Unformatted text preview: we find
∆V = (βw − βc) V0 ∆T
–6 = (207 × 10 –1 C° − 51 × 10 –6 C° ) π (9.5 × 10
–1 –3 m) (76 m)(54 C°) = 1.8 × 10
2 –4 m3 ______________________________________________________________________________
37. REASONING AND SOLUTION
a. The apparent weight of the sphere will be larger after it cools. This is because the sphere shrinks while cooling, displacing less water, and hence decreasing the buoyant force
acting on it.
b. The weight of the submerged sphere before cooling is
W0 = ρ gV0 − ρwgV0 The weight of the submerged sphere after cooling is
W = ρ gV0 − ρwgV The weight difference is ∆W = W − W0 = −ρwg∆V where ∆V = V − V0. The volume change is ∆V = β V0∆T. Now, ∆W = −ρwgβ V0∆T. The
sphere’s original volume is
V0 = (4/3)π r 3 = (4/3)(π)(0.50 m)3 = 0.52 m3 The coefficient of volumetric expansion β for aluminum is given in Table 12.1, so we have ∆W = −(1.0 × 103 kg/m3)(9.80 m/s2)[69 × 10–6 (C°)–1](0.52 m3)(−5.0 × 101 C°) = 18 N 648 TEMPERATURE AND HEAT ______________________________________________________________________________
38. REASONING When the mercury is heated, it expands a distance ∆L into the capillary,
creating a small cylinder of mercury in the capillary with a volume ∆V = π r 2∆L. The radius
r of the capillary is assumed constant, because we are ignoring the thermal expansion of the
glass. Because the increase ∆V in the volume of the mercury is due to thermal expansion,
−1 we can also express it as ∆V = βV0 ∆T (Equation 12.3), where β = 182 × 10−6 ( Co ) is the
coefficient of volume expansion for mercury (see Table 12.1), V0 is the volume of the
mercury before heating, and ∆T = 1.0 C° is the increase in the temperature of the mercury.
SOLUTION Solving ∆V = π r 2∆L for ∆L yields ∆L = ∆V
π r2 (1) Substituting ∆V = βV0 ∆T (Equation 12.3) into Equation (1), we obtain
−1 ∆V β V0 ∆T 182 × 10−6 ( Co ) ( 45 mm3 )(1.0 Co )
∆L = 2 =
=
= 9.0 mm
2
πr
π r2
π (1.7 × 10−2 mm ) 39. SSM REASONING In order to keep the water from expanding as its temperature
increases from 15 to 25 °C, the atmospheric pressure must be increased to compress the
water as it tries to expand. The magnitude of the pressure change ∆P needed to compress a
substance by an amount ∆V is, according to Equation 10.20, ∆ P = B( ∆V / V0 ) . The ratio
∆ V / V0 is, according to Equation 12.3, ∆V / V0 = β ∆T . Combining these two equations
yields
∆P = B β∆T
SOLUTION Taking the value for the coefficient of volumetric expansion β for water from
Table 12.1, we find that the change in atmospheric pressure that is required to keep the
water from expanding is
−1
∆P = (2.2 ×109 N/m 2 ) 207 ×10−6 ( C° ) (25 °C − 15 °C) ( ) 1 atm = 4.6 × 106 Pa = 45 atm
5 1.01×10 Pa ______________________________________________________________________________ Chapter 12 Problems 649 40. REASONING The change ∆V in volume is given by Equation 12.3 as ∆V = βV0∆T, where β is the coefficient of volume expansion, V0 is the initial volume, and ∆T is the increase in
temperature. The increase in volume of the mercury is given directly by this equation, with
V0 being the initial volume of the interior space of the brass shell minus the initial volume of
the steel ball. If the space occupied by the mercury did not change with temperature, the
spillage would simply be the increase in volume of the mercury. However, the space
occupied by the mercury does change with temperature. Both the brass shell and the steel ball
expand. The interior volume of the brass shell expands as if it were solid brass, and this
expansion provides more space for the mercury to occupy, thereby reducing the amount of
spillage. The expansion of the steel ball, in contrast, takes up space that would otherwise be
occupied by mercury, thereby increasing the amount of spillage. The total spillage, therefore,
is ∆VMercury − ∆VBrass + ∆VSteel.
SOLUTION Table 12.1 gives the coefficients of volume expansion for mercury, brass, and
steel. Applying Equation 12.3 to the mercury, the brass cavity, and the steel ball, we have Spillage = ∆VMercury − ∆VBrass + ∆VSteel
= β MercuryV0, Mercury ∆T − β BrassV0, Brass ∆T + βSteelV0, Steel ∆T
−1 ( )( ) = 182 ×10−6 ( C° ) 1.60 ×10−3 m3 − 0.70 × 10−3 m3 (12 C° ) 1444444444444
4
2444444444444
4
3
Mercury
−1 ( ) − 57 ×10−6 ( C° ) 1.60 × 10−3 m3 (12 C° ) 4
14444444 244444444
3
Brass
−1 ( ) + 36 × 10−6 ( C° ) 0.70 ×10−3 m3 (12 C° ) = 1.2 ×10−6 m3 4
14444444 244444444
3
Steel 41. SSM WWW REASONING The cavity that contains the liquid in either Pyrex
thermometer expands according to Equation 12.3, ∆ Vg = βgV0 ∆T . On the other hand, the volume of mercury expands by an amount ∆ Vm = β mV0 ∆T , while the volume of alcohol
expands by an amount ∆ Va = β a V0 ∆T . Therefore, the net change in volume for the mercury
thermometer is ∆ Vm − ∆Vg = (βm − βg )V0 ∆T 650 TEMPERATURE AND HEAT while...
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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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