11-1

USING ENERGY 11 Q11.1. Reason: The friction between your hands increases the kinetic energies in molecules of your hands, and this exhibits itself as an increase in thermal energy of your hands. The temperature of your hands goes up. Assess: Thermal energy is related to molecular kinetic energies.

Q11.2. Reason: The water entering the hydroelectric plant converts its gravitational potential energy to kinetic energy as it runs through the plant. This turns the turbines in the plant, transferring its kinetic energy to the rotational kinetic energy of the turbines. The turbines convert the mechanical kinetic energy of the rotation into electrical energy, which is then converted back to mechanical energy in the pump, which transfers mechanical energy into the kinetic energy and potential energy of the water being pumped through the fountain. Assess: With each energy transformation, energy is lost to thermal energy.

Q11.3. Reason: Sitting down to breakfast, you ingest food which is partly used at the time and partly converted to stored energy. When you get up and walk to your bike you convert that energy to the mechanical energy of walking. The energy is further converted to mechanical energy when you ride your bike.

Q11.4. Reason: Riding a bike, you do not change your height. As you run you constantly propel yourself off the ground. Extra energy is required for this.

Q11.5. Reason: The chicken’s efficiency in using the corn’s energy is not 100%, so some would be lost if you feed the corn to the chicken. Eat the chicken first, and eat the corn later. Assess: Your own body isn’t 100% efficient in using the corn either, but at least you wouldn’t lose twice.

Q11.6. Reason: We know from the previous chapter that kinetic energy increases as the square of the velocity. As velocity increases, kinetic energy increases. As the engine turns faster, the rate at which friction creates waste thermal energy also increases. Also, as the speed of the car increases the drag force on the car increases. All of this requires a faster rate of energy consumption of the energy available as chemical energy in the fuel.

Q11.7. Reason: Refer to Equation 11.12, the first law of thermodynamics. (a) The space shuttle gets hotter, and its thermal energy increases, but it isn’t heated in the sense of being in thermal contact with a hotter object. Instead, it gets hot by the other method of changing the thermal energy: work. As the shuttle slams into air molecules in the upper atmosphere, work is done on them by the shuttle and work is done on the shuttle by them. The atoms in the shuttle jiggle more vigorously. (b) Since energy must be conserved but the shuttle is slowing down, that kinetic energy decrease must be balanced by an increase in another form of energy. In this case the shuttle’s initial kinetic energy is decreased while the thermal energy of the shuttle is increased. Assess: In the microscopic view, even heating by contact with a hotter object is work as the faster molecules collide with (and do work on) the slower molecules. But work and heat appear as separate terms in Equation 11.12.

Q11.8. Reason: The temperature of the rod remains constant and no work is being done. Some small amount of heat is lost to the environment, but the energy transferred to the ice must be nearly 100 J, from the first law of thermodynamics.

Q11.9. Reason: Since the blocks are at the same temperature, the average kinetic energy of the atoms that make up the block is the same. However, since the 3 kg block contains more atoms, it contains more thermal energy. If the blocks are placed in contact, the average kinetic energy of the atoms in each will remain the same,

11-2 Chapter 11

since there is no source of a higher average kinetic energy. The total energy of the combined system is the sum of the energies of both. Assess: Thermal energy is directly related to average kinetic energy of the atoms or molecules that make up the system.

Q11.10. Reason: From Equation 11.8, if the temperature of a gas doubles, the average kinetic energy of the atoms in the gas doubles also. Assess: Average kinetic energy of atoms in a gas is directly proportional to temperature.

Q11.11. Reason: Since the bottles have the same temperature, the average kinetic energies of the atoms that are in each bottle is the same. Since both bottles have the same amount of atoms, the total thermal energy must be the same also. We can also see this from Equation 11.9. Assess: Thermal energy in a gas is directly proportional to the number of atoms in the gas and its temperature.

Q11.12. Reason: In cooling a pan of water in the fridge, the thermal energy of the water decreases while the water does no work.

Q11.13. Reason: Compressing a gas in an insulated container will increase the thermal energy in the gas without any heat transfer. Work is being done on the gas, so the thermal energy must increase.

Q11.14. Reason: Expanding a gas in an insulated container will decrease its thermal energy while no heat can be transferred from the system. Expanding a gas does negative work on the gas.

Q11.15. Reason: Compressing a gas in a container that isn’t insulated will increase its thermal energy since work is being done on the gas. Since the container is not insulated, some of this energy will escape as heat.

Q11.16. Reason: Decompressing a gas in a noninsulated container will decrease the thermal energy of the gas, since negative work is being done on it. Since container is not insulated, it is possible for some heat to enter the container.

Q11.17. Reason: We can obtain this situation if we have a process where the amount of work done by a system is equal to the heat that comes into the system. This can be true for a gas that is heated and allowed to expand freely and do work as it expands.

Q11.18. Reason: Even though heat isn’t added, the thermal energy increases because of the work done on the piston. This can raise the temperature enough to ignite the cotton. th > 0E∆ and = 0.Q It must be done quickly so the piston won’t have time to cool off. Assess: This jibes with everyday experience.

Q11.19. Reason: The piston does work on the gas which increases the thermal energy; and it does so quickly enough that heat doesn’t escape the system fast enough to keep the gas below ignition temperature. Assess: Diesel engines have a hard time starting in cold weather for this reason.

Q11.20. Reason: The green ink will diffuse into the water, mixing with the water until eventually the water is uniformly filled with green ink. In terms of entropy the ink does not have a lot of entropy at the beginning of the process as at the end. As time goes on the disorder, or entropy of the system goes up, due to the mixing of the ink with the water. Assess: Entropy is a measure of disorder. The second law says that entropy never decreases.

Q11.21. Reason: As you do work on the rubber band, W is positive. The heat you feel from the rubber band is leaving the rubber band, so Q is negative. Assess: For work done on a system W is positive. For heat leaving a system Q is negative.

