Unformatted text preview: lver fraction = 1
3 Gold fraction = 2
3 Recognizing that one third of the rod is silver and two thirds is gold and taking values for the
coefficients of linear expansion for silver and gold from Table 12.1, we have
∆L 1
2
1 2 = αSilver ∆T + α Gold ∆T = αSilver + α Gold ∆T
L0,Silver + L0, Gold
3
3
3 3 −1 1 −1 2 = 19 × 10−6 ( C° ) + 14 × 10−6 ( C° ) ( 26 C° ) = 4.1×10−4 3 3 91. SSM REASONING AND SOLUTION The volume V0 of an object changes by an
amount ∆ V when its temperature changes by an amount ∆ T ; the mathematical relationship
is given by Equation 12.3: ∆V = β V0 ∆T . Thus, the volume of the kettle at 24 °C can be
found by solving Equation 12.3 for V0 . According to Table 12.1, the coefficient of
volumetric expansion for copper is 51 × 10–6 (C°)–1 . Solving Equation 12.3 for V0 , we
have ∆V
1.2 ×10 –5 m3
=
= 3.1×10 –3 m3
–6
–1
β ∆T [51×10 (C°) ](100 °C – 24 °C)
______________________________________________________________________________
V0 = 680 TEMPERATURE AND HEAT 92. REASONING AND SOLUTION We require that the heat gained by the cold water equals
the heat lost by the hot water, i.e.,
Qcw = Qhw
Therefore,
ccwmcw∆Tcw = chwmhw∆Thw
Since the specific heat capacity is that of water in each case, ccw = chw, then
mcw(36.0 °C – 13.0 °C) = mhw(49.0 °C – 36.0 °C)
Then
mcw = (0.57)mhw
We also know that mcw + mhw = 191 kg. Substituting for mcw and solving for mhw, we have
191 kg
= 121 kg
1.57
______________________________________________________________________________
mhw = 93. SSM REASONING AND SOLUTION From the vapor pressure curve that accompanies
Problem 75, it is seen that the partial pressure of water vapor in the atmosphere at 10 °C is
about 1400 Pa, and that the equilibrium vapor pressure at 30 °C is about 4200 Pa. The
relative humidity is, from Equation 12.6, Percent 1400 Pa relative = × 100 = 33%
humidity 4200 Pa ______________________________________________________________________________ 94. REASONING AND SOLUTION If the voltage is proportional to the temperature
difference between the junctions, then V1
V
=2
∆T1 ∆T2 or ∆T2 = V2
∆T
V1 1 Thus,
T2 − 0.0 °C =
Solving for T2 yields −3 1.90×10 V
(110.0 °C − 0.0 °C)
−3
4.75×10 V T2 = 44.0 °C . ______________________________________________________________________________ Chapter 12 Problems 681 95. REASONING Since all of the heat generated by friction goes into the block of ice, only this
heat provides the heat needed to melt some of the ice. Since the surface on which the block
slides is horizontal, the gravitational potential energy does not change, and energy
conservation dictates that the heat generated by friction equals the amount by which the
2
kinetic energy decreases or QFriction = 1 Mv0 − 1 Mv 2 , where v0 and v are, respectively, the
2
2
initial and final speeds and M is the mass of the block. In reality, the mass of the block
decreases as the melting proceeds. However, only a very small amount of ice melts, so we
may consider M to be essentially constant at its initial value. The heat Q needed to melt a
mass m of water is given by Equation 12.5 as Q = mLf, where Lf is the latent heat of fusion.
Thus, by equating QFriction to mLf and solving for m, we can determine the mass of ice that
melts.
SOLUTION Equating QFriction to mLf and solving for m gives
2
QFriction = 1 Mv0 − 1 Mv 2 = mLf
2
2 m= ( 2
M v0 − v 2 2 Lf ) = ( 42 kg ) ( 7.3 m/s )2 − (3.5 m/s )2 = 2.6 ×10−3 kg
2 ( 33.5 × 104 J/kg ) We have taken the value for the latent heat of fusion for water from Table 12.3. 96. REASONING To freeze either liquid, heat must be removed to cool the liquid to its freezing
point. In either case, the heat Q that must be removed to lower 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. The amount ∆T by which the temperature is lowered is the initial
temperature T0 minus the freezing point temperature T. Once the liquid has been cooled to
its freezing point, additional heat must be removed to convert the liquid into a solid at the
freezing point. The heat Q that must be removed to freeze a mass m of liquid into a solid is
given by Equation 12.5 as Q = mLf, where Lf is the latent heat of fusion. The total heat to be
removed, then, is the sum of that specified by Equation 12.4 and that specified by
Equation 12.5, or QTotal = cm (T0 − T) + mLf. Since we know that same amount of heat is
removed from each liquid, we can set QTotal for liquid A equal to QTotal for liquid B and
solve the resulting equation for Lf, A − Lf, B.
SOLUTION Setting QTotal for liquid A equal to QTotal for liquid B gives cAm (T0 − TA) + mLf, A = cBm (T0 − TB) + mLf, B 682 TEMPERATURE AND HEAT Noting that the mass m can be eliminated algebraically from this result and solving for
Lf, A − Lf, B, we find
Lf, A − Lf, B = cB (T0 − TB ) − cA (T0 − TA )
= 2670 J/ ( kg ⋅ C° ) 25.0 °C − ( −96.0 °C ) − 1850 J/ ( kg ⋅ C° ) 25.0 °C − ( −68.0 °C ) = 1.51×105 J/kg 97. REA...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.
 Spring '13
 CHASTAIN
 Physics, The Lottery

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