Phase Changes - Red Hook Central School District

Thermal Properties, Heat
Capacity, Specific Heat.
Thermal Properties of Matter
When objects absorb heat (Q) their T rises.
When objects release/lose (Q) heat their T
Different materials undergo different DT for
same heat absorbed/released.
Thermal Capacity (C)= amt of heat E (Q)
needed to raise a body’s temperature 1oC.
• Water has high heat C,
• takes lots of E to change its T.
• Metals have low capacities.
A little heat E gives a lot of DT
Land heats and cools (DT)
more quickly than water
Temp is measure of average KE of particles.
1 kg of different materials contain different:
• # of particles, KE must be spread over varying
numbers of particles.
• Weights of particles, it takes more E to move
heavier particles.
• Bond strengths, some materials have particles that
are easy to move.
Why does water have such a high
heat capacity?
Water molecules form strong
intermolecular bonds with each other;
Metals have weak intermolecular bonds.
Define Thermal Capacity
Thermal Capacity (C) = amt of heat (Q) needed
to raise an object’s temperature 1oC.
C = DQ .
Units … J/oC
Q energy in Joules
DT temp change in oC or K.
Ex 1: If the thermal capacity of water is 5000
J/oC, how much heat is required to raise its
temperature from 20 – 100oC?
(5000)(80) = 400 kJ.
Ex 2: How much heat is lost from a
block of metal of C = 800 J/oC when it
cools from 60 – 20oC?
32 kJ
Heat Capacity vs. Specific Heat
• Heat Capacity C, relates to objects w/o
correcting for mass.
Small C
Large C
Specific Heat Capacity, c includes term for
mass of a material. Amt of E required to raise
T of 1kg mass of substance 1oC, or K.
c =
Q = mcDT
m = mass in kg
c = spec heat const J/kgoC
DT = Tf – Ti.
Q = energy in Joules
DT is positive if substance absorbs E (Tf > Ti ),
negative for E released (Tf < Ti ), .
Both Iron same c.
Same c
Same c
Specific heats some substances.
Easy to heat
Way hard to heat
Ex 3. A 45-g piece of aluminum is
heated from 25 – 55 oC. How much E
is required?
• Q = mc DT
• Q = (0.045 kg)(897 J/kgoC)(55-25) oC =
• 1200 J.
4. The SH of water is ~4200 J/kg oC. How
much heat will be required to raise the
temperature of 300 g from 20 – 60oC?
• 50.4 kJ
5. A metal block has a mass of 1.5 kg
loses 20kJ of heat. Its temperature
drops from 60 – 45o. What is the
specific heat capacity of the metal?
• 888.9 ~ 900 J/kgoC.
Experimental Methods to Determine SH.
Calorimeters – mix substances (usu. w/water)
observe Temp change. Use consv E to calculate.
Mixing substances obeys E conservation :
Q lost by one substance = Q gained by another
- mcDT1 = mcDT2.
mcDT1 + mcDT2 = 0 .
Throw hot metal into cold water then
Q lost by metal = Q gained by water
For solids with unknown c, liq with
known c.
6. A 0.05 kg metal bolt is heated to an
unknown temperature. It is then dropped
into a calorimeter containing 0.15 kg of
water of 21.0oC. The bolt and the water
reach a final temperature of 25.0oC. If the
SH of the metal is 899 J/kgoC, find the initial
temperature of the metal.
Ignore the heat lost to the cup.
• 81oC
Heating Coil in Liquid
How can we use E conservation to find c
Electric E lost by coil = E gained by liquid.
+ the cup
7. A heating coil with a voltage potential of 6
V and current of 3 A is immersed in .7
kg of water. How long will it take to heat
the water from 20oC to 35oC? Assume
the cup absorbs no heat.
• 2442 sec
• 41 min
Read Hamper 50– 57
Do pg 56- 57 All
Write out all work and
Phases/States of Matter
The Kinetic Particle
Theory of Matter
1. All matter is composed of small
particles (atoms, molecules, or ions).
2. They are in constant, random motion.
3. The particles are so small they can
be treated as if they are points
with no volume.
Phase Properties
Phase Proximity
far apart
moderate rotational
a lot
No fixed shape.
No fixed shape.
Fixed shape.
Very strong
Phases/States of Matter
Molecular Distance & Motion
Phase Changes of Matter
• Q absorbed - phase changes:
– Vaporization: liquid  vapour
– Melting: solid  liquid
Sublimation: solid  vapour
• Q released by phase changes:
– Condensation: vapour  liquid
– Fusion: liquid  solid
Deposition: vapour  solid
Heat & Phase Change
• When substances absorb or lose heat E the T changes until:
– It reaches “critical” T, T stops changing substance
undergoes phase change.