Q11.22. Reason: Increasing the temperature of the thermal reservoir increases the efficiency of an engine. After an amount of time it may be cheaper to have purchased the more expensive unit.

Using Energy 11-3

Q11.23. Reason: No engine can ever have more efficiency. Any higher efficiency would violate the second law of thermodynamics.

Q11.24. Reason: Some of the energy of braking is recovered in the battery and can be used again as electric energy to drive the car. In a gasoline-fueled vehicle the energy of braking is not recovered, but simply heats the brakes and then the air and is not recoverable. Assess: Flywheel systems recover some energy in braking by storing it as rotational kinetic energy which can then be re-directed to moving the car.

Q11.25. Reason: Yes, the entropy increases as the directed electromagnetic energy is transformed to randomly oriented thermal energy. Assess: High entropy is typical of random thermal energy.

Q11.26. Reason: The second law of thermodynamics is not violated in this case because it took energy to freeze the water into ice. The freezer expended energy to reduce the entropy. Assess: The ice cubes aren’t isolated, as specified in the second law.

Q11.27. Reason: A 100% efficient electric heater doesn’t violate the second law of thermodynamics because entropy increases as expected. Assess: The law prohibits moving heat from a cold object to a hot object with 100% efficiency, but not the other way around.

Q11.28. Reason: The source for energy in a person is the chemical energy in the food. Since the person is walking, part of that energy is converted to kinetic energy. Since no engine (even a person) can convert energy without loss of useful energy, some of the energy is converted to thermal energy. Some of the energy is converted to energy to maintain body temperature also. The answer is A.

Q11.29. Reason: Walking takes less energy for the same distance run. The time it takes to run the same distance as the walk is less. Therefore the both the energy used and power required for the run is greater. The answer is C.

Q11.30. Reason: As the temperature goes up, the average kinetic energy of the atoms that make up the gas is increasing. Since the average is increasing, the total also increased. The answer is B.

Q11.31. Reason: Thermal energy is given by Equation 11.9. Since the energy deceases by half and the energy is directly proportional to temperature, the temperature must be halved also. The answer is B.

Q11.32. Reason: The first law of thermodynamics is violated by this device since it converts a lesser amount of thermal energy to a greater amount of work output. This device also violates the second law of thermodynamics since it states that it is not possible to make a heat engine that converts thermal energy to an equivalent amount of work, and this device turns thermal energy to twice the amount of work! The answer is C.

Q11.33. Reason: The second law only applies to heat energy flowing spontaneously cold to hot.

Q11.34. Reason: The efficiency is H C

H

= Q QeQ− and the maximum efficiency is C

maxH

= 1 .TeT

− Set these equal

to each other and solve for C.Q C = 20 C = 293KT and H = 450 C = 723K.T

H C C

H H

= 1Q Q TQ T−

−

C C CC H

H H H

293K1 = 1 = = (100MJ) = 40MJ723K

Q T TQ QQ T T

− − ⇒

So the answer is D. Assess: Power plants do heat up rivers like this.

11-4 Chapter 11

Q11.35. Reason: Cmax

H C

COP = TT T−

and C

in

COP = .QW

Set these equal to each other and solve for inW .

Ci

C

H C

1.0 J= = = 0.13 J263 K297 K 263 K

nQW T

T T− −

The answer is C. Assess: So it takes 0.13 J to pump 1.0 J. Problems P11.1. Prepare: We will use conservation of energy to calculate the energy generated by the engine that is converted entirely to kinetic energy and then use the definition of efficiency to calculate the total energy generated by the engine. Solve: The conservation of energy states that the work done by the car engine is equal to the change in the kinetic energy. Thus,

21out 2W K mv= ∆ =

Using the definition of thermal efficiency,

( )( )2162out out

HH

1500 kg 15 m/s1.7 10 J

0.10W We QQ e

= ⇒ = = = ×

That is, the burning of gasoline transfers into the engine 1.7 × 106 J of energy. Assess: Note that the vast majority of the energy generated by the car’s engine is converted to heat. This is typical for engines in cars.

P11.2. Prepare: Use the definition of efficiency in Equation 11.2 where e is given as 60% and the electrical energy is 600 J. Solve: (a)

electric energy 0.60chemical energy

e = =

Solve this for chemical energy and plug in the values. electric energy 600 Jchemical energy 1000 J

0.60e= = =

(b) The second device uses twice as much chemical energy, or 2000 J, and generates only half the electric energy, or 300 J.

electric energy 300 J 0.15 15chemical energy 2000 J

e = = = = %

Assess: One would expect that requiring twice the chemical energy to generate half the electric energy would mean that the efficiency is down by a factor of four; that is precisely what we see.

P11.3. Prepare: Efficiency is given by Equation 11.2

e =what you get

what you had to pay

In this case the 34 0 10 W−. × of electrical energy is “what you get” as visible light, and the 11 2 10 W−. × of light energy is “what you had to pay.” Solve: The efficiency calculation gives

3

1

4 0 10 W 0 033 3 31 2 10 W

e−

−

. ×= = . = . %

. ×

Assess: Photovoltaic (PV) cells, also known as solar cells, are notoriously inefficient, and this is a typical value for traditional cells. However, advances in technology have been made and efficiencies of up to 20% are available; some researchers are aiming for 40% efficiency.

Using Energy 11-5

P11.4. Prepare: = 0.20e Solve: Each LED provides (1.0W)(0.20) = 0.20W of visible light. We therefore need eight of them to give 1.6 W of visible light power. Since each of them uses 1.0 W then the total power necessary is 8.0 W. Assess: This is a factor of five better than the incandescent bulb.

P11.5. Prepare: This is simply a unit conversion problem. Remember that a food Calorie is 1000 physics calories.

