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T = 123 °C .
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85. REASONING The change in length ∆L of the pipe is proportional to the coefficient of
linear expansion α for steel, the original length L0 of the pipe, and the change in temperature
∆T . The coefficient of linear expansion for steel can be found in Table 12.1.
SOLUTION The change in length of the pipe is
−1 ∆L = α L0 ∆T = 1.2 × 10−5 ( C° ) ( 65 m ) 18 °C − ( −45 °C ) = 4.9 × 10−2 m (12.2) Chapter 12 Problems 677 86. REASONING AND SOLUTION The heat required to evaporate the water is Q = mLv, and
to lower the temperature of the jogger we have Q = mjc∆T. Equating these two expressions
and solving for the mass m of the water, we have
m= m= m jc ∆T
Lv ( 75 kg ) [3500 J/(kg ⋅ C°)] (1.5 C° ) = 0.16 kg
2.42 ×106 J/kg
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87. SSM REASONING From the principle of conservation of energy, the heat lost by the
coin must be equal to the heat gained by the liquid nitrogen. The heat lost by the silver coin
is, from Equation 12.4, Q = ccoin m coin ∆Tcoin (see Table 12.2 for the specific heat capacity
of silver). If the liquid nitrogen is at its boiling point, –195.8 °C, then the heat gained by the
nitrogen will cause it to change phase from a liquid to a vapor. The heat gained by the
liquid nitrogen is Q = mnitrogen Lv , where mnitrogen is the mass of liquid nitrogen that vaporizes, and Lv is the latent heat of vaporization for nitrogen (see Table 12.3).
SOLUTION Qlost by = Qgained by
coin nitrogen c coin mcoin ∆ Tcoin = mnitrogen Lv Solving for the mass of the nitrogen that vaporizes
m nitrogen =
= c coin m coin ∆ Tcoin
Lv
[235 J/(kg ⋅ C °)](1.5 × 10 –2 kg)[25 °C – (–195.8 °C)] = 3.9 × 10 −3 kg 2.00 × 10 J/kg
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5 88. REASONING Since the container is being ignored and since we are assuming negligible
heat exchange with the environment, the principle of conservation of energy applies in the
following form: heat gained equals heat lost. In reaching equilibrium the colder aluminum
gains heat in warming to 0.0 ºC, and the warmer water loses heat in cooling to 0.0 ºC. In
either case, the heat Q that must be supplied or removed to change the temperature of a
substance of mass m by an amount ∆T is given by Equation 12.4 as Q = cm∆T, where c is the
specific heat capacity. In using this equation as we apply the energyconservation principle,
we must remember to express the change in temperature ∆T as the higher minus the lower
temperature. The water that freezes into ice also loses heat. The heat Q lost when a mass m 678 TEMPERATURE AND HEAT of water freezes is given by Equation 12.5 as Q = mLf, where Lf is the latent heat of fusion.
By including this amount of lost heat in the energyconservation equation, we will be able to
calculate the mass of water that is frozen.
SOLUTION Using the energyconservation principle and Equations 12.4 and 12.5 gives cAluminum mAluminum ∆TAluminum = cWater mWater ∆TWater +
144444
244444
3 144 2444
4
3
Heat gained by aluminum Heat lost by water mIce L
1444 f, Water 3
2444
Heat lost by water that freezes Solving for mIce, taking values for the specific heat capacities from Table 12.2, and taking the
latent heat for water from Table 12.3, we find that
mIce = cAluminum mAluminum ∆TAluminum − cWater mWater ∆TWater
Lf, Water 9.00 × 102 J/ ( kg ⋅ C° ) ( 0.200 kg ) 0.0 °C − ( −155 °C ) =
4
33.5 ×10 J/kg 4186J/ ( kg ⋅ C° ) (1.5 kg )( 3.0 °C − 0.0 °C ) −
= 0.027 kg
33.5 ×104 J/kg 89. REASONING AND SOLUTION The value for the coefficient of thermal expansion of
steel is given in Table 12.1. The relation, ∆L = α L0∆T, written in terms of the diameter d of
the rod, is
∆d
0.0026 cm
∆T =
=
= 110 C°
(12.2)
α d0 12 × 10−6 ( C° )−1 ( 2.0026 cm ) ______________________________________________________________________________ 90. REASONING To determine the fractional decrease in length ∆L
, we need the
L0,Silver + L0, Gold decrease ∆L in the rod’s length. It is the sum of the decreases in the silver part and the gold
part of the rod, or ∆L = ∆LSilver + ∆LGold. Each of the decreases can be expressed in terms of
the coefficient of linear expansion α, the initial length L0, and the change in temperature ∆T,
according to Equation 12.2. Chapter 12 Problems 679 SOLUTION Using Equation 12.2 to express the decrease in length of each part of the rod,
we find the total decrease in the rod’s length to be ∆L = αSilver L0, Silver ∆T + α Gold L0, Gold ∆T
1442443 144
4
4
244
3
∆LSilver
∆LGold
The fractional decrease in the rod’s length is, then, αSilver L0, Silver ∆T + α Gold L0, Gold ∆T
∆L
=
L0,Silver + L0, Gold
L0,Silver + L0, Gold L0, Silver
L0, Gold
= αSilver ∆T + α Gold ∆T
L L + L0, Gold + L0, Gold 0,Silver2444 0,Silver2444
144
4
3
144
4
3
Si...
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 Spring '13
 CHASTAIN
 Physics, The Lottery

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