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âˆ†TLiquid 840 J/ ( kg â‹… CÂ° ) ( 83.0 CÂ° âˆ’ 53.0 CÂ° ) =
= 2500 J/ ( kg â‹… CÂ° )
( 53.0 CÂ° âˆ’ 43.0 CÂ°) 51. REASONING According to Equation 6.10b, the average power is the change in energy
divided by the time. The change in energy in this problem is the heat supplied to the water
and the coffee mug to raise their temperature from 15 to 100 ÂºC, which is the boiling point of
water. The time is given as three minutes (180 s). The heat Q that must be added to raise the
temperature of a substance of mass m by an amount âˆ†T is given by Equation 12.4 as
Q = cmâˆ†T, where c is the specific heat capacity. This equation will be used for the water and
the material of which the mug is made.
SOLUTION Using Equation 6.10b, we write the average power P as
P= Change in energy QWater + QMug
=
Time
Time The heats QWater and QMug each can be expressed with the aid of Equation 12.4, so that we
obtain 656 TEMPERATURE AND HEAT P= QWater + QMug
Time = cWater mWater âˆ†T + cMug mMug âˆ†T
Time 4186 J/ ( kg â‹… CÂ° ) ( 0.25 kg )(100.0 Â°C âˆ’ 15 Â°C ) = + 920 J/ ( kg â‹… CÂ° ) ( 0.35 kg )(100.0 Â°C âˆ’ 15 Â°C ) 180 s = 650 W The specific heat of water has been taken from Table 12.2. 52. REASONING Two portions of the same liquid that have the same mass, but different initial
temperatures, when mixed together, will yield an equilibrium mixture that has a temperature
lying exactly midway between the two initial temperatures. That is, the final temperature is
T = 1 (T0A + T0B ) . This occurs only when the mixing occurs without any exchange of heat
2
with the surroundings. Then, all of the heat lost from the warmer portion is gained by the
cooler portion. Since each portion is identical except for temperature, the warmer portion
cools down by the same number of degrees that the cooler portion warms up, yielding a
mixture whose final equilibrium temperature is midway between the two initial temperatures.
There are two ways to apply the logic above to the problem at hand. Portions A and B have
the same mass m, so they will yield a combined mass of 2m. Thus, we can imagine portions
A and B
mixed together to
yield an equilibrium temperature of
1T
1 94.0 Â°C + 78.0 Â°C = 86.0 Â°C . This mixture has a mass 2m, just like the
+ T0B ) = 2 (
)
2 ( 0A
mass of portion C, so when it is mixed with portion C, the final equilibrium temperature will
be 1 ( 86.0 Â°C + 34.0 Â°C ) = 60.0 Â°C . The other way to apply the logic is to mix half of portion
2
C with portion A and half with portion B. This will produce two mixtures with different
temperatures and each having a mass 2m, which can then be combined. The final equilibrium
temperature of this combination is again 60.0 ÂºC. Either way, the guiding principle is that the
heat gained by the cooler liquid is equal to the heat lost by the warmer liquid. SOLUTION Since we are assuming negligible heat exchange with the surroundings, the
principle of conservation of energy applies in the following form: heat lost equals heat
gained. In reaching equilibrium the warmer portions lose heat and the cooler portions gain
heat. In either case, the heat Q that must be supplied or removed to change the temperature of
a substance of mass m by an amount âˆ†T is given by Q = cmâˆ†T (Equation 12.4), where c is
the specific heat capacity. The final temperature is 50.0 ÂºC. We have, then, cC mC ( 50.0 Â°C âˆ’ 34.0 Â°C ) = cA mA ( 94.0 Â°C âˆ’ 50.0 Â°C ) + cB mB ( 78.0 Â°C âˆ’ 50.0 Â°C )
4
3
14444244443 1444444444 24444444444
Heat gained Heat lost The specific heat capacities for each portion have the same value c, while mA = mB = m. With
these substitutions, we find Chapter 12 Problems 657 cmC ( 50.0 Â°C âˆ’ 34.0 Â°C ) = cm ( 94.0 Â°C âˆ’ 50.0 Â°C ) + cm ( 78.0 Â°C âˆ’ 50.0 Â°C )
mC ( 50.0 Â°C âˆ’ 34.0 Â°C ) = m ( 94.0 Â°C âˆ’ 50.0 Â°C ) + m ( 78.0 Â°C âˆ’ 50.0 Â°C )
Solving this equation for the mass mC of portion C gives
mC = m ( 94.0 Â°C âˆ’ 50.0 Â°C ) + m ( 78.0 Â°C âˆ’ 50.0 Â°C )
50.0 Â°C âˆ’ 34.0 Â°C = 4.50 m 53. REASONING When heat Q is supplied to the silver bar, its temperature changes by an
amount âˆ†T. The relation between Q and âˆ†T is given by Equation 12.4 as Q = c m âˆ†T , where
c is the specific heat capacity of silver and m is the mass. Solving this equation for m yields
m= Q
c âˆ†T (1) When the temperature of the bar changes by an amount âˆ†T, the change âˆ†L in its length is
given by Equation 12.2 as âˆ†L = Î± L0 âˆ†T . Solving this equation for âˆ†T gives
âˆ†T = âˆ†L
Î± L0 (2) where L0 is the initial length and Î± is the coefficient of linear expansion for silver.
Substituting the expression for âˆ†T in Equation (2) into Equation (1) gives
m= Î± Q L0
Q
Q
=
=
c âˆ†T
c âˆ†L âˆ†L c Î± L0 SOLUTION Taking the values for Î± and c from Tables 12.1 and 12.2, respectively, the
mass of the silver bar is
m= Î± Q L0
c âˆ†L = 19 Ã— 10âˆ’6 ( CÂ° ) âˆ’1 ( 4200 J )( 0.15 m ) 235 J/ ( kg â‹… CÂ° ) ( 4.3 Ã— 10âˆ’3 m ) = 1.2 Ã— 10âˆ’2 kg ______________________________________________________________________________ 658 TEMPERATURE AND HEAT 54. REASONING Each second, the water heater increases the temperature of the water p...
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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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