Solve: 6 64 19J1000 Cal 1000 kcal 1000000 cal 4 19 10 J 4 2 10 J1cal

.= = = . × ≈ . ×

Assess: A 1000 Cal burger has a lot of joules of energy! A small burger without the cheese and bacon could have significantly less than this, but still over a million joules.

P11.6. Prepare: As detailed in Table 11.3, the energy per second needed to carry on basic life processes totals to about 100 W for a 68 kg individual (we’ll assume that is the mass of an average human). That means they’ll need 100 J of energy every second. We need to figure out how many seconds in a day and then multiply by 100 J/s to give the total number of joules needed (energy = power × time). Then we’ll convert joules to calories and then to kcal (which is the nutritional Cal). Solve:

( ) 24 hr 60 min 60 s 100 J 1 cal 1 kcaldaily intake 1 d1 d 1 hr 1 min 1 s 4.19 J 1000 cal

2060 Cal 2100 Cal

=

= ≈

Assess: So a person of average weight (mass) needs just over two thousand Calories per day to maintain basic life processes. As the intermediate calculation showed (before the last two factors), that’s about 8600 000 joules per day. Active people of the same mass will need more caloric intake. Mass matters too: Larger people need greater daily caloric intake. On the other hand, if you regularly eat more energy than you use in basic life processes and activities then your body will store it as fat.

P11.7. Prepare: Various fuels and the corresponding energy in 1 g of each are listed in Table 11.1. Fats in foods such as “energy bars” have an energy content of 38 kJ per gram. Since we have 6 g of fat in our energy bar, we simply need to multiply 6 g by 38 kJ/g. Solve:

38 kJ6 g 228 kJ 230 kJ1 g

= ≈

1000 J230 kJ 230 000 J1 kJ

=

1 cal228 000 J 54 000 cal4 19 J

= .

1 kcal54 000 cal 54 kcal 54 Cal1000 cal

= =

Assess: Comparing with Table 11.2 shows that the fat in the energy bar does not provide as much energy as a fried egg; however, the energy bar may also contain carbohydrates in addition, and fat provides more energy per gram than carbohydrates do, so the total number of food calories in the energy bar might be 150. A 68 kg person needs just over 2000 Cal for basic life processes, so they would need to eat about 15 energy bars per day if that is all they ate. You may have learned in a health or nutrition class that 1 g of fat provides about 9 Cal of energy. Our calculations above bear this out: 54.4 Cal/6g 9 0 Cal/g.= .

11-6 Chapter 11

P11.8. Prepare: Various fuels and the corresponding energy in 1 g of each are listed in Table 11.1. Carbohydrates in foods such as “energy bars” have an energy content of 17 kJ per gram. Since we have 22 g of carbohydrates in our energy bar, we simply need to multiply 22 g by 17 kJ/g. Solve: Keep one extra digit in intermediate calculations but report the answers to only two significant figures.

17 kJ22g 374 kJ 370 kJ1g

= ≈

1000 J374 kJ 374 000 J 370 000 J1 kJ

= ≈

1cal374 000 J 89 300 cal 89 000 cal4.19 J

= ≈

1 kcal89 300 cal 89.3 kcal 89.3 Cal 89 Cal1000 cal

= = ≈

Assess: Comparing with Table 11.2 shows that the carbohydrates in the energy bar do not provide as much energy as a fried egg; however, the energy bar may also contain fat (see Problem 11.6) in addition, and fat provides more energy per gram than carbohydrates do, so the total number of food calories in the energy bar might be 150. A 68 kg person needs just over 2000 Cal for basic life processes, so they would need to eat about 15 energy bars per day if that is all they ate. You may have learned in a health or nutrition class that 1 g of carbohydrates provides about 4 Cal of energy. Our calculations above bear this out: 89.3 Cal/22 g = 4.0 Cal/g.

P11.9. Prepare: Various fuels and the corresponding energy in 1 g of each are listed in Table 11.1. Carbohydrates in foods such as “energy bars” have an energy content of 17 kJ per gram. Since we have 22 g of carbohydrates in our energy bar, we simply need to multiply 22 g by 17 kJ/g. Table 11.4 tells us that a 68 kg person needs to expend 380 J/s to walk at a speed of 5 km/hr. Solve: The energy in the carbohydrates in the bar is

517 kJ 1000 J22 g 370 kJ 370 000 J 3 7 10 J1g 1 kJ

= = = . ×

The time that the chemical energy will last at the rate of 380 J/s is 5

chem 3.7 10 J 970 s 16min 0 27hr380 W

EtP

∆ ×∆ = = = = = .

And the distance that can be covered during this time at 5 km/hr is

(5 km/hr)(0 27 hr) 1 4 kmx v t∆ = ∆ = . ≈ .

Assess: The answer seems to be in the right ball park; we didn’t get an answer of just a few cm nor an answer of many km—either of which we would be suspicious of given just one energy bar.

P11.10. Prepare: Table 11.4 tells us that a 68 kg person (we’ll assume this is your mass) needs to expend 480 J/s to pedal a bicycle at a speed of 15 km/hr. Table 11.1 helps us calculate the chemical energy stored in one gallon of gasoline (which has a mass of 3.2 kg).

8chem

1000 g 44 kJ 1000 J3.2 kg 1.4 10 J1 kg 1g 1 kJ

= ≈ ×

E

Solve: The time that the chemical energy will last at the rate of 480 J/s is 8

5chem 1.4 10 J 2.93 10 s 81 hr480 W

EtP

∆ ×∆ = = = × =

Using Energy 11-7

And the distance that can be covered during this time at 15 km/hr is

(15 km/hr)(81 hr) 1200 kmx v t∆ = ∆ = ≈

to two significant figures. Assess: The driving distance from Dallas, Texas to Denver, Colorado is just over 1200 km. This seems far for one gallon of gasoline, but you are going much slower than a car would (which increases your efficiency by decreasing the drag) and you are taking a lot less mass. Also remember that a car’s efficiency is probably less than 10% as shown in Integrated Example 11.16, while the efficiency of a human cycling is 20%–30%. To put it in units of mpg to compare to your car, you would be able to cycle 760 miles with the energy supplied by that one gallon of gas, which is 30 times better than a car that gets 25 mpg.

