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					HW Assignment 36: Chap 10: CQ 10, Problems 30, 40, 50                             40 pts

Conceptual Question:
CQ 10:10
After food is cooked in a pressure cooker, why is it very important to cool the container with
cold water before attempting to remove the lid?

Answer:
At high temperature and pressure, the steam inside exerts large forces on the pot and cover.
Strong latches hold them together, but they would explode apart if you tried to open the hot
cooker.

Problems:
Problem 10:30:
A tank having a volume of .1 m3 contains helium gas at 150 atm. How many balloons can the
tank blow up if each filled balloon is a sphere 0.300 m in diameter at an absolute pressure of 1.2
atm?

Solution:
                                                            4 3
       The volume of helium in each balloon is Vb             r.
                                                             3

       The total volume of the helium at P2  1. at will be
                                               20 m

                     P         150 at 
               V 2   1  V1  
                      P2 
                                        m
                                 1. at 
                                   20 m 
                                            
                                           0. m
                                            100       3
                                                            12. m
                                                                5     3




       Thus, the number of balloons that can be filled is

                     V2       12. m 3
                                   5
               N                               884 baloons
                                                         l
                     V b  4 3  0. m
                                     150   3
Problem 10:40:
The temperature near the top of the atmosphere on Venus is 240 K. a) find the rms speed of
hydrogen (H2) at that point in Venus's atmosphere. b) Repeat for carbon dioxide c) It has been
found that if the rms speed exceeds 1/6 the planet's escape velocity, the gas eventually leaks out
of the atmosphere and into outer space. If the escape velocity on Venus is 10.3 km/s, does
hydrogen escape? Does carbon dioxide escape?
Solution:
                  1             3
From KEm olecule  m v2  kBT
                  2             2
                                          3kB T
the rms speed of a molecule is vrm s  v2 
                                            m
                                   ar
                                m ol m ass M
The mass of the molecule is m            
                                    NA      NA
                                                2.  103 kg m ol
                                                  00
a)      For hydrogen  H 2  , m                                    3.  1027 kg
                                                                       32
                                              6.  10 m ol
                                               02    23
                                                              e
                                                          ecul m ol

                                     
                                    3 1.  1023 J K  240 K 
                                       38                          
                 
At T  240 K , vrm s
                       H2
                            
                                              3.  10
                                               32            27
                                                                   kg
                                                                                1. km s
                                                                                  73

                                                44.  103 kg m ol
                                                   0
(b)     For carbon dioxide, m                                         7.  1026 kg , and
                                                                         31
                                              6.  10 m ol
                                               02    23
                                                               e
                                                            ecul m ol

                                                                                      
                                                               3 1.  1023 J K  240 K 
                                                                  38
             at T  240 K ,                
                                          vrm s
                                                  CO 2
                                                         
                                                                        7.  10
                                                                         31      26
                                                                                       kg
                                                                                                0.
                                                                                                  369 km s


                               3vescape
                            10. km s
(c)     Since, on Venus,              1. km s , we should expect that
                                        71
                       6         6
 hydrogen w iles
             l cape butcarbon di de w ilnot . Indeed, it is found that carbon dioxide is
                                oxi    l
the predominant component in the atmosphere of Venus and hydrogen is present only in
combination with other elements.


Problem 10:50:
A vertical cylinder of cross sectional area 0.5 m2 if fitted with a tight fitting frictionless piston of
mass 5 kg. If there are 3 mol of an ideal gas in the cylinder at 500 K, determine the height h at
which the piston will be in equilibrium under its own weight.
Solution:
When gas the supports the piston in equilibrium, the gauge pressure of the gas is
         F m g  5. kg  9. m s 
                                     2
                      0       80
Pgauge                                9.  102 Pa , and the absolute pressure is
                                           8
         A     A         0. m 2
                          050
                  
P  Patm  Pgauge  1.  105  9.  102 Pa
                     013          8                      
The ideal gas law gives the volume as V  nRT P , so the height of the cylindrical space is
             V nRT      3. m ol  8. J m ol K   500 K 
                          0      31
        h                                                 2. m
                                                               4
             A PA          
                     1.  10 Pa  9.  102 Pa 0. m 2
                      013     5
                                      8               050                                
HW Assignment 37: Chap 11: CQ 16 and Prob. 6, 18, 23                                       40 pts

CQ 11:16
Energy is added to ice, raising its temperature from -10 to 5 degrees C. A larger amount of
energy is added to the same mass of liquid water, raising its temperature from 15 to 20 degrees
C. From these results, we can conclude that a) overcoming the latent heat of fusion of ice
requires an input of energy b) the latent heat of fusion of ice delivers some energy to the system,
c) the specific heat of ice is less than that of water d) the specific heat of ice is greater than that
of water.

Answer:
c) the specific heat of ice is less than that of water.


Problem 11:6
As part of an exercise routine a 50 kg person climbs 10 m up a vertical rope. How many food
Calories are expended in a single climb up the rope? (1 food Calorie = 1000 calories).

Solution:
The internal energy converted to mechanical energy in one ascent of the rope is
            Q  PEg  m gh . Since 1 Cal i  1000 cal i  4186 J es ,
                                        or e         or es         oul


                                                       1 Cal i 
                                                            or e
                   Q   50. kg  9. m s2  10. m  
                           0        80          0                  1. Cal i
                                                                      17  or e
                                                       4186 J 


Problem 11:18
When a driver brakes an automobile the friction between brake drums and the brake shoes
converts the car's kinetic energy to thermal energy. If a 1500 kg automobile traveling at 30 m/s
comes to a halt, how much does the temperature rise in each of the four 8 kg iron brake drums?
(specific heat of iron is 448 J/kg-C)

Solution:
The kinetic energy given up by the car is absorbed as internal energy by the four brake drums (a
total mass of 32 kg of iron). Thus, KE  Q  m drum scFe  T  or


                            2 1500 kg 30 m s    47C
                                               2
                      2     1
               m carvi
               1
        T    2
                         
             m drum scFe    32 kg 448 J kg  °C 
Problem 11:23
What mass of steam that is initially at 120 degrees C is needed to warm 350 g of water and its
300 g aluminum container from 20 to 50 degrees C?

