# Calorimetry ```Physics 12
Objectives
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Define specific heat capacity.
Solve problems involving specific heat
capacities.
Explain the difference between solid, liquid,
and gaseous phases.
Explain in terms of molecular behavior why
temperature does not change during a phase
change.
Define latent heat.
Solve problems involving latent heats.
Activities
Worksheet
 Lab demonstration: Determining the
specific heat capacity of a substance
 Lab demonstration: Determining the
latent heat of a substance
 Design lab: Design a homemade
calorimeter and test it

Specific heat capacity

If heat flows into an object, its
temperature rises.

What factors might affect the magnitude
of the temperature change?
Specific heat capacity

The amount of energy Q required to change
the temperature of a given material is
proportional to the mass m of the material
and to the temperature change ΔT, shown by
the simplistic expression:
Q = mc ΔT
where c is the characteristic of the material
called its specific heat capacity
Specific heat capacity

A high specific heat capacity means that
more energy is required to achieve the
same temperature change, i.e. it is more
“difficult” to raise the temperature of that
material.

If a material is a good heat conductor (e.g.
metals) would you expect it to have a high
or low specific heat capacity?
Specific heat capacity
Specific heat capacities of specific substances
Substance
Specific heat capacity, c / Jkg1&deg;C-1
aluminum
900
copper
390
iron or steel
450
130
wood
1700
water (ice)
2100
water (liquid)
4186
water (steam)
2010
human body
(average)
3470
Specific heat capacity
Q/J
m / kg
c / Jkg-1C-1
?
?
constant
increased
?
increased
constant
increased
constant
constant
constant
constant
constant
constant
increased
ΔT / &deg;C or
&deg;K
constant
increased
?
?
constant
Specific heat capacity
(a)
How much heat input is needed to raise
the temperature of an empty 20-kg vat
made of iron from 10C to 90C?
(b)
What if the vat is filled with 20 kg of
water?
Specific heat capacity
(a)
(b)
720 kJ
7400 kJ
Specific heat capacity
You accidentally let an empty iron frying pan
get very hot on the stove (approx. 200C).
What happens when you dunk it into a few
inches of cool water in the bottom of the
sink?
Will the final temperature be midway between
the initial temperatures of the water and
pan? Will the water start boiling?
(Assume the mass of the water is roughly the
same as the mass of the pan.)
Calorimetry
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In discussing heat and thermodynamics,
we shall often refer to systems.
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What is the difference between open,
closed, and isolated systems?
Calorimetry
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Open system
Mass and energy may leave and enter
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Closed system
Energy may leave and enter but mass may not
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Isolated system
Neither mass nor energy may leave or enter
Calorimetry
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We will often make the assumption that
the systems we are dealing with are
isolated.
Why is this necessary?
Calorimetry

In an isolated system, heat lost by one part of
the system is equal to the heat gained by
another part:
heat lost = heat gained
or
energy out of one part = energy into another
part
Calorimetry
If 200 cm3 of tea at 95C is poured into a 150g glass cup initially at 25C, what will be
the common final temperature T of the tea
and cup when thermal equilibrium is
reached, assuming no heat flows to the
surroundings?
Calorimetry
T = 86C
Would this be the case in the “real world”?
Calorimetry
The exchange of energy (as shown in the
previous example) is the basis for the
technique known as calorimetry.
 Calorimetry is the quantitative
measurement of heat exchange.
 A calorimeter is used.
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Calorimetry
A simple water calorimeter
Calorimetry
An engineer wishes to determine the
specific heat of a new metal alloy. A 0.150kg sample of the alloy is heated to 540C. It
is then quickly placed in 400 g of water at
10.0C, which is contained in a 200-g
aluminum calorimeter cup. The final
temperature of the system is 30.5C.
Calculate the specific heat of the alloy.
Calorimetry
c = 500 Jkg-1C-1
Calorimetry