P11.11. Prepare: A typical efficiency for climbing stairs is about 25%, so we can assume that 25% of the chemical energy in the candy bar is transformed to increased potential energy.

5g

1 kcal 1000 cal 4 2 J(0 25)(400 Cal) 4 2 10 J1Cal 1 kcal 1cal

U .

∆ = . = . ×

Solve: Since gU mg y∆ = ∆ , the height gained is

5g

24 2 10 J 710 m

m s(60 kg)(9 8 )/y

mg∆ . ×

∆ = = =.

If we assume that each flight of stairs has a height of 2.7 m (as is done in Example 11.5), this gives

710 mNumber of flights 260 flights2.7 m

= ≈

Assess: This is more than enough to get to the top of the Empire State Building twice—all fueled by one candy bar! This is a remarkable result.

P11.12. Prepare: In weightlifting, a barbell curl is an exercise in which the barbell is held down at arms’ length against the thighs and then raised in semi-circular motion until the forearms touch the biceps. We’ll assume that the weightlifter expends metabolic energy when he lifts the 30 kg bar, but not as he lowers it. We’ll also assume 25% efficiency. Table 11.2 tells us that a typical slice of pizza has a metabolic energy content of 300 Cal or 1260 kJ. The weightlifter will use the 1260 kJ (at 25% efficiency) to lift the 30 kg bar, increasing its potential energy. In one repetition he’ll increase the potential energy by ∆Ug = mg∆y = mg(0.060 m), and in n repetitions by nmg (0.060 m), where n is what we want to know. Solve:

26

(energy from pizza) (efficiency) ym s m(1.26 10 J) (0.25) (30 kg) ( ) (0.60 )/9.80

nmgn

∆=

× =

Solving for n gives 1790 repetitions, which should be reported to two significant figures as 1800 repetitions. Assess: That’s a lot of curls! Exercising in this way to “burn off” an extra slice of pizza is almost impossible; people can’t do 1800 reps in a row, and there isn’t time before the next meal anyway. And n would be four times larger if the weightlifter were 100% efficient!

P11.13. Prepare: The work done in one repetition is the force (the weight of the barbell) multiplied by the distance lifted h, since the two vectors are in the same direction. To figure the energy needed for 20 repetitions we need to multiply the answer from part (a) by 20. However, we also need to take into account that the weightlifter only uses the energy in her food at 25% efficiency.

e =what you get

what you had to pay

where “what you get” is the total work done on the barbell in 20 reps, and “what you had to pay” is the total energy expended (which is what we want to know).

11-8 Chapter 11

Solve: (a) 2(40 kg)(9 8 )(0 5 m) 196 J 200 Jm/sW mgh= = . . = ≈

(b) ( J rep)(20 reps day)what you get 196 / /what you had to pay 15680 J/day 16000 J/day

25 0.25= = = ≈

%

(c) The number of 400 Calorie donuts needed to supply 15680 J is simply 15680 J/(400 Cal/donut).

15680 J/day 1 Cal 1 kcal 1 Cal 0 0094 donuts/day400 Cal/donut 1 kcal 1000 cal 4.19 J

= .

Assess: Straightforward problems like part (a) help develop our intuition about how big a joule is. 200 J is not a lot of energy compared to the chemical energy consumed in a meal. 16 000 J is still not a large number of joules compared to the 420 000 J of chemical energy in a fried egg. About a hundredth of a donut is enough to provide the energy for 20 reps of a 40 kg bench press.

P11.14. Prepare: To get from Kelvin to Celsius subtract 273. To get from Celsius to Fahrenheit multiply by 1.8 and add 32. Solve:

700 K = 427 C = 800 F 90 K = 183 C = 297 F− −

Assess: The range of temperatures on Mercury is staggering.

P11.15. Prepare: Doubling the kinetic energy corresponds to doubling the absolute temperature, so we find the equivalent on the Kelvin scale. Solve:

20 C = 293 K

Double this to get 586K and subtract 273 to get back to 313 C.

Assess: Don’t double the Celsius temperature. It seems that 313 C is more than double 20 C but it works right when converted to the absolute Kelvin scale.

P11.16. Prepare: We must convert the temperature to the Kelvin scale before reducing it by 10%. 20 C = 293 K. Solve:

(293 K)(0.90) 273 = 9.3 C− −

Assess: It is important to do these calculations in an absolute temperature scale.

P11.17. Prepare: Solve th B3=2

E Nk T∆ ∆ for .T∆

Solve:

th23 23

B

2 2 10 J= = = 4.8 K = 4.8 C3 3 (1.0 10 )(1.38 10 )

ETNk −

∆∆

× ×

f i= = 0 C 4.8 C = 4.8 CT T T+ ∆ +

Assess: We expected the temperature to rise from the added thermal energy.

P11.18. Prepare: Solve th B3=2

E Nk T∆ ∆ for .T∆

Solve:

th22 23

B

2 2 4.3 J= = = 9.4 K = 9.4 C3 3 (2.2 10 )(1.38 10 )

ETNk −

∆ −∆ − −

× ×

f i= = 20 C 9.4 C = 11 CT T T+ ∆ −

Using Energy 11-9

Assess: We expected the temperature to drop from the removed thermal energy.

P11.19. Prepare: We will use the first law of thermodynamics, Equation 11.12. Solve: The first law of thermodynamics is

∆Eth = W + Q ⇒ −200 J = 500 J + Q ⇒ Q = −700 J

The negative sign means a transfer of energy from the system to the environment. Assess: Because W > 0 means a transfer of energy into the system, Q must be less than zero and larger in magnitude than W so that Eth f < Eth i.

