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Unformatted text preview: t given by Q = cm∆T
(Equation 12.4), where c is the specific heat of water (see Table 12.2), and ∆T is 37 C°, the
difference between body temperature (37 °C) and the temperature of ice water (0 °C) . We
will use Equation 12.4 to calculate the required mass m of ice water, and Equation 11.1 to
find the corresponding volume V. SOLUTION The volume V of the water is
V= m ρ (11.1) Solving Q = cm∆T (Equation 12.4) for the mass m, we obtain
m= Q
c ∆T (1) Chapter 12 Problems 653 Substituting Equation (1) into Equation 11.1 yields
Q
V = c∆T = ρ Q
ρ c∆T (2) The amount Q of heat is given as 430 kcal, which must be converted to joules via the
equivalence 1 kcal = 4186 J: 4186 J 6
Q = 430 kcal = 1.8 × 10 J 1 kcal ( ) Equation (2), then, gives the volume of ice water in m3:
V= Q
1.8 ×106 J
=
= 0.012 m3
ρ c∆T 1.0 ×103 kg/m3 4186 J/ kg ⋅ Co 37 Co ( ) ( )( ) To convert this result to liters, we use the equivalence 1 liter = 1.0 ×10−3 m3 : ( ) 1 liter
V = 0.012 m3 = 12 liters 1.0 ×10−3 m3 47. REASONING The metabolic processes occurring in the person’s body produce the heat that
is added to the water. As a result, the temperature of the water increases. The heat Q that
must be supplied to increase 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 increase ∆T
in temperature is the final higher temperature Tf minus the initial lower temperature T0.
Hence, we will solve Equation 12.4 for the desired final temperature.
SOLUTION From Equation 12.4, we have
Q = cm∆T = cm (Tf − T0 ) Solving for the final temperature, noting that the heat is Q = (3.0 × 105 J/h)(0.50 h) and
taking the specific heat capacity of water from Table 12.2, we obtain ( )
( 3.0 × 105 J/h ( 0.50 h )
Q
Tf = T0 +
= 21.00 °C +
= 21.03 °C
cm 4186 J/ ( kg ⋅ C° ) 1.2 × 103 kg ) 654 TEMPERATURE AND HEAT 48. REASONING The change ∆T in temperature is determined by the amount Q of heat added,
the specific heat capacity c and mass m of the material, according to Q = cm∆T (Equation
12.4). The heat supplied to each bar and the mass of each bar are the same, but the changes
in temperature are different. The only factor that can account for the different temperature
changes is the specific heat capacities, which must be different. We will apply
Equation 12.4 to each bar and thereby determine the unknown specific heat capacity.
SOLUTION The heat supplied to each bar is given by Q = cm∆T (Equation 12.4). The
amount of heat QG supplied to the glass is equal to the heat QS supplied to the other
substance. Thus,
QG = QS or cG m ∆TG = cS m ∆TS (1) We know that cG = 840 J/(kg⋅C°) from Table 12.2. Solving Equation (1) for cS and
eliminating the mass m algebraically, we obtain ∆T
cS = cG G ∆T
S 88 °C – 25 °C = [840 J/(kg ⋅ C°)] = 235 J/(kg ⋅ C°) 250.0 °C – 25 °C ______________________________________________________________________________
49. SSM REASONING Let the system be comprised only of the metal forging and the oil.
Then, according to the principle of energy conservation, the heat lost by the forging equals
the heat gained by the oil, or Qmetal = Qoil . According to Equation 12.4, the heat lost by the
forging is Qmetal = cmetal mmetal (T0metal − Teq ) , where Teq is the final temperature of the
system at thermal equilibrium. Similarly, the heat gained by the oil is given by
Qoil = coil moil( Teq − T0oil ) .
SOLUTION Qmetal = Qoil
c metal mmetal ( T0metal − Teq ) = c oil moil ( Teq − T0oil ) Solving for T0metal , we have T0metal =
or c oil moil ( Teq − T0oil )
c metal mmetal + Teq [2700 J/(kg ⋅ C°)](710 kg)(47 °C − 32 °C)
+ 47 °C = 940 °C
[430 J/(kg ⋅ C°)](75 kg)
______________________________________________________________________________
T0metal = Chapter 12 Problems 655 50. REASONING Since the container of the glass and the liquid is being ignored and since we
are assuming negligible heat exchange with the environment, the principle of conservation of
energy applies. In reaching equilibrium the cooler liquid gains heat, and the hotter glass loses
heat. We will apply this principle by equating the heat gained to the heat lost. 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.
SOLUTION Applying the energyconservation principle and using Equation 12.4 give
cLiquid m∆TLiquid = cGlass m∆TGlass
14 244
4
3
144
244
3
Heat gained by liquid Heat lost by glass Since it is the same for both the glass and the liquid, the mass m can be eliminated
algebraically from this equation. Solving for cLiquid and taking the specific heat capacity for
glass from Table 12.2, we find cLiquid = cGlass...
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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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