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Unformatted text preview: AÂ° ( 70.0 AÂ° ) The temperature on the B scale is, therefore, T = + 20.0 Â°B + 85.4 BÂ° = 105.4 Â°B
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5. SSM REASONING AND SOLUTION
a. The Kelvin temperature and the temperature on the Celsius scale are related by
Equation 12.1: T = Tc + 273.15, where T is the Kelvin temperature and Tc is the Celsius
temperature. Therefore, a temperature of 77 K on the Celsius scale is
Tc = T â€“ 273.15 = 77 K â€“ 273.15 K = â€“196 Â° C b. The temperature of â€“196 Â°C is 196 Celsius degrees below the ice point of 0 Â°C. Since
9
1 CÂ° = 5 FÂ° , this number of Celsius degrees corresponds to 9 FÂ° = 353 FÂ°
196 CÂ° 5 1 CÂ° Subtracting 353 Fahrenheit degrees from the ice point of 32.0 Â°F on the Fahrenheit scale
gives a Fahrenheit temperature of â€“321 Â° F .
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6. REASONING AND SOLUTION The difference between these two averages, expressed in
Fahrenheit degrees, is 98.6 Â°F âˆ’ 98.2 Â°F = 0.4 FÂ°
Since 1 CÂ° is equal to 9
FÂ°, we can make the following conversion
5 1 CÂ° = 0.2 CÂ°
0.4 FÂ° 9 FÂ° 5 ______________________________________________________________________________ 7. REASONING The relationship between the pressure P and temperature T of a gas kept at a
constant volume is illustrated by the pressureversustemperature graph in Figure 12.4. The
pressures and temperatures at any two points on this graph can be used to calculate the slope
âˆ†P/âˆ†T of the graph. If we choose one of the points to be the absolute zero of temperature
measurement, then the slope can be expressed as Chapter 12 Problems âˆ†P P âˆ’ P0
=
âˆ†T T âˆ’ T0 631 (1) where P and T are the pressure and temperature, respectively, of any given point on the
graph, and P0 = 0 Pa, T0 = âˆ’273.15Â° correspond to the pressure and temperature of gas at the
absolute zero of temperature measurement. We will use the given data in Equation (1) to
determine the slope of the graph from the temperature T1 = 0.00 Â°C at the higher pressure
P1 = 5.00Ã—105 Pa. Once the slope âˆ†P/âˆ†T is known, we can use Equation (1) to calculate the
temperature T2 when the gas is at the lower pressure P2 = 2.00Ã—105 Pa.
SOLUTION Substituting P0 = 0 Pa into Equation (1) yields âˆ†P P âˆ’ ( 0 Pa )
P
=
=
âˆ†T
T âˆ’ T0
T âˆ’ T0 (2) The slope is the same, no matter which pair of points is used to calculate it, so from
Equation (2) we have that P
P2
âˆ†P
1
=
=
âˆ†T T1 âˆ’ T0 T2 âˆ’ T0 (3) Solving Equation (3) for T2, we obtain P (T2 âˆ’ T0 ) = P2 (T1 âˆ’ T0 )
1 or T2 âˆ’ T0 = P2
P
1 (T1 âˆ’ T0 ) or T2 = P2
P
1 (T1 âˆ’ T0 ) + T0 Therefore, the temperature T2 is T2 = 8. ( 2.00 Ã—105 Pa ) o
o
o
0.00 C âˆ’ ( âˆ’273.15 C ) + ( âˆ’273.15 C ) = ( 5.00 Ã—105 Pa ) âˆ’164 oC REASONING The absolute zero point on the Kelvin scale occurs at âˆ’273.15 Â°C. This
temperature is below the methane freezing point of âˆ’182.6 Â°C by an amount that is 90.6 CÂ°
[âˆ’182.6 Â°C âˆ’ (âˆ’273.15 Â°C)]. Before we can locate the absolute zero point on the Methane
scale, we must convert 90.6 CÂ° into an equivalent number of Methane degrees (MÂ°). To
determine the equivalence between Methane degrees and Celsius degrees, we will calculate
the temperature difference between the freezing and boiling points of methane on Titan in
Celsius degrees, and equate this to 100 MÂ°, which is the difference between the two points
on the Methane scale. 632 TEMPERATURE AND HEAT SOLUTION The temperature difference between the freezing and boiling points of
methane on Titan is, by definition, 100 MÂ°. The freezing point and boiling point, in degrees
Celsius, are âˆ’182.6 Â°C and âˆ’155.2 Â°C, respectively. Thus, we have the equivalence between
Celsius degrees and Methane degrees: ( ) 100.0 Mo = âˆ’155.2 o C âˆ’ âˆ’182.6 o C = 27.4 Co (1) Therefore, if the absolute zero point on the Kelvin scale is 90.6 CÂ° below methaneâ€™s
freezing point on the Methane scale, the difference in Methane degrees is ( ) 100.0 Mo = 331 Mo
90.6 Co 27.4 Co If the freezing point of methane is 0 Â°M, and the absolute zero point of the Kelvin scale is
331 MÂ° below this, then 0 K corresponds to âˆ’331 Â°M 9. SSM REASONING AND SOLUTION The Rankine and Fahrenheit degrees are the
same size, since the difference between the steam point and ice point temperatures is the
same for both. The difference in the ice points of the two scales is 491.67 â€“ 32.00 = 459.67.
To get Rankine from Fahrenheit this amount must be added, so TR = TF + 459.67 . ______________________________________________________________________________
10. REASONING The change in temperature is the final temperature T minus the initial
temperature T0, or âˆ†T = T âˆ’ T0 . Thus, T = T0 + âˆ†T . When the bat is heated, its length
changes by an amount given by Equation 12.2 as âˆ†L = Î± L0 âˆ†T , where Î± is the coefficient
of linear expansion, and L0 is the batâ€™s initial length. Solving this expression for âˆ†T and
substituting the result into T = T0 + âˆ†T will allow us to find the final temperature of the bat.
SOLUTION Solving âˆ†L = Î± L0 âˆ...
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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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