P11.20. Prepare: Equation 11.12 gives the thermal energy change th ,E W Q∆ = + and Figure 11.16 helps us figure out the signs. The 600 J of heat energy transferred to the system will be a positive Q while the 400 J of work that the system does means that W will be negative. Solve:

th ( 400 J) 600 J 200 JE∆ = − + =

Assess: We must remember that when the system does work W is negative.

P11.21. Prepare: Equation 11.12 gives the thermal energy change th ,E W Q∆ = + and Figure 11.16 helps us figure out the signs. The 300 J of heat energy transferred to the system will be a positive Q, and the thE∆ is a positive 150 J. Solve:

th 150 J 300 J 150 JW E Q= ∆ − = − = −

Assess: Since W is negative, it means that the work is done by the system instead of on it.

P11.22. Prepare: Equation 11.12 gives the thermal energy change th ,W Q∆ = +E and Figure 11.16 helps us figure out the signs. The 10 J of heat removed the system (the gas sample) means Q is negative, while the 20 J of work that the piston does on the system is a positive W. Solve:

th 20 J ( 10 J) 10 J∆ = + =E ±

Since the change in thermal energy is positive, the temperature of the gas increases. Assess: We must remember that when work is done on the system W is positive.

P11.23. Prepare: The efficiency of an engine is given by Equation 11.13. Solve: (a) The work done by the engine per cycle is

out H C 55 kJ 40 kJ 15 kJW Q Q= − = − =

(b) During each cycle, the heat transferred into the engine is H 55 kJ,Q = and the heat exhausted is C 40 kJ.Q = The thermal efficiency of the heat engine is

C

H

40 kJ1 1 0.2755 kJ

QeQ

= − = − =

Assess: We could have also gotten the answer to part (b) from part (a), out H/ (15 J)/(55 J) 0.27.e W Q= = =

P11.24. Prepare: We will need to calculate the energy of the hot reservoir to determine the efficiency. Solve: During each cycle, the work done by the engine is out 20 JW = and the engine exhausts C 30 JQ = of heat energy. Because out H C,W Q Q= −

H out C 20 J 30 J 50 JQ W Q= + = + =

Thus, the efficiency of the engine is

C

H

30 J1 1 0.4050 J

QeQ

= − = − =

11-10 Chapter 11

Assess: This makes sense, since about half the energy of the engine is lost as heat.

P11.25. Prepare: Assume that the heat engine follows a closed cycle. The efficiency of an engine is given by Equation 11.13. Solve: The engine’s efficiency is

out out

H C out

200 J 0.25 25600 J 200 J

W WeQ Q W

= = = = = %+ +

Assess: This is a reasonable efficiency.

P11.26. Prepare: We will use Equation 11.13, which defines the efficiency in terms of the work done by the engine and in terms of the heat energy of the hot and cold reservoirs. Solve: (a) The engine has a thermal efficiency of 40 0.40e = % = and a work output of 100 J per cycle. The heat input is calculated as follows:

outH

H H

100 J0.40 250 JWe QQ Q

= ⇒ = ⇒ =

(b) Because out H C,W Q Q= − the heat exhausted is

C H out 250 J 100 J 150 JQ Q W= − = − =

Assess: This makes sense since about half the energy provided by the hot reservoir is used to do work.

P11.27. Prepare: The efficiency of a Carnot engine depends only on the absolute temperatures of the hot and cold reservoirs. Solve: (a) The efficiency of a Carnot engine is

CCarnot

H

1 TeT

= −( )

C0.60 1427 273 K

T⇒ = −

+ C 280 K 7 CT⇒ = = °

(b) C H max(1 ) (673 K)(0.40) 269 K 3.8 CT T e= − = = = − Assess: A real engine would need a lower temperature than 7°C to provide 60% efficiency because no real engine can match the Carnot efficiency.

P11.28. Prepare: The maximum possible efficiency for a heat engine is provided by the Carnot engine. Solve: The maximum efficiency is

( )( )

Cmax Carnot

H

273 20 K1 1 0.6644

273 600 KTe eT

+= = − = − =

+

Because the heat engine is running at only 30% of the maximum efficiency, ( ) max0.30 0.20.e e= = The amount of heat that must be extracted is

outH

1000 J 5000 J0.20

WQe

= = =

Assess: This is reasonable, since most of the energy in this engine is lost to the cold reservoir.

P11.29. Prepare: C = 20 C 273 = 293 K.T + The maximum efficiency of a heat engine is Cmax

H

= 1 TeT

− .

Solve for H.T Solve:

CH

max

293 K= = = 733 K = 460 C1 1 0.60

TTe− −

Assess: The hot reservoir has to be really quite hot to achieve that efficiency.

Using Energy 11-11

P11.30. Prepare: The temperatures must be converted to the Kelvin scale. 650 C = 923 K and

30 C = 303 K. Solve:

Cmax

H

303 K= 1 = 1 = 0.67 = 67923 K

TeT

− − %

Assess: 67% is quite respectable (better than photovoltaic cells) but requires a high H.T

P11.31. Prepare: The COP of a refrigerator is given by the equation before Equation 11.15. Solve: The coefficient of performance of the refrigerator is

C H in

in in

50 J 20 J 1.520 J

Q Q WKW W

− −= = = =

P11.32. Prepare: The air conditioner is a heat pump whose job is to keep the inside of the building cool. The temperatures of the hot and cold sides must be expressed in kelvin. In this problem C 80 F 300 K= ° =T and

H =95 F=308 K.°T Solve: We use Equation 11.15 to compute the maximum coefficient of performance as follows:

Cmax

H C

300 KCOP =37.5 38308 K 300 K

≈− −

= =TT T

Assess: A coefficient of performance of 38 means that we pump 38 J of thermal energy for an energy cost of 1 J. This is clearly much better than the COP of 3.2 given as typical in the problem, reflecting the practical limitations including insulation, etc.

