Thermal Behavior of Matter and Heat Engines - Department of ...

Unit 11
Thermal Behavior of Matter and Heat Engines
11.1 Thermal expansion
11.2 Heat and mechanical work
11.3 Heat capacity
11.4 Specific heat
11.5 Latent Heat
11.6 Conduction
11.7 Convection
11.8 Radiation
11.9 Heat engines and the Carnot cycle
11.1 Thermal expansion
Most substances expand when heated. The expansion of liquid, such as mercury or
alcohol, results in a column of liquid of variable height within the glass neck of the
thermometer.
11.11 Linear expansion
The change in the length of rod is directly proportional to the change in temperature.
The constant of proportionality is referred to as α and L 0 is the initial length of the
rod.
∆
0
L = α L ∆T
SI unit for α: K −1 = ( C o ) −1 .
Alternatively, we can refer α as the fractional change of length per unit change of
L 1
temperature, e.g. α = ⎛ ⎜ ∆ ⎞
⎟ .
⎝ L0
⎠ ∆ T
1

Substance Coefficient of Expansion, α (K −1 )
Lead
Aluminum
Copper
Iron (Steel)
Concrete
Window glass
Quartz
29 × 10 −6
24 × 10 −6
17 × 10 −6
12 × 10 −6
12 × 10 −6
11 × 10 −6
0.50 × 10 −6
Table: Coefficients of thermal expansion
An interesting application of thermal expansion is in the behavior of a bimetallic strip.
As the name suggests, a bimetallic strip consists of two metals bonded together to
form a linear strip of metal. Since two different metals will, in general, have different
coefficients of linear expansion, α, the two sides of the strip will change lengths by
different amounts when heated or cooled. Refer to the figure.
(a) A bimetallic strip composed of metals A and B. If metal B has a larger coefficient
of linear expansion than metal A, it will shrink more when cooled (b), and expand
more when heated (c). A bimetallic strip can be used to construct a thermostat (d). If
the temperature falls, the strip bends downward and closes the electrical circuit, which
then operates a heater. When the temperature rises, the strip deflects in the opposite
direction, breaking the circuit and turning off the heater.
Thermal expansion, though small, is far from negligible in many everyday situations.
This is especially true when long objects such as railroad tracks, bridges, or pipelines
are involved. Bridges and elevated highways, must include expansion joints to
prevent the roadway from buckling when it expands in hot weather. Similarly,
2

pipelines typically include loops that allow for expansion and contraction when the
temperature changes.
11.12 Area expansion
Since the length of an object changes with temperature, it follows that its area changes
as well. The initial area of the square is A = L 2 . If the temperature of the square is
changed by ∆T, and the length of each side of the square by ∆L,
2
2 2 2
2 2 2
A'
= ( L + ∆L)
= ( L + α L∆T
) = ( L + 2αL
∆T
+ α L ∆T
) .
2 2
Neglect the higher order term, e.g. α L ∆T
2
2
, since α ∆T
2
relatively small, we have
2 2
A'
= L + 2αL
∆T
= A + 2αA∆T
That is
∆A
≈ 2 αA∆T
.
Though the calculation works out from square, it is applicable to area of any shape.
Example
A washer has hole in the middle. As the washer is heated, does the hole (a) expand,
(b) shrink, or (c) stay the same?
3

Answer:
To make a washer from a disk of metal, we can cut along a circular curve, and remove
the inner disk. If we now heat the system, both the washer and the inner disk expand.
On the other hand, if we had left the inner disk in place and heated the original disk it
would also expand. Removing the heated inner disk would create an expanded
washer, with an expanded hole in the middle. We obtain the same result whether we
remove the inner disk and then heat, or heat first and then remove the inner disk.
Thus, heating the washer causes both it and its hole to expand, and they both expand
with the same coefficient of linear expansion.
11.13 Volume expansion
Similar to the argument in the expansion of area with temperature, the volume
changes with temperature. The initial volume of the cube is V = L 3 . If the temperature
of the cube is changed by ∆T, and the length of each side of the cube by ∆L,
3
3 3 3
2 3 2 3 3 3
V ' = ( L + ∆L)
= ( L + α L∆T
) = ( L + 3αL
∆T
+ 3α
L ∆T
+ α L ∆T
) .
2 3 2
3 3
Neglect the higher order term, e.g. α L ∆T and α L ∆T
3
, we have
3 3
V ' = L + 3αL
∆T
= V + 3αV∆T
That is
∆V
≈ 3 αV∆T
.
Though the calculation works out from cube, it is applicable to volume of any shape.
We define the coefficient of volume expansion, β where ∆V
≈ β V∆T
, and β = 3α.
Example
A copper flask with a volume of 150 cm 3 is filled to the brim with olive oil. If the
temperature of the system is increased from 6.0 o C to 31 o C, how much oil spills from
the flask? The coefficient of expansion of olive oil is given as 0.69 × 10 −3 K −1 .
Heat by 25 C o
4

