Measuring the Quantity of Heat
Physics
January 21, 2013
Coach Stephens
Review
• On the previous page, we learned what heat does to an
object when it is gained or released.
– Heat gains or losses result in changes in temperature, changes
in state, or the performance of work.
– Heat is a transfer of energy.
– When gained or lost by an object, there will be corresponding
energy changes within that object.
– A change in temperature is associated with changes in the
average kinetic energy of the particles within the object.
– A change in state is associated with changes in the internal
potential energy possessed by the object.
– And when work is done, there is an overall transfer of energy
to the object upon which the work is done.
– In this part of Lesson 2, we will investigate the question How
does one measure the quantity of heat gained or released by
an object?
Specific Heat Capacity
Suppose that several objects composed of different
materials are heated in the same manner. Will
the objects warm up at equal rates?
The answer: most likely not. Different materials
would warm up at different rates because each
material has its own specific heat capacity.
http://www.wolframalpha.com/widgets/view.jsp?id
=1e747ddbea997a1b933aaf58a7953c3c
Specific Heat Capacity
• The specific heat capacity refers to the
amount of heat required to cause a unit of
mass (say a gram or a kilogram) to change its
temperature by 1°C.
• Standard metric units are
Joules/kilogram/Kelvin (J/kg/K).
• More commonly used units are J/g/°C.
Examples
• The specific heat capacity of solid aluminum (0.904
J/g/°C) is different than the specific heat capacity of
solid iron (0.449 J/g/°C).
• This means that it would require more heat to
increase the temperature of a given mass of
aluminum by 1°C compared to the amount of heat
required to increase the temperature of the same
mass of iron by 1°C.
• In fact, it would take about twice as much heat to
increase the temperature of a sample of aluminum a
given amount compared to the same temperature
change of the same amount of iron.
• This is because the specific heat capacity of
aluminum is nearly twice the value of iron.
Molar Heat Capacity
• Heat capacities are listed on a per gram or per kilogram basis.
• Occasionally, the value is listed on a per mole basis, in which
case it is called the molar heat capacity.
• The fact that they are listed on a per amount basis is an
indication that the quantity of heat required to raise the
temperature of a substance depends on how much substance
there is.
• Any person who has boiled a pot of water on a stove,
undoubtedly know this truth.
Continued…
• Water boils at 100°C at sea level and at slightly lowered
temperatures at higher elevations.
• To bring a pot of water to a boil, its temperature must first
be raised to 100°C.
• This temperature change is achieved by the absorption of
heat from the stove burner.
• One quickly notices that it takes considerably more time to
bring a full pot of water to a boil than to bring a half-full of
water to a boil.
• This is because the full pot of water must absorb more
heat to result in the same temperature change.
• In fact, it requires twice as much heat to cause the same
temperature change in twice the mass of water.
per K or per °C
• Specific heat capacities are also listed on a per K or a per °C basis.
• The fact that the specific heat capacity is listed on a per degree basis
is an indication that the quantity of heat required to raise a given
mass of substance to a specific temperature depends upon the
change in temperature required to reach that final temperature.
• In other words, it is not the final temperature that is of importance, it
is the overall temperature change.
• It takes more heat to change the temperature of water from 20°C to
100°C (a change of 80°C) than to increase the temperature of the
same amount of water from 60°C to 100°C (a change of 40°C).
• In fact, it requires twice as much heat to change the temperature of
a given mass of water by 80°C compared to the change of 40°C.
• A person who wishes to bring water to a boil on a stovetop more
quickly should begin with warm tap water instead of cold tap water.
Specific ENERGY Capacity
• This discussion of specific heat capacity deserves one final
comment.
• The term specific heat capacity is somewhat of a misnomer.
• The term implies that substances have a capacity to contain
heat.
• As has been previously discussed, heat is not something that is
contained in an object.
• Heat is something that is transferred to or from an object.
• Objects contain energy in a variety of forms.
• When that energy is transferred to other objects of different
temperatures, we refer to transferred energy as heat or
thermal energy.
• While it's not likely to catch on, a more appropriate term
would be specific energy capacity.
Relating the Quantity of Heat to the
Temperature Change
• Specific heat capacities provide a means of
mathematically relating the amount of
thermal energy gained (or lost) by a sample of
any substance to the sample's mass and its
resulting temperature change.
• The relationship between these four
quantities is often expressed by the following
equation:
Q = m•C•ΔT
Q = m•C•ΔT
• where Q is the quantity of heat transferred to or from
the object,
• m is the mass of the object,
• C is the specific heat capacity of the material the object
is composed of, and
• ΔT is the resulting temperature change of the object.
• As in all situations in science, a delta value for any
quantity is calculated by subtracting the initial value of
the quantity from the final value of the quantity.
• In this case, ΔT is equal to Tfinal - Tinitial.
• When using the above equation, the Q value can turn
out to be either positive or negative.
Positive vs. Negative Results
• As always, a positive and a negative result
from a calculation has physical significance.
• A positive Q value indicates that the object
gained thermal energy from its surroundings;
this would correspond to an increase in
temperature and a positive ΔT value.