Solution:
In order to come to equilibrium at 50°C, the steam must: cool to 100°C, condense, and then cool
       (as condensed water) to 50°C. Thus, the conservation of energy equation is

               m steam cteam 120C  100C   Lv  c 100C  50C 
                        s                             w               
                                                                    
                                                     m w c  m cupc l  50C- C 
                                                           w        A         20



       or      m steam 
                             m                 
                                   c  m cupcA l  30C 
                                  w w
                                                                .
                           csteam  20C   Lv  cw  50C 

       This gives

                               0. kg  4186 J kg  C    0. kg  900 J kg  C    30C 
                               350                               300                      
               m steam                                                                                 ,
                            2010 J kg  C   20C    2.  10 J kg    4186 J kg  C   50C 
                                                            26    6




       and     m steam  2.  102 kg  21 g
                          1
HW Assignment 38: Chap 11: CQ 12 and Prob. 36, 42, 66                                          40 pts

CQ 11:12
You need to pick up a very hot cooling pot in your kitchen. You have a pair of hot pads.
Should you soak them in cold water or keep them dry in order to pick up the pot most
comfortably?

Answer:
Keep them dry. The air pockets in the pad conduct energy slowly. Wet pads absorb some
    energy in warming up themselves, but the pot would still be hot and the water would
    quickly conduct a lot of energy to your hand.



Problem 11:36
A box with a total surface area of 1.2 sq. m and a wall thickness of 4 cm is made of insulating
material. A 10 W electric heater inside the box maintains the inside temperature at 15 degrees
C above the outside temperature. Find the thermal conductivity k of the insulating material.

Solution:
Since the air temperature inside the box remains constant, the power input from the heater
                                                                           T 
             must equal the energy transfer to the exterior. Thus, Ã  kA       , giving
                                                                           L 

                    Ã  L  10. W   4.  102 m 
                               0        00                    2
               k                                 2.  10 W m  C
                                                        22
                             20  2 
                    A  T  1. m      15. C
                                           0       




Problem 11:42
The surface temperature of the Sun is about 5800 K. Taking the radius of the Sun to be 6.96 x
108 m, calculate the total energy radiated by the Sun each second. Assume e = 0.965.


Solution:
With an emissivity of e 0. , temperature of T  5800 K , and radius of r 6.  108 m , the
                           965                                              96
       total power radiated by the spherical Sun is

                                                
                                            W
                                                                    
                                                                     
                                                   4 6.96  10 m   0.965  5 800 K 
                                                                   2
               P   AeT 4   5.67  108 2 4                   8                         4

                                         m K                     


       or      P  3.77 1026 W
Problem 11:66
A wood stove is used to heat a single room. The stove is cylindrical in shape with a diameter of
40 cm and a length of 50 cm and operates at a temperature of 400 F. a) If the temperature of the
room is 70 F determine the amount of radiant energy delivered to the room by the stove each
second if the emissivity is 0.920. b) If the room is square with walls that are 8 ft high and 25
feet long, determine the R value needed in the walls and ceiling to maintain the inside
temperature at 70 F if the outside temperature is 32 F.

Solution:
(a)    The surface area of the stove is A stove  A ends  A cylindrical  2 r2    2 r  , or
                                                                                            h
                                                                          i
                                                                         s de

                  A stove  2  0. m              2  0. m    0. m   0. m 2
                                               2
                                  200                      200       500      880

                                                                 5
             The temperature of the stove is Ts                    400°F  32. F  204C  477 K while that
                                                                               0
                                                                 9
                                         5
             of the air in the room is Tr 70. F  32. F  21. C  294 K . If the emissivity of
                                              0        0         1
                                         9
             the stove is e 0. , the net power radiated to the room is
                              920

                                
                  Ã   A stovee Ts4  Tr4   
                      
                     5.  108 W m 2  K 4 0. m
                       67                    880                 2
                                                                        0.   477 K 
                                                                           920 
                                                                                            4
                                                                                                  294 K  
                                                                                                           4
                                                                                                             

             or à  2.  103 W
                     03

       (b) The total surface area of the walls and ceiling of the room is
              A  4A wal  A ceii  48. f  25. f    25. f   1.  103 f 2
                                                                   2
                       l        lng   00 t      0 t         0 t      43       t
             If the temperature of the room is constant, the power lost by conduction through
             the walls and ceiling must equal the power radiated by the stove. Thus, from
             thermal conduction equation, Ã  A  Th  Tc  R i , the net R value needed in the
             walls and ceiling is


                  Ri 
                          A  Th  Tc 
                                          
                                            1.  10
                                              43           3
                                                            f 2   70. F  32. F  1054 J  1 h 
                                                             t          0°    0°
                                                                                                 
                               Ã                           2.  10 J s
                                                            03        3
                                                                                    1 Bt   3600 s
                                                                                         u


             or                                                             Ri  7. f 2 F  h Bt
                                                                                   78 t            u             in US units


                                                       2
                       ft 2  F  h  1m         1o C 3600s 1Btu           m2  C  s
or          Ri  7.78                          o                1.37             in SI units
                           Btu        3.281 ft  1.8 F  1h    1054 J             J

				
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