oC
4.18
0.897
Iron
Tungsten
Lead
0.45
0.134
0.129
THE EQUATION RELATING THE AMOUNT OF
HEAT REQUIRED TO RAISE THE
TEMPERATURE OF A GIVEN AMOUNT OF
MATERIAL IS GIVEN AS:
Q = m x C x DT
Where m = mass
C = heat capacity
DT = change in temperature
Q = heat transferred
What quantity of heat is required to raise the
temperature of 450 grams of water from 15°C
to 85°C? The specific heat capacity of water
is 4.18 J/g/°C.
Q = m x C x DT
m = 450 g
C = 4.18 J/g/oC
DT = 85 – 15 = 70o
q
Q = 450g x 4.18 J/g/oC x 70o
= 131670 J = 131.7 kJ
A 12.9 gram sample of an unknown metal at
26.5°C is placed in a Styrofoam cup containing
50.0 grams of water at 88.6°C. The water cools
down and the metal warms up until thermal
equilibrium is achieved at 87.1°C. Assuming all
the heat lost by the water is gained by the
metal and that the cup is perfectly insulated,
determine the specific heat capacity of the
unknown metal. The specific heat capacity of
water is 4.18 J/g/°C.
IF THEY COME TO THERMAL EQUILIBRIUM,
THE HEAT LOST BY THE WATER = THE
HEAT GAINED BY THE METAL
Qwater = 50 g x 4.18 x (88.6 – 87.1)
= 50 x 4.18 x 1.5 = 313.5 J
Qmetal = 12.9 x Cm x (87.1 – 26.5)
313.5 = 12.9 x Cm x 60.6
Or Cm = 313.5/782 = 0.402 J/g/oC
FOR CHANGES IN STATE, YOU HAVE
SIMILAR SITUATIONS. TO MELT
SOMETHING, YOU HAVE TO PUT ENERGY IN.
TO FREEZE SOMETHING, ENERGY HAS TO BE
LOST OR REMOVED.
Qfusion = m x DHfusion
FOR VAPORIZATION (BOILING) AND
CONDENSATION
Qvap = m x DHvap
Elise places 48.2 grams of ice in her beverage.
What quantity of energy would be absorbed by
the ice (and released by the beverage) during
the melting process? The heat of fusion of
water is 333 J/g.
Qfusion = m x DHfusion
= 48.2 g x333 J/g
= 16051 J = 16.1 kJ
What is the minimum amount of liquid water at
26.5 degrees that would be required to
completely melt 50.0 grams of ice? The
specific heat capacity of liquid water is 4.18
J/g/°C and the specific heat of fusion of ice
is 333 J/g.
FIRST, LET’S MELT THE ICE.
Qfusion = 50 x 333 J/g = 16650 J
NOW, THE WATER CAN COOL FROM
26.5o TO 0o
Qwater = m x 4.18 x 26.5
So 16650 = m x 111
Or m = 16650/111 = 150 g
IF YOU ARE GOING TO MAKE CALCULATIONS
INVOLVING HEATING AND/OR COOLING AND PHASE
CHANGES, THINK ABOUT THE STEPS NEEDED. EACH
STRAIGHT LINE STEP NEEDS A SEPARATE
CALCULATION.
HEAT CAPACITIES FOR WATER IN
VARIOUS STATES:
Solid Water: C=2.00 J/g/°C
Liquid Water: C = 4.18 J/g/°C
Gaseous Water: C = 2.01 J/g/°C
HOW MUCH ENERGY WOULD BE REQUIRED
TO CONVERT WATER AT -20o C TO WATER
AT 120o C?
THINK ABOUT YOUR STEPS.