# PHYS 1401 General College Physics I

```PHYS 1401 General College Physics I
(43172 - MTWR 5:30-7:50 PM)
Home Work 12:
(pages 366-369)
1, 5, 9, 13, 15, 29, 35, 43, 57, 67
1.
The difference between these two averages, expressed in Fahrenheit degrees, is
98.6 F  98.2 F  0.4 F
Since 1 C&deg; is equal to
9
F&deg;, we can make the following conversion
5
 1 C 
(0.4 F) 
 0.2 C
(9/5) F
5.
The temperature of –273.15 &deg;C is 273.15 Celsius degrees below the ice point of 0
&deg;C. This number of Celsius degrees corresponds to
(9/5) F 
(273.15 C) 
 491.67 F
1 C 
Subtracting 491.67 F&deg; from the ice point of 32.00 &deg;F on the Fahrenheit scale gives
a Fahrenheit temperature of –459.67 F
9.
Using the value for the coefficient of thermal expansion of steel given in Table
12.1, we find that the linear expansion of the aircraft carrier is

L = L0 T = (12  10–6 C–1)(370 m)(21 &deg;C  2.0 &deg;C)  0.084 m
(12.2)
13.
The value for the coefficient of thermal expansion of steel is given in Table 12.1.
The relation, L =  L0T, written in terms of the diameter d of the rod, is
d
0.0026 cm
T 

 110 C
(12.2)
6
 d0
12  10 C 1 2.0026 cm 


15.
Assuming that the rod expands linearly with heat, we first calculate the quantity
L/ T using the data given in the problem.
L
8.47  10–4 m

 1.13  10 –5 m/C
T 100.0 C – 25.0 C
Therefore, when the rod is cooled from 25.0 &deg;C, it will shrink by
L  (1.13  10 –5 m/C ) T
 (1.13  10 –5 m/C ) (0.00 C – 25.0 C)  –2.82  10 –4 m
29.
According to Equation 12.3, V  VT . Taking the coefficient of volumetric
expansion  for water from Table 12.1, we find that
Vair
 V T

3.7  10–3 (C)–1
 air 0
 air 
 18
Vwater water V0 T water 207  10 –6 (C)–1
35.
In order to keep the water from expanding as its temperature increases from 15 to
25 &deg;C, the atmospheric pressure must be increased to compress the water as it tries to
expand. The magnitude of the pressure change P needed to compress a substance by
an amount V is, according to Equation 10.20, P  B(V / V0 ) . The ratio V / V0 is,
according to Equation 12.3, V / V0  T . Combining these two equations yields
P  B T
SOLUTION Taking the value for the coefficient of volumetric expansion  for
water from Table 12.1, we find that the change in atmospheric pressure that is required to
keep the water from expanding is
1
P  (2.2  109 N/m 2 )  207 106  C   (25 C  15 C)




1 atm


 4.6 106 Pa 
  45 atm
5
 1.0110 Pa 

43. 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 &times; 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

57.
The heat required is Q = mLf + cmT, where m = V. See Table 12.2 for the
specific heat c and Table 12.3 for the latent heat Lf. Thus,
Q = VLf + cVT
Q = (917 kg/m3)(4.50  10–4 m)(1.25 m2){33.5  104 J/kg
+ [2.00  103 J/(kgC&deg;)](12.0 C&deg;)}  1.85 10 5 J
67.
From inspection of the graph that accompanies this problem, a pressure of
3.5 106 Pa corresponds to a temperature of 0 &deg;C. Therefore, liquid carbon dioxide
exists in equilibrium with its vapor phase at 0 C when the vapor pressure is
3.5 106 Pa .
```