Answer:
The change in temperature ∆T = 31 C o – 6 C o = 25 C o = 25 K.
As the system is heated, both the flask and the oil expand.
The change in volume of the olive oil:
The change in volume of the flask:
∆V oil
= βV∆T
=
=
∆
−3
−1
3
3
( 0.69×
10 K )(150cm
)(25 K)
2.6cm
V flask
= βV∆T
= 3αV∆T
−6
−1
3
= 3(17 × 10 K )(150cm
)(25 K)
=
0.19cm
3
The difference in volume expansions is the volume of oil that spills out:
3
3
3
∆ Voil − ∆V
flask
= 2.6cm
− 0.19cm
= 2.4cm
.
11.2 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 o C to 15.5 o C. With these experiments, Joule
was able to show that 1 kcal = 4186 J, or equivalently, that one calorie is the
5

equivalent of 4.186 J of mechanical work. This is referred to as the mechanical
equivalent of heat:
1 cal = 4.186 J.
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 ,000cal
= 305,000cal
⎜ = 1.28×
10 J
1 cal
⎟
.
⎝ ⎠
To consume the milkshake, the work done against gravity = Q = mgH, where H is the
6
Q 1.28×
10 J
height of stairs. The height H is given by H = =
= 1763m
.
2
mg (74.0 kg)(9.81m
/ s )
The number of stairs is obtained:
1760m = 8815 stairs .
0.200m
11.3 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:
Q
C = ∆ T
The heat capacity C has SI unit: J/K = J/C o .
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.
6

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.
Q 505 J
∆ T = = = 0. 121K
.
C 4186 J / K
(b)
Since heat is removed from the system, and Q = −1010 J, the change in
temperature ∆T:
Q −1010
J
∆ T = = = −0.
241K
.
C 4186 J / K
11.4 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.
Q
c =
m∆T
The specific heat c has SI unit: J/(kg⋅K) = J/(kg⋅C o ).
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.
7

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 o C 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: Q = m c ( T − T ) .
The heat flow into the aluminum: Q = m c T − T ) .
W
W
a
W
W
a a
(
a
Conservation of energy implies
Q = Q , which gives T = 31 o C.
W
a
Method 2:
The heat flow out of water: Q = m c T − T ) .
W
W
W
(
W
The heat flow into the aluminum: Q = m c T − T ) .
a
a a
(
a
Conservation of energy implies Q Q = 0 , which gives T = 31 o C.
W
+ a
8

11.5 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, L f . The latent heat required to
convert a liquid to a gas is the latent heat of vaporization, L v . The latent heat needed to
convert a solid directly to a gas is the latent heat of sublimation, L s .
Material Latent heat of fusion (J/kg) Latent heat of vaporization (J/kg)
Water 33.5 × 10 4 22.6× 10 5
Ammonia 33.2 × 10 4 13.7× 10 5
Copper 20.7 × 10 4 47.3× 10 5
Ethyl alcohol 10.8 × 10 4 8.55× 10 5
Gold 6.28 × 10 4 17.2× 10 5
Nitrogen 2.57 × 10 4 2.00× 10 5
Lead 2.32 × 10 4 8.59× 10 5
Oxygen 1.39 × 10 4 2.13× 10 5
Example
Both water at 100 o C and steam at 100 o C 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?
9

Answer:
As the water or steam comes into contact with skin, it cools down to the temperature
of skin, e.g. 39 o C. For the case of steam, it needs to cool down to 100 o C water first
and then further cool down to 39 o C. That means, the body skin absorb more heat from
the steam than water, and the steam burn is worse.
11.5 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 crosssectional
area A, with one end at the temperature T 1 and the other at the temperature
T 2 > T 1 . 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 = T 2
− 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.
10

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 o C and the temperature outside is 0.0 o C?
Answer:
Since the thermal conductivity of glass is 0.84 W/(m⋅K) and the
⎛ ∆T
⎞
formula for heat flow is Q = kA⎜
⎟t
, hence,
⎝ L ⎠
2 ⎛ 21K
⎞
8
Q = [ 0.84W
/( m ⋅ K)](1.0
m)
⎜ (24 × 60 × 60 s)
= 3.0 × 10 J
0.0050 m
⎟
.
⎝ ⎠
11

