Unit 13
Heat
13.1
Heat and mechanical work
13.2
Heat capacity
13.3
Specific heat
13.4
Latent Heat
13.5
Conduction
13.6
Convection
13.7
Radiation
13.1 Heat and mechanical work
In this section we consider the connection between heat and
mechanical work. We also discuss the conservation of energy as it
regards heat. Recall that heat is the energy transferred from one object
to another. The equivalence between work and heat was first explored
quantitatively by James Prescott Joule (1818−1889), the British
physicist. In one of his experiments, Joule observed the increase in
temperature in a device similar to that shown in figure. Here, a mass
m falls through a certain distance h, during which gravity does the
mechanical work mgh. As the mass falls it turns the paddles in the
water, which results in a slight warming of the water. By measuring
the mechanical work, mgh, and the increase in the water’s
temperature, ∆T, Joule was able to show that energy was indeed
conserved. It had been converted from gravitational potential energy
to heat, and an increased temperature.
Before Joule’s work, heat was measured in a unit called the calorie
(cal). In particular, one kilocalorie (kcal) was defined as the amount
of heat needed to raise the temperature of 1 kg of water from 14.5 oC
to 15.5 oC. With these experiments, Joule was able to show that 1 kcal = 4186 J, or
equivalently, that one calorie is the equivalent of 4.186 J of mechanical work. This is referred
to as the mechanical equivalent of heat:
1 cal = 4.186 J.
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In studies of nutrition a different calorie is used. It is the Calorie, with a capital C, where 1 C
= 1 kcal.
Example
A 74.0-kg person drinks a thick, rich, 305-C milkshake. How
many stairs must this person climb to work off the shake? Let
the height of a stair be 20.0 cm.
Answer
We should first convert the energy of the milkshake to joules:
 4.186 J 
6
=
Q 305, 000 cal
= 305, 000 cal 
=
 1.28 × 10 J .
 1 cal 
To consume the milkshake, the work done against gravity = Q = mgH, where H is the height
of stairs. The height H is given by=
H
The number of stairs is obtained:
1.28 × 106 J
Q
=
= 1763 m .
mg (74.0 kg )(9.81 m / s 2 )
1760 m
= 8815 stairs .
0.200 m
13.2 Heat capacity
Suppose we add the heat Q to a given object, and its temperature increases by the amount ∆T.
The heat capacity, C in this object is defined as follow:
C=
Q
∆T
The heat capacity C has SI unit: J/K = J/Co.
Note that,
Q is positive if ∆T, is positive; that is, if heat is added to a system.
Q is negative if ∆T, is negative; that is, if heat is removed from a system.
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Example
The heat capacity of 1.00 kg of water is 4186 J/K. What is the temperature change of the
water if (a) 505 J of heat is added to the system, or (b) 1010 J of heat is removed?
Answer
(a)
We calculate the ∆T for Q = 505 J.
∆T =
(b)
505 J
Q
=
= 0.121 K .
C 4186 J / K
Since heat is removed from the system, and Q = −1010 J, the change in temperature
∆T:
−1010 J
Q
∆T = =
=−0.241 K .
C 4186 J / K
13.3 Specific heat
Since heat capacity varies with the amount of substance, even the substances are of the same
type. A new quantity, the specific heat, c, is then defined. The specific heat depends on the
substance, and not on the amount of the substance.
c=
Q
m∆T
The specific heat c has SI unit: J/(kg⋅K) = J/(kg⋅Co).
For example the specific heat of water is
c water = 4186 J /(kg ⋅ K ) .
Specific heats for common substances are listed in table. Note that the specific heat of water
is by far the largest of any common material. This is just another of the many unusual
properties of water. Having such a large specific heat means that water can give off or take in
large quantities of heat with little change in temperature. It is for this reason that if you bite a
pie that is just out the oven, you are much more likely to burn you tongue on the fruit filling
(which has a high water content) than on the much drier crust.
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Substance
Specific heat, c, [J/(kg⋅K)]
Water
4186
Ice
2090
Steam
2010
Aluminum
900
Glass
703
Iron (steel)
448
Copper
387
Silver
234
Gold
129
Lead
128
Table: Specific heats at atmospheric temperature and pressure
Example
550 g of water at 32 oC is poured into a 210-g aluminum can with an initial temperature of 15
o
C. Find the final temperature of the system, assuming no heat is exchanged with the
surroundings.
Answer
Method 1:
The heat flow out of water:
=
QW mW cW (TW − T ) .
The heat flow into the aluminum:
=
Qa ma ca (T − Ta ) .
Conservation of energy implies QW = Qa , which gives T = 31 oC.
Method 2:
The heat flow out of water:
=
QW mW cW (T − TW ) .
The heat flow into the aluminum:
=
Qa ma ca (T − Ta ) .
Conservation of energy implies QW + Qa =
0 , which gives T = 31 oC.
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13.4 Latent heats
The latent heat, L, is the heat that must be added to or removed from one kilogram of a
substance to convert it from one phase to another. During the conversion process, the
temperature of the system remains constant. Latent heat is a positive quantity.
Mathematically, Q = mL, where the SI unit of L is J/kg. The latent heat to melt a substance is
referred to as the latent heat of fusion, Lf. The latent heat required to convert a liquid to a gas
is the latent heat of vaporization, Lv. The latent heat needed to convert a solid directly to a gas
is the latent heat of sublimation, Ls.
Material
Latent heat of fusion (J/kg)
Latent heat of vaporization (J/kg)
Water
33.5 × 104
22.6× 105
Ammonia
33.2 × 104
13.7× 105
Copper
20.7 × 104
47.3× 105
Ethyl alcohol
10.8 × 104
8.55× 105
Gold
6.28 × 104
17.2× 105
Nitrogen
2.57 × 104
2.00× 105
Lead
2.32 × 104
8.59× 105
Oxygen
1.39 × 104
2.13× 105
Example
Both water at 100oC and steam at 100oC can cause serious burn. Is a burn produced by steam
likely to be (a) more severe, (b) less severe, or (c) the same as a burn produced by water?
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Answer
As the water or steam comes into contact with skin, it cools down to the temperature of skin,
e.g. 39oC. For the case of steam, it needs to cool down to 100oC water first and then further
cool down to 39oC. That means, the body skin absorb more heat from the steam than water,
and the steam burn is worse.
13.5 Conduction
Perhaps the most familiar form of heat exchange is conduction, which is the flow of heat
directly through a physical material. For example, if you hold one end of a metal rod and put
the other end in a fire, it doesn’t take long before you begin to feel warmth on your end. The
heat you feel is transported along the rod by conduction. In microscopic point of view, when
you placed one end of the rod into the fire the high
temperature at that location caused the molecules to vibrate
with an increased amplitude. These molecules in turn jostle
their neighbors, and cause them to vibrate with greater
amplitude as well. Eventually the effect travels from
molecule to molecule across the length of the rod, resulting
in the macroscopic phenomenon of conduction.
Consider now, how much heat flows as a result of conduction? To answer this question we
consider the simple system. Here we show a rod of length L and cross-sectional area A, with
one end at the temperature T1 and the other at the temperature T2 > T1. Experiments show that
the amount of heat Q that flows through this rod:

Increases in proportion to the rod’s cross-sectional area, A;

Increases in proportion to the temperature difference, ∆T = T2 − T1 ;

Increases steadily with time, t;

Decreases with the length of the rod, L.
Combining these observations in a mathematical expression gives:
 ∆T 
Q = kA
t .
 L 
The constant k is referred to as the thermal conductivity of the rod.
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Substance
Thermal conductivity, k, [W/(m⋅K)]
Silver
417
Copper
395
Gold
291
Aluminum
217
Steel, low carbon
66.9
Lead
34.3
Stainless steel
16.3
Ice
1.6
Concrete
1.3
Glass
0.84
Water
0.60
Wood
0.10
Wool
0.040
Air
0.0234
Table: Thermal conductivities
Example
One of the windows in a house has the shape of a square 1.0 m on
a side. The glass in the window is 0.50 cm thick. How much heat
is lost through this window in one day if the temperature in the
house is 21 oC and the temperature outside is 0.0 oC?
Answer
Since the thermal conductivity of glass is 0.84 W/(m⋅K) and the
 ∆T
formula for heat flow is Q = kA 
 L

 t , hence,

 21 K 
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Q = [0.84 W /(m ⋅ K )](1.0 m) 2 
 (24 × 60 × 60 s ) = 3.0 × 10 J .
 0.0050 m 
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Example (Challenging)
Two rods of equal length (0.525 m) are arranged in series.
The rods have a square cross section, 1.50. cm on a side. Find
(a) the temperature at the lead-copper junction, and (b) the
amount of heat that flows through the rods in 1.00 s.
Assume that no heat is exchanged between the rods and the
surroundings, except at the ends.
Answer
(a)
Let the temperature at the lead-copper junction be T.
The heat flow through the lead rod:
 T − 2.00 oC 
Ql = kl A 
t .
L


The heat flow through the copper rod:
 106 oC − T
Qc = kc A 
L


t .