P11.33. Prepare: The COP of a refrigerator is given by the equation just before Equation 11.15. Solve: (a) The heat extracted from the cold reservoir is calculated as follows:

C CC

in

4.0 200 J50 J

Q QCOP QW

= ⇒ = ⇒ =

(b) The heat exhausted to the hot reservoir is

H C in 200 J 50 J 250 JQ Q W= + = + =

P11.34. Prepare: The heat pump’s job is to heat the inside of the house by pumping thermal energy from the (colder) outside to the (warmer) inside. The temperatures of the hot and cold sides must be expressed in kelvin. In this problem C 20 C 253 KT = − ° = and

H 20 C 293 K.T = ° = Solve: We use Equation 11.16 to compute the maximum coefficient of performance as follows:

Hmax

H C

293 KCOP 7 325 7 3293 K 253 K

TT T

= = = . ≈ .− −

Assess: The COP describes the ratio of thermal energy pumped to electrical power consumption; therefore a coefficient of performance of 7.3 means that we pump 7.3 J of thermal energy for an energy cost of 1 J. This value is the maximum possible for the given temperatures. In practice a typical heat pump has a COP of about three. An electrical resistance heater has a COP of one. Can you see why?

P11.35. Prepare: The efficiency of a Carnot engine (eCarnot) depends only on the temperatures of the hot and cold reservoirs. On the other hand, the thermal efficiency (e) of a heat engine depends on the heats QH and QC. Solve: (a) According to the first law of thermodynamics, H out C.Q W Q= + For engine (a), QH = 50 J, QC = 20 J and Wout = 30 J, so the first law of thermodynamics is obeyed. For engine (b), QH = 10 J, QC = 7 J and Wout = 4 J, so the first law is violated. For engine (c) the first law of thermodynamics is obeyed.

11-12 Chapter 11

(b) For the three heat engines, the maximum or Carnot efficiency is

CCarnot

H

300 K1 1 0.50600 K

TeT

= − = − =

Engine (a) has

C out

H H

30 J1 0.6050 J

Q WeQ Q

= − = = =

This is larger than eCarnot, thus violating the second law of thermodynamics. For engine (b),

outCarnot

H

4 J 0.4010 J

We eQ

= = = <

so the second law is obeyed. Engine (c) has a thermal efficiency that is

Carnot10 J 0.3330 J

e e= = <

so the second law of thermodynamics is obeyed. Assess: The only engine that doesn’t violate the first or second law is engine (c).

P11.36. Prepare: For a refrigerator QH = QC + Win, and the coefficient of performance and the Carnot coefficient of performance are

C

in

QCOPW

= CCarnot

H C

TCOPT T

=−

Solve: Please refer to Figure P11.36. (a) For refrigerator (a) H C in (60 J 40 J 20 J),Q Q W= + = + so the first law of thermodynamics is obeyed. For refrigerator (b) 50 J 40 J 10 J,= + so the first law of thermodynamics is obeyed. For the refrigerator (c) 40 J ≠ 30 J 20 J+ , so the first law of thermodynamics is violated. (b) For the three refrigerators, the maximum coefficient of performance is

CCarnot

H C

300 K 3400 K 300 K

TCOPT T

= = =− −

For refrigerator (a),

CCarnot

in

40 J 220 J

QCOP COPW

= = = <

so the second law of thermodynamics is obeyed. For refrigerator (b),

CCarnot

in

40 J 410 J

QCOP COPW

= = = >

so the second law of thermodynamics is violated. For refrigerator (c),

Carnot30 J 1.520 J

COP COP= = <

so the second law is obeyed. Assess: Both the first and second laws of thermodynamics must be obeyed by a refigerator. The only refrigerator which does not violate either is refrigerator (a).

P11.37. Prepare: We’ll show all the possibilities and then directly count the probability that all three balls will be in Box 1.

Using Energy 11-13

Solve:

There are eight possible arrangements, and in only one of them are all three balls in Box 1. Therefore, the probability of that happening is 1/8 0 125 13= . ≈ % . Assess: It is not very probable to have all three balls in Box 1 or all in Box 2; much more probable is to have two balls in one box and one ball in the other. This trend becomes more pronounced as the number of balls increases.

P11.38. Prepare: We see from Table 11.2 that a typical slice of pizza has an energy content of 300 Cal = 1260 kJ. In addition, we consult Table 11.4 to see that walking at 5 km/hr requires about 380 W of metabolic power for a 68 kg individual (hence that assumption for your mass as well). The speed of 5 km/hr (a typical walking speed) won’t directly enter the rest of the calculation; we only needed it to correspond with the entry in Table 11.4. We will also estimate your efficiency at 25%, as is typical of walking, running, and cycling. It goes without saying (but we will anyway) that you could nudge these estimates either way if you want to do the computation for walking speeds faster or slower than 5 km/hr, or for a person of mass slightly greater or less than 68 kg, or for pizza slices that are larger or smaller than average. You should still be able to produce an answer that is in the right ballpark, correct to one or two significant figures. Solve: Use energy power time,= × modified to account for the efficiency of 25 .e = % Here the energy (in pizza) is what you had to “pay,” and the power × time is what you get. So energy power time.× = ×e Call the number of pizza slices (what we want to know) n. The total energy you eat is n times 1.26 kJ. A couple of simple pre-conversions will make the following clearer and cleaner. 380 W 380 J/s= and 1 hr = 3600 s.

energy power time× = ×e

6(1.26 10 J) 25 380 J/s 3600 s× × % = ×n

Solving for n:

6

380 J/s 3600 s 4.3 slices(1.26 10 J) 0.25

×= =

× ×n

This should be reported to one significant figure as four slices since the some of the data were given to just one significant figure. Assess: Four slices of pizza sounds about right for a 1 hr walk, doesn’t it?

11-14 Chapter 11

P11.39. Prepare: We ignore the energy needed for horizontal motion and simply convert the chemical energy to gravitational potential energy. Solve:

2(60 kg)(9.8 m/s )(500 m)= = = 1.176 MJ0.25

mghEe

The total energy divided by the energy per burrito will give the number of burritos.