Example
Two metal rods are to be used to conduct
heat from a region of 100 o C to a region
at 0 o C. The rods can be placed in
parallel, as shown in figure. Is the heat
conducted in the parallel arrangement (a)
greater than, (b) less than, or (c) the same as the heat conducted with the rods in
series?
Answer:
The parallel arrangement conducts more heat for two reasons.
(i) The cross-sectional area available for heat flow is twice as large for the
parallel rods.
(ii) The parallel configuration has both rods the full temperature difference.
Example
Two 0.525-m rods, one lead the other copper, are
connected between metal plates held at 2.00 o C and
106 o C. The rods have a square cross section, 1.50
cm on a side. How much heat flows through the
two rods in 1.00 s? Assume that no heat is
exchanged between the rods and the surroundings, except at the ends.
Answer:
The heat flow in one second through the lead rod:
Q
l
⎛ ∆T
⎞
= kl
A⎜
⎟t
⎝ L ⎠
2
⎛ 104 K ⎞
= [34.3W
/( m ⋅ K)](0.0150
m)
⎜ (1.00 s)
0.525m
⎟
⎝ ⎠
= 1.53 J
The heat flow in one second through the copper rod:
12

Q
c
⎛ ∆T
⎞
= kc
A⎜
⎟t
⎝ L ⎠
= [395W
/( m ⋅ K)](0.0150m)
= 17.6 J
2
⎛ 104 K ⎞
⎜ (1.00 s)
0.525m
⎟
⎝ ⎠
Total heat flow in one second:
Q + Q
= 1 .53 J + 17.6 J = 19.
l c
1
J
Example
The two rods in the above example is arranged in series.
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:
The heat flow through the copper rod:
Q
l
Q
c
⎛ T − 2.00
= kl
A
⎜
⎝ L
C
⎟ ⎞
t .
⎠
o
⎛106
C − T
kc
A
t
L ⎟ ⎞
=
⎜
.
⎝ ⎠
Since the heat flow in the lead equal to the heat flow in the copper.
o
o
⎛ T − 2.00 C ⎞ ⎛106
C − T
kl
A
t kc
A
t
L
L ⎟ ⎞
⎜
⎟ =
⎜
⎝
⎠ ⎝ ⎠
o
o
We have k ( T − 2.00 C)
= k (106 C − T ) , hence
l
c
o
o
(106 C)
kc
+ (2.00 C)
kl
o
T = = 97.7 C .
k + k
c
l
(b) 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)
o
Lead:
o
⎛ T − 2.00 C ⎞
Ql = kl
A
⎜
t = 1. 41J
L
⎟
⎝
⎠
13

o
⎛106
C − T ⎞
Copper: Qc = kc
A
⎜
t = 1. 41J
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.
11.6 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. As with the room heater, the warm portions of the
fluid 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.
Alternating land and sea breezes are example of convection in the atmosphere. Refer
to the figures. (a) During the day, the sun warms the land more rapidly than the water.
This is because the land, which is mostly rocks, has a lower specific heat than the
water. The warm land heats the air above it, which becomes less dense and rises.
Cooler air from over the water flows in to take its place, producing a “sea breeze”. (b)
At night, the land cools off more rapidly than the water – again because of its lower
specific heat. Now it is the air above the relatively warm water that rises and is
replaced by cooler air from over the land, producing a “land breeze”.
14

11.7 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
SI unit of power is W, the Watt.
The constant σ in this expression is a fundamental physical constant, the Stefan-
Boltzmann constant:
−8
2 4
σ = 5.67 × 10 W /( m ⋅ K ) .
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.
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 T s , the net
power radiated by the object is
P
4 4
= eσ A(
T − ) .
net
T s
4
Example
A man has his surface area of skin 1.15 m 2 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.
15

Answer:
(a) In the dressing room:
P
net
= eσA(
T
4
− T
4
s
= (0.900)[5.67 × 10
)
−8
W /( m
2
⋅ K
4
2
)](1.15m
) × [(303 K)
4
− (293 K)
4
] = 62.1W
(b)
P
net
In the outdoor:
= eσA(
T
4
− T
4
s
= (0.900)[5.67 × 10
)
−8
W /( m
2
⋅ K
4
2
)](1.15m
) × [(303 K)
4
− (273 K)
4
] = 169W
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.
11.9 Heat engines and the Carnot cycle
A heat engine is a device that converts heat into work. The classic example of this
type of device is steam engine. The basic elements of a steam engine are a boiler,
where heat converts water to steam, and a
piston that can be displaced by the
expanding steam. In some engines, the
steam is simply exhausted into the
atmosphere after it has expanded against
the piston. More sophisticated engines
send the exhaust steam to a condenser,
where it is cooled and condensed back to
liquid water, then recycled to the boiler.
Heat engines have in common are:
(i) A high temperature region, or reservoir,
that supplies heat to the engine (the boiler
in the steam engine);
(ii) A low temperature reservoir where
“waste” heat is released (the condenser in
the steam engine); and
(iii) An engine that operates in a cycle fashion.
16

