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through from T0 = 11 °C to a final temperature T by transferring an amount of heat
Q = cm∆T (Equation 12.4), where m is the mass of the water heated in one second, ( c = 4186 J/ kg ⋅ Co ) is the specific heat of water (see Table 12.2 in the text), and ∆T = T − T0 is the difference between the higher and lower temperatures of the water. Solving Q = cm∆T for the temperature difference ∆T = Q
, we have that the final
cm temperature of the water is
T = T0 + ∆T = T0 + Q
cm (1) The power P of the water heater is 28 kW, or 28×103 J/s. This means that the water heater
increases the internal energy of the water by 28×103 J each second. Therefore, the maximum
amount of heat the water can absorb in one second is Q = 28×103 J. The last quantity in
Equation (1) that we must find is the mass m of water passing through the heater in one
second. We will determine the value of m from the volume flow rate, the number of showers
in simultaneous use, and the density of water.
SOLUTION There are four showers in operation simultaneously, so the total volume flow
rate of water through the water heater is ( ) Volume flow rate = 4 14 ×10−5 m3 /s = 56 ×10−5 m3/s (2) Therefore, the heater must heat a volume V = 56 ×10−5 m3 of water each second. The
corresponding mass m of the water can be found from m = ρV (Equation 11.1) by using a
value of ρ = 1.00×103 kg/m3 for the density of water (see Table 11.1 in the text).
Substituting Equation 11.1 into Equation (1), then, yields the maximum possible hot water
temperature: T = T0 + 55. Q
28 ×103 J
= 11 oC +
= 23 oC
c ρV 4186 J/ kg ⋅ Co 1.00 ×103 kg/m3 56 ×10−5 m3 ( )( )( ) SSM REASONING When heat Q is supplied to the block, its temperature changes by an
amount ∆T. The relation between Q and ∆T is given by
Q = c m ∆T (12.4) where c is the specific heat capacity and m is the mass. When the temperature of the block
changes by an amount ∆T, the change ∆V in its volume is given by Equation 12.3 as Chapter 12 Problems 659 ∆V = β V0 ∆T , where β is the coefficient of volume expansion and V0 is the initial volume of the block. Solving for ∆T gives ∆T = ∆V
β V0 Substituting this expression for ∆T into Equation 12.4 gives ∆V Q = c m ∆T = c m β V0 SOLUTION The heat supplied to the block is ∆V 750 J/ ( kg ⋅ C° ) (130 kg ) (1.2 ×10 m )
Q = cm
= 4.0 × 105 J
=
β V0 6.4 ×10−5 ( C° )−1 ( 4.6 ×10−2 m3 ) ______________________________________________________________________________
−5 3 56. REASONING Because the container is insulated, no heat is transferred to the surroundings.
Therefore, in order to reach equilibrium at temperature Teq, the oil must absorb an amount of
heat Qoil equal to the heat Qwater given up by the water. Neither the water nor the oil
undergoes a phase change, so we will use Q = cm∆T (Equation 12.4) to determine the
amount of heat exchanged between the liquids. In Equation 12.4, c is the specific heat, m is
the mass and ∆T is the temperature difference that each liquid undergoes. Thus, we have Qoil = coil moil ∆Toil and Qwater = cwater mwater ∆Twater (1) For the water, the difference between the higher and lower temperatures is
∆Twater = 90.0 oC − Teq . With this substitution, the expression for Qwater becomes ( Qwater = cwater mwater 90.0 oC − Teq ) (2) For the oil, the temperature change ∆Toil is related to the increase ∆V in its volume by
∆V = βV0 ∆Toil (Equation 12.3), where β is the coefficient of volume expansion of the oil, and V0 is the volume of the oil before the water is added. Solving Equation 12.3 for ∆Toil,
we obtain
∆V
∆Toil =
(3)
βV0
Substituting Equation (3) into the first of Equations (1) yields 660 TEMPERATURE AND HEAT Qoil = coil moil ∆V (4) βV0 The mass moil of the oil in the container is related to its density ρ and volume V0 according
to ρ = moil V0 (Equation 11.1). Solving Equation 11.1 for the mass of the oil yields
moil = ρV0 , which we substitute into Equation (4): Qoil = coil ρ V0 ∆V β V0 = coil ρ ∆V (5) β SOLUTION Equation (5) gives the amount of heat absorbed by the oil, so it must be equal
to Equation (2), which gives the amount of heat lost by the water. Therefore, we have coil ρ ∆V β ( = cwater mwater 90.0 oC − Teq ) 90.0 oC − Teq = or coil ρ ∆V β cwater mwater (6) Solving Equation (6) for the equilibrium temperature, we obtain Teq = 90.0 oC − coil ρ ∆V β cwater mwater ( )( )( ) 1970 J/ kg ⋅ Co 924 kg/m3 1.20 ×10−5 m3 = 90.0 C − = 32.1 oC
721×10−6 (Co ) −1 4186 J/ kg ⋅ Co ( 0.125 kg ) o 57. ( ) SSM REASONING Heat Q1 must be added to raise the temperature of the aluminum in
its solid phase from 130 °C to its melting point at 660 °C. According to Equation 12.4,
Q1 = cm∆T . The specific heat c of aluminum is given in Table 12.2. Once the solid
aluminum is at its melting point, additional heat Q2 must be supplied to change its phase
from solid to liquid. The additional heat required to m...
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 Spring '13
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 Physics, The Lottery

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