1.176 MJ = 0.80 burritos1470 kJ / burrito

Assess: So a typical hiker could climb a 500 m hill on the energy provided by just less than one frozen burrito.

P11.40. Prepare: We consult Table 11.2 to see that we will need to assume a large size meal of burger, fries, and drink which has an energy content of 1350 Cal = 5670 kJ. In addition, we consult Table 11.4 to see that swimming at a fast crawl requires about 800 W of metabolic power for a 68 kg individual. We will also estimate the efficiency of the swimmer to be 25% (see Problem 11.43). Solve: Use energy = power × time, modified to account for the efficiency of 25e = % . Here the energy (in the burger, fries, and drink) is what you had to “pay,” and the power × time in swimming is what you get. So energy × e = power × time.

energy (5670 kJ)(0.25)time 1770 s 30 minpower 800 J/s

×= = = =

e

Assess: Fast swimming is strenuous exercise and uses up the metabolic energy fairly quickly, so 30 min for a large junk food meal is about right.

P11.41. Prepare: We want to compare the efficiencies of three modes of human motion. Note that the running and cycling speeds are both 15 km/hr, but the walking speed is only 1/3 that—5 km/hr. But since the walker walks for three times (1 hr) the runner runs and cycler cycles (1/3 hr), the distance traveled is the same in all three cases.

(15 km/hr) (1/3 hr) (5 km/hr)(1 hr) 5 kmx v t∆ = ∆ = = =

So to compare efficiencies we can compute the energy needed to go 5 km in each case. The problem assumes the person has a mass of 68 kg because that is the assumption made in table 11.4, where we find that, for such an individual, running at 15 km/hr uses 1150 J/s, walking at 5 km/hr uses 380 J/s, and cycling at 15 km/hr (as illustrated in Example 11.4) use 480 J/s. Also remember that 1 hr 3600 s,= so 20 min 1/3 hr 1200 s.= = Solve: Power is the rate of using energy, / .P E t= ∆ ∆ (a) Running 5 km (at 15 km/hr for 1/3 hr):

6(1150 J/s)(1200 s) 1 4 10 JE P t∆ = ∆ = = . ×

(b) Walking 5 km (at 15 km/hr for 1 hr): 6(380 J/s)(3600 s) 1 4 10 JE P t∆ = ∆ = = . ×

which is the same as the answer in part (a) to two significant figures. (c) Cycling 5 km at (15 km/hr for 1/3 hr):

5(480 J/s)(1200 s) 5 8 10 JE P t∆ = ∆ = ≈ . ×

which is the same answer obtained in Example 11.4, and which is less than half the answers in parts (a) and (b). This means that cycling is the most efficient, by far, of the three modes of human motion since cycling uses less than half the energy to go 5 km than running or walking does. Assess: Is this result expected? Yes, because cycling is not a weight-bearing exercise like running and walking. The bike holds up the body and the cyclist’s energy can go into forward motion rather than repeatedly launching the body up.

P11.42. Prepare: In part (a) we will estimate the efficiency of the cyclist to be 25%. That is, we’ll assume that 25% of the 480 W is used for forward motion against air resistance.

Using Energy 11-15

In part (b) recall that power = force × speed. Solve: (a) (0.25)(480 W)=120 W

(b) power 120 W 3600 s 1 kmforce 28.8 N 29 Nspeed 15 km/hr 1 hr 1000 m

= = = ≈

Assess: 25% efficiency is about right for cycling, and 480 W is the value given for cycling at 15 km/hr in Table 11.4. Neither the 120 W of power for forward propulsion nor the 29 N of force seems like a lot, but these figures agree with Figure 11.7.

P11.43. Prepare: Read all the parts of a multipart problem to see the general direction of thought and what preliminary calculations will be needed to get to the final answer. We convert the time to seconds: 10 minutes and 49 seconds is a total of 649s 650s.≈ Solve: (a) Do a quick calculation for total height: ( )( )86 floors 3.7 m/floor 320 m .y= = ∆

2g (60 kg)(9 8 )(320 m) 187 kJ 190 kJm/sU mg y∆ = ∆ = . = ≈

This is true for the winner and any other 60 kg person who finished the race. (b) We’ll use the efficiency formula where “what you get” is the change in potential energy of the person, and “what you had to pay” is the total energy the winner expended during the race.

what you get 187 kJwhat you had to pay = =748 kJ 750 kJ25 0.25

= ≈%

(c) This is just unit conversion from the answer in part (b).

1000 J 1 cal 1 kcal748 kJ 179 kcal 180 kcal 180 Cal1 kJ 4 19 J 1000 cal

= ≈ = .

(d) Since the winner was 25% efficient, and as we have already noted, 25% of the energy went into increasing the potential energy of the racer. The rest (75%) went into thermal energy.

(179 Cal)(0 75) 134 Cal 130 Cal. = ≈

(e) To compute the winner’s metabolic power we need to divide the total metabolic energy “burned up” by the time it took.

748 kJ 1200 W650 s

EPt

∆= = ≈

∆

Assess: The winner of this race obviously made a good effort. In fact, the result of 1200 W is about one and a half horsepower.

P11.44. Prepare: In part (a) the swimmer’s metabolic power is 800 W but the efficiency is only 25%, so the power that is used for forward propulsion is 0.25 800 W 200 W.× = The time it takes to swim 50 m is given as 22s. In part (b) we divide the 4400 J—half for the arms, and half for the legs. We further divide the arms’ half by 30 to get the energy expenditure per arm stroke. In part (c) the force will vary over the stroke, but we are only asked to determine the average force. The distance the hand moves through the water in a stroke (half a circle) is (0.90m) 2.83m.rπ π= ≈ The work done is the force multiplied by the distance, and the work done is the energy expended. Solve: (a) (200 W)(22 s) 4400 J= ∆ = =E P t (b) 2200 J/30 73 J= per arm stroke.

