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Heat must also be added to boil water into steam. Following the same type of reasoning as
in the case of melting, we conclude that the steam has the greater mass.
The amount of heat Q that must be supplied to change the phase of m kilograms of a
substance is given by Equation 12.5 as Q = mL, where L is the latent heat of the substance.
Since the latent heat of vaporization Lv for water is greater than the latent heat of fusion Lf
(see Table 12.3), the change in mass is greater when liquid water turns into steam at 100 °C
than when ice turns into liquid water at 0 °C.
SOLUTION The change in mass ∆m associated with a change in rest energy ∆E0 is given
by Equation 28.5 as ∆m = ∆E0/c2. The change in rest energy is the heat Q that must be
added to change the phase of the water, so that Q = mL. Thus, the change in mass is
∆m = Q/c2 = mL /c2.
a. According to Table 12.3, the latent heat of fusion for water is Lf = 3.35 × 105 J/kg. The
change in mass associated with the ice to liquid-water phase change at 0 °C is, then, ∆m = mLf
c 2 = ( 2.00 kg ) ( 3.35 × 105 J/kg ) (3.00 × 10 8 m/s ) 2 = 7.44 × 10−12 kg b. According to Table 12.3, the latent heat of vaporization for water is Lf = 2.26 × 106 J/kg.
The change in mass associated with the liquid-water to steam phase change at 100 °C is
∆m = mLv
c 2 = ( 2.00 kg ) ( 2.26 × 106 J/kg ) (3.00 × 108 m/s ) 2 = 5.02 × 10−11 kg Chapter 28 Problems 1495 As expected, the change in mass for the liquid-to-steam phase change is greater than that for
the ice-to-liquid phase change.
29. SSM REASONING According to the work-energy theorem, Equation 6.3, the work that
must be done on the electron to accelerate it from rest to a speed of 0.990c is equal to the
kinetic energy of the electron when it is moving at 0.990c.
SOLUTION Using Equation 28.6, we find that 1 2
–1 KE = mc 1 – (v 2 / c 2 ) 1 8
–1 = 5.0 × 10 –13 J
kg)(3.00 × 10 m/s) 1 – (0.990c)2 / c 2 ______________________________________________________________________________
= (9.11 × 10 –31 30. REASONING AND SOLUTION The energy E0 produced in one year is the product of the
power P generated and the time t, E0 = P t. This energy is equivalent to an amount of mass
m given by Equation 28.5 as E0 = mc2. Thus, we have that mc 2 = P t or m= Pt
7 The mass of nuclear fuel consumed in one year (3.15 × 10 s) is P t ( 3.0 × 109 W ) ( 3.15 × 107 s )
m= 2 =
= 1.1 kg
( 3.00 × 108 m/s )
31. REASONING The rate R that the quasar is losing mass is equal to the mass m it loses
divided by the time t during which the loss occurs, or R = m / t . The mass that the quasar
loses is equivalent to a certain amount of energy E0; this equivalency is expressed by
m = E0 / c 2 (Equation 28.5), where c is the speed of light. According to Equation 6.10b, the energy radiated is equal to the average power P times the time, E0 = Pt. The average
power is the rate at which the quasar radiates energy. 1496 SPECIAL RELATIVITY SOLUTION Combining R = m / t and m = E0 / c 2 , we find that E0
ct Since the energy radiated by the quasar is E0 = Pt , the rate at which the quasar loses mass
can be written as
1 E0 1 P t P
1.0 × 1041 W
= 1.1× 1024 kg/s 2 t 2
c ct c
( 3.00 ×108 m/s )
R= 32. REASONING The total relativistic energy E is related to the rest energy E0 = mc2 and the
speed v according to Equation 28.4:
mc 2 E= v2
c = E0 (28.4) v2
c A greater total energy E does not necessarily mean that an object has a greater speed. It is
not the total energy alone that matters, but the ratio E/E0 of the total energy to the rest
energy. According to Equation 28.4 this ratio is E
E0 1 (1) v2
c Larger values for this ratio mean that the speed v is greater. When the speed is greater,
square root term in the denominator on the right-hand side of Equation (1) is smaller, so the
reciprocal of the square root term is larger.
By considering the ratio given in Equation (1), we can rank the speeds of the objects. This
ratio is listed in the following table:
Object Total Energy (E) Rest Energy (E0) E/E0 A 2.00ε ε 2.00 B 3.00ε ε 3.00 C 3.00ε 2.00ε 1.50 Chapter 28 Problems 1497 The ratio E/E0 is greatest for object B and smallest for object C. Thus, the ranking of the
B (largest), A, C
SOLUTION Solving Equation (1) for the speed v, we obtain
E0 2 1
c2 or 1
2 0 1− v
c2 or 1 ( E / E0 ) 2 = 1− v2
c 2 or v = c 1− 1 ( E / E0 )2 Applying this result to each object, we find
vA = c 1 − vB = c 1 − vC = c 1 − 1 ( ) 2
E / E0
A 1 ( ) 2
E / E0
( E / E0 )C = c 1− 1
2.002 = c 1− 1
3.002 = c 1− 1
1.502 These results are consistent with our expected ranking. In particular, note that the largest
total energy for object C does not imply that its speed is th...
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