– gain/loss of additional heat energy results in the phase
transformation of matter from one phase to another:
Solid  liquid  gas
Substance absorbs E to break molc bonds.
Chem PE increases av. KE does not change.
• Consider melting: as heat is absorbed T goes up
particles vibrate more. The KE increases. At the
melting point, particles vibrate enough to slip from
their fixed positions breaking intermolecular bonds.
More and more particles break solid bonds until all
are slipping past each other in the liquid phase.
• The PE of the system increases but the av KE stays
constant at the melting (ice) & boiling points (steam).
Phase equilibrium
Multiple phases exist at
1 T.
Molecules leave and
enter phases at equal
Ice / liquid water: 0 °C
Liquid water / steam:
100 °C
Ice water
Latent heat is a physical property that describes
how much E is required to transform the mass of a
substance from one phase to another.
Latent heat of fusion Lf –
heat absorbed during melting
or released during freezing.
Latent heat of vaporization Lv –
heat absorbed during vaporization,
released during condensation.
Latent Heat
describes heat energy (Q) gained/lost to mass
of substance (m) that undergoes a phase
change is:
Q = m.L
m = mass in kg
Q – E absorbed/released in J.
L - latent heat (the SI units are J/kg)
Lf or Lv.
Some Thermal Values
Ex 1: How much E is released cooling 10g of
water from steam at 133oC to liquid at 53oC?
Cstm = 2.01x103 J/kgoC
Cwat = 4186 J/kgoC
Lv = 2.26 x 106J/kg
• Q to cool steam 133o – 100o
mcDT stm
• Q to condense at 100o
• Q to cool liquid 100 – 53o
mcDT wat
• To cool steam
• (0.01kg)(2.01x103J/kgoC)(33oC) = 633 J
• To condense at 100o
• (0.01 kg)(2.26x106J/kg) =
2.26 x 104 J
• To cool water
• (0.01kg)(4186 J/kgoC)(47o) =
1967 J
• Total E
2.52x104 J
Ex 2. How much energy is released when 1.5
kg of water vapor at 100oC is placed in a
freezer and converted to ice at -7oC.
• Specific Heat
– c ice = 2.1 x 103 J/K kg.
– cWat = 4.2 x 103J/K kg.
• Lf = 3.34 x 105 J/kg
• Lv = 22.5 x 105 J/kg
4.5 x
• IB Set Specific, Latent
Heat 2014
Heating Curves
show temperature change with
absorbed energy for a substance.
Heating/Cooling Curve
Temp vs. Energy Q
Sketch This
What is the slope on this portion of the graph?
Q = mcDT
DT = Q or
1 (Q)
The slope is 1
Temp vs. time graph
Heating time related to Q absorbed.
Use slope gradient with known specific heat & mass of
substance to calculate power output of heater. Then use
power output to calculate latent heat. P = mc (slope)
Example Cooling Curve
• Hypothetical substance.
• Phase Changes for ice IB Prob.
• Kerr pg 91 #3- 6, 8, 11.
Determining Lf for Ice
• Mixing Method:
Add ice known mass & T water.
Let it melt.
Thermal E lost by water =
thermal E gained by ice.
-mc(DT)w =
mcDTice + mLf
+ mcDT liq ice.
Go to evaporation.
Molecules leave surface & enter vapor
phase at any T.
The molecules in a liquid have a
distribution of speeds. The average
speed determines the T of the liquid.
The fastest molecules have enough
energy to overcome the attraction
between the molecules in the liquid.
They evaporate and become vapor.
Evaporative cooling: as the fastest
molecules evaporate, they leave behind
the coolest molecules and the average
temperature of the liquid decreases.
Example: sweat
Volatile liquids = rapid evaporation.
Evaporation rate depends on:
• Surface area.
• Ambient Temperature.
• Vapor Pressure – hi
– Vap press = hi volatility
• Humidity of Surroundings
• Boiling occurs when the average motion of
particles is fast enough to overcome the
forces holding them close together.
• This happens evenly throughout a boiling
liquid. You will see bubbles form.
• The temperature is uniform throughout.
Boiling vs. Evaporation
• In the cases of both boiling and evaporation,
intermolecular forces between particles are
• The greater the space between the particles, the
weaker the intermolecular force.
• To break the bond between two particles, one
particle has to be moving fast enough to
overcome the pull of the other, until it gets so
far away that pull is diminished.
Elem Chem Gas Solid Liq.