Monday, September 26, 2011

Wiley Plus Chapter 18

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4. At what temperature (degrees Celsius) is the Fahrenheit scale reading equal to (a) 4 times that of the Celsius and (b) 1/5 times that of the Celsius?

12.An aluminum-alloy rod has a length of 8.3939 cm at 13.000°C and a length of 8.4047 cm at the boiling point of water. (a) What is the length (in cm) of the rod at the freezing point of water? (b) What is the temperature (in Celsius) if the length of the rod is 8.4384 cm? Give your answers to five significant figures.

27.Calculate the minimum amount of energy, in joules, required to completely melt 228 g of silver initially at 18.5°C. The melting point of silver is at 962°C. Its specific heat capacity is 236 J/kg·K and its latent heat of fusion is 105 kJ/kg.

43. A sample of gas expands from 1.0 m3 to 4.0 m3 while its pressure decreases from 40 Pa to 10 Pa. How much work is done by the gas if its pressure changes with volume via (a) path A, (b) path B, and (c) path C in Figure 18-36?

53.Consider a slab of face area A and thickness L. Suppose that L = 17 cm, A = 68 cm2, and the material is copper. If the faces of the slab are maintained at temperatures TH = 123°C and TC = 38°C, and a steady state is reached, find the conduction rate through the slab. The thermal conductivity of copper is 401 W/m·K.

ANSWERS
 5. (a) Let the reading on the Celsius scale be x and the reading on the Fahrenheit scale
be y. Then y95x32. If we require y = 2x, then we have 9 2x x 32         x (5)(32) 160 C

which yields y = 2x = 320°F.

(b) In this case, we require y12x and find

1 9 (10)(32)
x x 32         x 24.6 C
2 5 13

which yields y = x/2 = –12.3°F.

12. (a) The coefficient of linear expansion   for the alloy is

L 10.015cm 10.000cm 5
1.88 10 /C . 
LT (10.01cm)(100 C 20.000 C)

Thus, from 100°C to 0°C we have

52
L L T (10.015cm)(1.88 10 /C )(0 C 100 C) =  1.88   10 cm. 

The length at 0°C is therefore L  = L +  L = (10.015 cm – 0.0188 cm) = 9.996 cm.

(b) Let the temperature be Tx. Then from 20°C to Tx we have

5
L 10.009cm  10.000cm =  L T (1.88 10 /C )(10.000cm) T, 

                  giving  T = 48 °C. Thus, Tx = (20°C + 48 °C )= 68°C.





1
 


27. The melting point of silver is 1235 K, so the temperature of the silver must first
be raised from 15.0° C (= 288 K) to 1235 K. This requires heat

4
Q cm(TfTi) (236J/kg K)(0.130kg)(1235 C 288 C) 2.91 10 J. 

Now  the silver at  its melting point must be melted.  If LF  is  the heat of  fusion  for
silver, this requires
34
Q mLF 0.130kg 105 10 J/kg 1.36 10 J. 

The total heat required is ( 2.91   104 J + 1.36   104 J ) = 4.27   104 J.


43. (a) One part of path A represents a constant pressure process. The volume changes
from 1.0 m3 to 4.0 m3 while the pressure remains at 40 Pa. The work done is

3 3 2
WA p V (40Pa)(4.0m 1.0m ) 1.2 10 J.

(b) The other part of the path represents a constant volume process. No work is done
during this process. The total work done over the entire path is 120 J. To find the
work done over path B we need to know the pressure as a function of volume. Then,
we can evaluate the integral W =   p dV. According to the graph, the pressure is a
linear  function  of  the  volume,  so we may write  p =  a  +  bV, where  a  and  b  are
constants. In order for the pressure to be 40 Pa when the volume is 1.0 m3 and 10 Pa
when the volume is 4.00 m3 the values of the constants must be a = 50 Pa and b = –10
Pa/m3. Thus, 

p = 50 Pa – (10 Pa/m3)V
and

44
24
WB p dV 50 10V dV 50V 5V 200J 50J 80J  +  5.0J = 75J. 
11 1

(c) One part of path C represents a constant pressure process in which the volume
changes from 1.0 m3 to 4.0 m3 while p remains at 10 Pa. The work done is

33
WC p V  (10Pa)(4.0m 1.0m ) 30J. 

2
 
The other part of the process is at constant volume and no work is done. The total
work is 30 J. We note that the work is different for different paths.

          53. The rate of heat flow is given by

TT
P kAHC, 
cond
L

where k is the thermal conductivity of copper (401 W/m·K), A is the cross-sectional
area (in a plane perpendicular to the flow), L is the distance along the direction of
flow between the points where the temperature is TH and TC. Thus,

42
401W/m K 90.0 10 m 125 C 10.0 C
3
P 1.66 10 J/s. 
cond
0.250m

The thermal conductivity is found in Table 18-6 of the text. Recall that a change in
Kelvin temperature is numerically equivalent to a change on the Celsius scale.
 
  

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