Since the heat flow in the lead equal to the heat flow in the copper.
 T − 2.00 oC 
 106 oC − T 
kl A 
 t = kc A 
t
L
L




We have kl (T − 2.00 oC ) = kc (106 oC − T ) , hence
=
T
(b)
(106 oC ) kc + (2.00 oC ) kl
= 97.7 oC .
k c + kl
We substitute T into the formula of heat flow for each rod. Note that the time interval
we are going to calculate is one second (i.e. t = 1 s), L = 0.525 m and A = (0.015)2 m2.
 T − 2.00 oC 
Lead:
=
Ql k=
 t 1.41 J
l A
L


 106 oC − T 
Copper: Qc k=
=
 t 1.41 J
c A
L


The heat flow in this series arrangement in one second is 1.41 J. Note also that the heat flow
through the rods in series, 1.41 J, is much less than the heat flow through the rods in parallel,
19.1 J. This confirms our conclusion in the last example.
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13.6 Convection
Suppose you want to heat a small room. To do so, you bring a portable electric heater into the
room and turn it on. As the heating coils get red-hot they heat the air in their vicinity, and as
this air warms, it expands, becoming less dense. Because of its lower density, the warm air
rises, to be replaced by cold dense air descending from overhead. This sets up a circulating
flow of air that transports heat from the heating coils to the air throughout the room. Heat
exchange of this type is referred to as convection.
In general, convection occurs when a fluid is unevenly heated. An
example occurs in a kettle of water over stove, the warm portions of
the fluid at the bottom of kettle rise because of their lower density
and cool portions sink because of their higher density. Thus, in
convection, temperature differences result in a flow of fluid. It is this
physical flow of matter that carries heat throughout the system.
13.7 Radiation
All objects give off energy as a result of radiation. The energy radiated by an object is in the
form of electromagnetic waves, which include visible light as well as infrared and ultraviolet
radiation. Thus, unlike convection and conduction, radiation has no need for a physical
material to mediate the energy transfer, since electromagnetic waves can propagate through
empty space – that is, through a vacuum.
The energy radiated per time by an object – that is, the radiated power, P – is proportional to
the surface area, A, over the radiation occurs. It also depends on the temperature of the object.
This behavior is described in the Stefan-Boltzmann law:
P = eσAT 4
SI unit of power is W, the Watt.
The constant σ in this expression is a fundamental physical constant, the Stefan-Boltzmann
constant:
σ = 5.67 × 10 −8 W /(m 2 ⋅ K 4 ) .
The other constant is the emissivity, e. It is a dimensionless number between 0 and 1 that
indicates how effective the object is in radiating energy. A value of 1 means that the object is
a perfect radiator.
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Experiments show that objects absorb radiation from their surroundings according to the
same law, the Stefan-Boltzmann law, by which they emit radiation. Thus, if the temperature
of an object is T, and its surroundings are at the temperature Ts, the net power radiated by the
object is
Pnet = eσA(T 4 − Ts4 ) .
Example
A man has his surface area of skin 1.15 m2 and a surface
temperature 303 K. Find the net radiated power from this
person (a) in a dressing room where the temperature is 293 K,
and (b) outside, where the temperature is 273 K. Assume an
emissivity of 0.900 for the person’s skin.
Answer
(a)
In the dressing room:
=
Pnet eσ A(T 4 − Ts4 )
= (0.900)[5.67 × 10−8 W /(m 2 ⋅ K 4 )](1.15 m 2 ) × [(303 K ) 4 − (293 K ) 4=
] 62.1W
(b)
In the outdoor:
=
Pnet eσ A(T 4 − Ts4 )
= (0.900)[5.67 × 10−8 W /(m 2 ⋅ K 4 )](1.15 m 2 ) × [(303 K ) 4 − (273 K ) 4=
] 169 W
In the warm room the net radiated power is roughly that of a small lightbulb (about 60 W);
outdoors the net radiated power has more than doubled, and is comparable to that of a 150-W
lightbulb.
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