• A negative Q value indicates that the object
released thermal energy to its surroundings;
this would correspond to a decrease in
temperature and a negative ΔT value.
Example Problems
• Knowing any three of these four quantities allows an
individual to calculate the fourth quantity.
• A common task in many physics classes involves solving
problems associated with the relationships between
these four quantities.
• As examples, consider the two problems on the next
slide.
• The solution to each problem is worked out for you.
Example #1
• 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.
Like any problem in physics, the solution begins by
identifying known quantities and relating them to the
symbols used in the relevant equation.
In this problem, we know the following:
m = 450 g
C = 4.18 J/g/°C
Tinitial = 15°C
Tfinal = 85°C
Continued…
• We wish to determine the value of Q - the quantity
of heat.
• To do so, we would use the equation Q = m•C•ΔT.
• The m and the C are known;
• the ΔT can be determined from the initial and final
temperature.
T = Tfinal - Tinitial = 85°C - 15°C = 70.°C
With three of the four quantities of the relevant
equation known, we can substitute and solve for Q.
Q = m•C•ΔT = (450 g)•(4.18 J/g/°C)•(70.°C)
Q = 131670 J
Q = 1.3x105 J = 130 kJ (rounded to two significant
digits)
Example #2
• 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.
Continued…
• Compared to the previous problem, this is a much
more difficult problem. In fact, this problem is like two
problems in one.
• At the center of the problem-solving strategy is the
recognition that the quantity of heat lost by the water
(Qwater) equals the quantity of heat gained by the metal
(Qmetal).
• Since the m, C and ΔT values of the water are known,
the Qwater can be calculated. This Qwater value equals the
Qmetal value.
• Once the Qmetal value is known, it can be used with the
m and ΔT value of the metal to calculate the Qmetal.
• Use of this strategy leads to the following solution:
Continued…
• Part 1: Determine the Heat Lost by the Water
Given:
m = 50.0 g
C = 4.18 J/g/°C
Tinitial = 88.6°C
Tfinal = 87.1°C
ΔT = -1.5°C (Tfinal - Tinitial)
• Solve for Qwater:
Qwater = m•C•ΔT = (50.0 g)•(4.18 J/g/°C)•(-1.5°C)
Qwater = -313.5 J (unrounded)
(The - sign indicates that heat is lost by the water)
Continued…
• Part 2: Determine the value of Cmetal
Given:
Qmetal = 313.5 J (use a + sign since the metal is gaining
heat)
m = 12.9 g
Tinitial = 26.5°C
Tfinal = 87.1°C
ΔT = (Tfinal - Tinitial )
• Solve for Cmetal:
Rearrange Qmetal = mmetal•Cmetal•ΔTmetal to obtain
Cmetal = Qmetal / (mmetal•ΔTmetal)
• Cmetal = Qmetal / (mmetal•ΔTmetal) = (313.5 J)/[(12.9
g)•(60.6°C)]
Cmetal = 0.40103 J/g/°C
Cmetal = 0.40 J/g/°C (rounded to two significant digits)
Heat & Changes of State
• The discussion above and the accompanying equation
relates the heat gained or lost by an object to the resulting
temperature changes of that object.
• As we have learned, sometimes heat is gained or lost but
there is no temperature change.
• This is the case when the substance is undergoing a state
change.
• So now we must investigate the mathematics related to
changes in state and the quantity of heat.
• To begin the discussion, let's consider the various state
changes that could be observed for a sample of matter.
• The table below lists several state changes and identifies
the name commonly associated with each process.
State Changes
• Process
Melting
Freezing
Vaporization
Condensation
Sublimation
Deposition
Change of State
Solid to Liquid
Liquid to Solid
Liquid to Gas
Gas to Liquid
Solid to Gas
Gas to Solid
State Changes
• In the case of melting, boiling and sublimation, energy would
have to be added to the sample of matter in order to cause the
change of state.
• Such state changes are referred to as being endothermic.
• Freezing, condensation and deposition are exothermic;
• energy is released by the sample of matter when these state
changes occur.
• So one might notice that a sample of ice (solid water)
undergoes melting when it is placed on or near a burner.
• Heat is transferred from the burner to the sample of ice;
• energy is gained by the ice causing the change of state.
• But how much energy would be required to cause such a
change of state?
• Is there a mathematical formula that might help in determining
the answer to this question? There most certainly is.
Energy Required
• The amount of energy required to change the state of a
sample of matter depends on three things.
– It depends upon what the substance is, on how much
substance is undergoing the state change, and upon what state
change that is occurring.
• For instance, it requires a different amount of energy to melt
ice (solid water) compared to melting iron.
• And it requires a different amount of energy to melt ice (solid
water) as it does to vaporize the same amount of liquid water.
• And finally, it requires a different amount of energy to melt
10.0 grams of ice compared to melting 100.0 grams of ice.
• The substance, the process and the amount of substance are
the three variables that affect the amount of energy required
to cause a specific change in state.
http://www.wolframalpha.com/widgets/view.jsp?id=48f7d3043b
c03e6c48a6f0ebc0f258a8