In addition, though not shown in the figure, heat engines have a working substance
(steam in the steam engine) that causes the engine to operate.
Notice that a certain amount of heat, Q h , is supplied to the engine from the high
temperature or “hot” reservoir. Of this heat, a fraction appears as work, W, and the
rest is given off as waste heat, Q c , at a relatively low temperature to the “cold”
reservoir. Let Q h and Q c denote magnitudes, so that both quantities are positive,
energy conservation can be written as follows: W
= Q h
− Q .
Of particular interest for any engine is its efficiency, e, which is simply the fraction of
the heat supplied to the engine that appears as work.
W
e = .
Q h
We have also the relations
e
W
Q
− Q
Q
h c
c
= = = 1 − .
Qh
Qh
Qh
The unit of e is dimensionless. Lord Kelvin suggested that the ratio of heats (Q h and
Qc), and temperatures (T h , T c ) for an engine of maximum efficiency e max are related
by the formula
Q
c
T =
c , which is also known as the Carnot relation. The maximum
Q
h
T
h
Tc
efficiency of a heat engine is rewritten as emax = 1− .
T
Example
Suppose you have a heat engine that can operate in one of the two different modes. In
mode 1, the temperatures of the two reservoirs are T c = 200 K and T h = 400 K; in
mode 2, the temperatures are T c = 400 K and T h = 600 K. Is the efficiency of mode 1
(a) greater than, (b) less than, or (c) equal to the efficiency of mode 2?
h
c
Answer:
Tc
The efficiency depends on the ratio of the two temperatures, e = 1 − rather than on
T
1 1
their difference. In this case, the efficiency of mode 1 is e
1
= 1−
= and the
2 2
efficiency of mode 2 is e
2
= 1−
3
2
=
lower temperatures, is more efficient.
1
. Thus, mode 1, even though it operates at the
3
h
17

When we stated the second law of thermodynamics,
we said that the spontaneous flow of heat is always
from high temperature to low temperature. The key
word here is “spontaneous”. It is possible for heat to
flow “uphill”, from a cold object to a hot one, but it
doesn’t happen spontaneously - work must be done
on the system to make it happen, just as work must
be done to pump water from a well. Refrigerators, air
conditioners, heat pumps are devices that use work
to make heat flow against its natural tendency.
By energy conservation, we have Q
= Q W .
h c
+
We define the coefficient of performance, COP, for a
refrigerator as an indicator of its effectiveness:
Qc
COP = .
W
Typical values for COP are in the range 2 to 6.
A heat pump can be considered as an air conditioner with the reservoirs switched. A
heat pump does a work W to remove an amount of heat from the cold reservoir of
outdoor air, then exhausts a heat Q h into the hot reservoir of air in the room. Just as
with the refrigerator and the air conditioner, the heat going to the hot reservoir is
Q
= Q W .
h c
+
In an ideal, reversible heat pump, we have
Q
c
T =
c holds, just as it does for a heat
engine. Thus, if you want to add a heat Q h to a room, the work that must be done to
accomplish this is
⎛ Q ⎞ ⎛ ⎞
= − =
⎜ −
c
T
⎟ =
⎜ −
c
W Q
⎟
h
Qc
Qh
1 Qh
1 .
⎝ Qh
⎠ ⎝ Th
⎠
Example
An ideal heat pump, one that satisfies the Carnot relation, is used to heat a room that
is at 293 K. If the pump does 275 J of work, how much heat does it supply to the
room if the outdoor temperature is (a) 273 K or (b) 263 K?
Q
h
T
h
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Answer:
(a) Since W
⎛ T ⎞
= ⎜ −
c
Q
⎟
h
1 , we have
⎝ Th
⎠
W 275 J
Qh = =
= 4030 J .
⎛ Tc
⎞ ⎛ 273 K ⎞
⎜1
− 1
T
⎟
⎜ −
h
293 K
⎟
⎝ ⎠ ⎝ ⎠
(b) Similarly, we have
W 275 J
Qh = =
= 2690 J .
⎛ Tc
⎞ ⎛ 263K
⎞
⎜1
− 1
T
⎟
⎜ −
h
293K
⎟
⎝ ⎠ ⎝ ⎠
As one might expect, the same amount of work provides less heat
when the outside temperature is lower. That is, more work must be done on a colder
day to provide the same heat to the inside air.
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