In order to determine the specific heat of a
particular substance, the following
expression is used:
Qlost = Qgained
m1c1ΔT1 = m2c2 ΔT2
where the two substances share a final temperature
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Thus, all other quantities except one must be
measured or known.
Phase change

Recall that matter most commonly exists
in three states: solid, liquid, and gas.
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What are the differences between these
three states (or phases) in terms of
molecular structure and motion?
Phase change
Comparison of the three common phases of matter (on Earth)
Shape
solid
liquid
gas
Volume
Particle
motion
Phase change
Comparison of the three common phases of
matter (on Earth)
Shape
Volume
Particle
motion
solid
definite
definite
liquid
indefinite
definite
vibrational
vibrational
rotational
translational
gas
indefinite
indefinite
same as liquid
but quicker
Phase change

When a material changes phase from solid
to liquid or liquid to gas, a certain amount
of energy is involved in this change of
phase.
Phase change
Temperature as a function of heat added to 10.0 g of ice
Phase change

The heat required to change a substance
from solid to liquid is called latent heat of
fusion.
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The heat required to change a substance
from liquid to gas is called the latent heat of
vaporization.

Values for latent heats will vary depending
on the substance.
Phase change
Latent heats
Substance
Heat of fusion /
kJkg-1
Heat of vaporization
/
kJkg-1
oxygen
ethyl alcohol
water
14
104
333
210
850
2260
iron
289
6340
Phase change

What factors might affect the amount of
energy needed to change the phase of a
substance?
Phase change

The heat involved in a change of phase Q
depends not only on the latent heat but also
on the total mass of the substance, i.e.
Q = mL
where m is the mass of the substance and L is the latent
heat
Phase change
How much energy does a freezer have to
remove from 1.5 kg of water at 20C to
make ice at –12C?
Phase change
6.6 x 105 J
Phase change
At a reception, a 0.50-kg chunk of ice at –
10C is placed in 3.0 kg of tea at 20C. At
what temperature and in what phase will
the final mixture be?
The tea can be considered as water. Ignore
any heat flow to the surroundings,
including the container.
Phase change
T = 5C
Phase change
The specific heat of liquid mercury is
140 Jkg-1C-1. When 1.0 kg of solid mercury
at its melting point of –39C is placed in a
0.50-kg aluminum calorimeter filled with
1.2 kg of water at 20.0C, the final
temperature of the combination is found
to be 16.5C. What is the heat of fusion of
mercury in Jkg-1?
Phase change
L = 11 x 103 Jkg-1
Objectives
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


Define specific heat capacity.
Solve problems involving specific heat
capacities.
Explain the difference between solid, liquid,
and gaseous phases.
Explain in terms of molecular behavior why
temperature does not change during a phase
change.
Define latent heat.
Solve problems involving latent heats.
Measuring specific heat
Data collection
mass of metal
initial temperature of metal
mass of calorimeter
mass of calorimeter + water
initial temperature of water
final temperature
Measuring specific heat
Data processing
Include propagation of uncertainty
mass of water
ΔT of water
ΔT of metal
Measuring specific heat
Data processing
Include propagation of uncertainty
mass of water
ΔT of water
ΔT of metal
Q = mcΔT
Q (lost) = Q (gained)
Measuring specific heat

Homework
Write a conclusion and evaluation of the lab
activity
Design a homemade calorimeter using the
materials available. Try to minimize the
amount of energy lost to the surroundings.
You will use this calorimeter for the next lab
activity. The calorimeter with the smallest
percent discrepancy gets bonus points.
Measuring latent heat of fusion
Data collection
mass of cup 1
mass of cup 1 + water
mass of cup 2
mass of cup 2 + ice
initial temperature of ice
initial temperature of water
final temperature
Measuring latent heat of fusion
Data processing
mass of water
ΔT of water
mass of ice
ΔT of ice
Measuring latent heat of fusion
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Homework