QUIZ 15.5
Chapter 15
Turn in 18, 28, 30, 36, 42, 53, 59, 64, 66
Hints will be provided soon; Note that 18, 36 and 42 already have hints of sort. Click here for Q15. More #36 and #42 hints are below. 
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18. See class notes and hints by clicking here.
28. was added at the last minute; I could not resist  given the wide ramifications involved. It is more difficult and will be assigned as extra credit. You are asked to find the input heat of an engine placed in series with another engine. The heat from the first engine (1) supplies the input heat to  the second (2). Thus we have our first relationship in a list of several (marked with letters below)  equations  we will need to solve for Qin1 /t defined to be  the input heat rate for engine 1:  

Qout1 /t = Qin2/t      (A).    NOTE THE DIVISION BY THE TIME t. 

We are dividing all symbols by the time t to indicate we are seeking rates of heat.

We want Qin1 /t = W1/(t*e1 )   (B)
where e1 is engine1's efficiency and W1 is engine1's  work. Here we are using the definition of efficiency as the ratio of work to input heat.

NOTE: Qin2/ t=  W2/(t*e2)      (C)
where e2 is engine2's efficiency and W2 is engine2's  work.  Again,  we are using the definition of efficiency as the ratio of work to input heat.

Using (A) , (C) can be rewritten as

Qout1 /t = W2/(t*e2)   (D)


Now, Qout1 /t = Qin1 /t - W1/t  from conservation of energy for any engine: Qin1 /t = Qout1 /t + W1/t .

Thus, Qin1/t - W1 /t= W2/(t*e2)  (E)

Now,  we are given that:

W1/t + W2/t = 1100 x10 6  (W). Use the symbol a for this total output power. (W) means watts = J/s.

W1/t + W2/t = a.       (F)

Solve  (B), (E) and (F) for Qin1/t  in J/s.  Then  divide your answer by 2.8x107 J/kg to get the rate of coal burning in kg/s.
30.  This is a discussion of the same refrigerator problem even though the temperature inside the box is colder---- 0 degrees Celsius or 32 Fahrenheit  to be exact.

1. Remember , the cycle begins at the motor, with compression of cold low pressure gas to a higher temperature, much higher than the kitchen temperature TH outside the box. Thus, with the help of heat exchangers, the hot gas releases  QH heat into the lower temperature kitchen.. 

2. The  hot, high  temperature gas has now cooled after losing the above mentioned heat. This low temperature gas condenses into liquid. 

3. The air pressure above the gas is then reduced by way  of a valve to the outside and the liquid evaporates into an extremely  cold gas at temperature, much  lower than the temperature TL inside of the ice box--the heat flows from inside the ice box into the gas, keeping the ice box interior at a low temperature  through  removal of heat QL. 

4. The cycle begins again when the low pressure,  low temperature gas reaches step 1's  compressor .

Use Formula 15-6c, but use it wisely. You are solving for TL in Kelvin (K) . In this problem you must be consistent with these units, from  beginning to end.
36. This problem would require calculus-- in particular, integration--- if you did it exactly. But 2B's prerequisite is  only trigonometry and pre-calculus, so you must approximate. Turn your attention to example 15-15 -- we are doing something fundamentally the same here: Finding the final common equilibrium temperature , Tf, then finding  average temperatures between Tf and the initial temperatures. In example 15-15- you must find two averages , one for each of the components---hot water and cold water--- that were mixed and came to final temperature Tf. In this problem , you only have to compute one average--between the initial    and  final water temperature. Compute the average of 0 and 100 degree Celsius. Then plug that average into the denominator of formula 15-8. Now, the numerator, or the heat added Q, equals the heat required to raise the temperature from 0 to 100  Celsius. That formula is in chapter 14---equations 14-2 and requires the given mass m = 1.0 kg, specific heat of water  and   change in temperature. 

Note: As shown in the examples, you must convert to Kelvin after you find the average in degrees Celsius.
42. Now we are talking about a virtual carbon copy of example 15-15. Please read this helper carefully.

A. compute Tf using  chapter 14 methods. See example 14-4.

B. Compute the average between 30 degrees Celsius and Tf . 

C. Compute the average between  between 60 degrees Celsius and Tf.

d. Find the sum of the entropy changes as in example 15-15. Your answer should be positive

Note: As shown in the examples, you must convert to Kelvin after you find the average in degrees Celsius.

53. Use the same notation as in problem 28, but in a simpler context. This is a 4 cylinder engine. 

At the outset you should convert to J/s by  multiplying 220 J by  45 cycles per second .  Thus the rate of work (in J/s) is   220*45 J/s = 9.9x103 J/s = W/t.  Remember to multiply by 4 to take into account the cylinders.
Note: (b) can be answered immediately:  0.25 = (W/t) /(QH/t ) . Solve for QH/t .  
(a) See above comments. 
(c) Time = t = 35x10 6 J/ (QH/t ), where QH/t   is "hinted" above. 
59. Qin = 3.0x104 (in kcal/gal)  AND work = 25 hp *(746  W/ hp) =  1.865x104 W. Find the number of gallons per hour (gal/h) consumed by the car using the given speed (in km/h) and the mileage (in km/h)  . Once you find  this, you can convert Qin into kcal/h. Convert Qin into W (J/s) and then  you can find the efficiency by dividing the work by Qin.
64. Work = 100 hp * 746 W/hp = 7.46x104 W and e = 0.15 = efficiency.
(a) Use formula  15-5 for the Carnot efficiency ,  then find the requested ratio.
(b) From the work and efficiency find Qin ( in W) and then  find Qout = Qin - work ( in W) . Then multiply your answers by 3600 seconds. .
66. See example 15-8  and related text.  Remember ---when the human body does work, the internal energy remains constant when the energy is replenished by food.  

Think of your work "power stroke " with the following model:

The inside  of your body is represented  by an ideal gas  sealed in a container. On top of the container is a  piston that can move up or down, increasing or decreasing the volume.  Underneath the container is a burner with an open flame that allows heat to enter the gas. See figure 15.1, but add a bunsen burner below the container supplying energy to the gas. 

The flame  represents  input heat from food you metabolize , not aborption from the outside. (Remember your body temperature is greater than the outside, so heat cannot flow from the outside to inside your body.) The gas does work on the piston as it rises.  In order for the internal energy, and thus temperature, not to decrease, input heat must be supplied by the flame (food). The resulting process is an "isothermal expansion"  i.e.  expansion at constant temperature.

Suppose  I do work without any heat being added from food. I would lose internal energy if I did work. Fat, the food in this case,  would compensate for reductions in  internal energy due to my work. You can think of it this way: As the problem reads, if I metabolize  1.0 kg of fat, then I could do 3.7x107 J of work without any loss in internal energy.  The rate of energy production for fat is  3.7 x107 J/kg. Suppose I do work at the rate 95 J/s as the problem states. To get the total work,  multiply 95 J/s by the number of seconds in 24 hours. 

(a) To get the amount of fat ( in kg) burned in one day, divide  the total work by 3.7 x107 J/kg  .
(b) Divide 3.7 x107 J  by 95 J/s to get  time t. 
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