(c) energy 73 Jforce 25.9 Ndistance 2.83 m

= = =

which should be reported to two significant figures as 26 N. Assess: Swimming is strenuous, so 4400 J is not too much in 22 s.

11-16 Chapter 11

73 J is about the energy needed to lift a 7.4 kg object vertically 1 m (the change in potential energy would be 73 J). This seems about as hard (energy intensive) as a swimming arm stroke. Think about holding a 2.7 kg object at rest in your hand. The force you exert on it is about 26 N, which seems comparable to the force on the water in a swimming stroke.

P11.45. Prepare: We assume we can add the metabolic power required to walk horizontally to the power needed to increase the gravitational potential energy. Solve:

380 W 380 W tan 7 =yP mg mgvt

∆= + = +

∆

2 1 h 1000 m380 W (68 kg)(9.8 m/s )(5 km/h) tan 7 = 493.6 W 490 W3600 s 1 km

+ ≈

Assess: We see that it takes an extra 113 W to go up the incline over what it would take to walk horizontally.

P11.46. Prepare: The student consumed an extra 1000 Cal = 4190 kJ. We also use kinematic equations to compute how high the student jumps because the student will transform the chemical energy to potential energy.

22 ii0 = 2 =

2vv a y yg

+ ∆ ⇒ ∆

Solve: Let n be the number of jumps.

2 2i

4190 kJ( )( ) = = = = = 44000 jumps1 1( )( ) ( )( ) (0.25)( (70 kg)(3.3 m/s) )2 2

E En e mgh E ne mg y e mv

∆ ∆∆ ⇒

∆

Assess: It would be difficult to jump that many times. A better option is to not eat the extra 1000 Cal in the first place, or to exercise regularly so your metabolism is higher.

P11.47. Prepare: What we want to get is the 75 W of useful power.

Solve: (a) what you get 75 Wwhat you had to pay = = = 300 W0.25e

(b) 3600 s= = (300W)(1h) = 1080 000 J 1.1 MJ1h

E P t

∆ ≈

1Cal(1080 000 J) = 258 Cal 260 Cal4190 J

≈

Assess: That’s as much energy as you’d get from a latte with whole milk.

P11.48. Prepare: The typical efficiency for a human is 25%. Solve: (a) human generatorpower output = (power input)( )( ) = (400W)(0.25)(0.8) = 80We e

(b) 400W = 5 people80W/person

Assess: If both efficiencies were 100% then it would only take one person.

P11.49. Prepare: We use ratios to solve the problem. Solve:

(a) 3.0 Cal 1000 g(68 kg) = 10200 Cal 10000 Cal20 g 1 kg

≈

(b) 10200 Cal 1 d 1 h 1000 cal 4.19 J= = = 495 W 490 W1 d 24 h 3600 s 1 Cal 1 cal

EPt

≈ ∆

Using Energy 11-17

This is 390 W more than, or about five times the 100 W stated in the chapter. Assess: We expected a number bigger than the 100 W quoted for a resting human.

P11.50. Prepare: We use ratios to solve the problem. Solve:

(a) 70000 Cal(68 kg) = 952 Cal 950 Cal5000 kg

≈

(b) 952 Cal 1 d 1 h 1000 cal 4.19 J= = = 46 W1 d 24 h 3600 s 1 Cal 1 cal

EPt

∆

This is 54 W less than, or just under half the 100 W stated in the chapter. Assess: The 46 W is less than a human, as we expect.

P11.51. Prepare: The conversion factor is 746 W = 1 hp. Solve: (a) Because of the 25% efficiency, the horse will need to eat 4 1 hp× worth of Calories.

746 J 1 cal 1 Cal 3600 s= = (4 1 hp)(1 h) = 2564 Cal 2600 Cal1 hp 4.19 J 1000 cal 1 h

E P t

∆ × ≈

(b) The mass needed would be the total energy divided by the energy per kilogram of hay.

61000 cal 4.19 J 1 MJ2563 Cal

1 Cal 1 cal 1 10 J= = 1.07 kg 1.1 kg

10 MJ/kgm

× ≈

Assess: A kilogram of food would be a lot for a human, but not so much for a horse.

P11.52. Prepare: Use t B3=2hE Nk T and =E P t∆ ∆ to find .T∆ See Example 11.8.

Solve:

26 23B B

60 s(125W)(10 min)1 min2 2 2= = = = 6.0 K

3 3 3 (6.0 10 )(1.38 10 J/K)E P tT

NK NK −

∆ ∆ ∆× ×

This T∆ is equal to 6.0 C. Assess: The temperature wouldn’t really go up that much in real life because thermal energy would be transferred to the walls instead of merely raising the temperature of the air.

P11.53. Prepare: We’ll use the equation that relates changes in energy with changes in temperature, Equation 11.10, which, fortunately, has N in it, which is what we want to know. We are not told explicitly that T∆ is negative, but we can conclude so since thermal energy was removed. Equation 11.10:

th32

∆ = ∆BE Nk T

Solve: Solve Equation 11.10 for N.

22 22th23

B

2 2( 30 J)N 7.246 10 atoms 7.2 10 atoms3 3(1.38 10 J/K)( 20 K)−

∆ −= = = × ≈ ×

∆ × −E

k T

Assess: This is a large number, but typical in problems like this. Avogadro’s number is 236 02 10 ,. × so our result is less than the number of carbon atoms in 12 g of carbon.

P11.54. Prepare: The theoretical maximum efficiency of a heat engine is given in terms of the hot reservoir temperature and the cold reservoir temperature by Equation 11.14.

11-18 Chapter 11

Cmax

H

1 TeT

= −

Here we are given H 400KT = and 0.20e = and asked to find the maximum possible C.T This maximum occurs when max.e e= Solve Equation 11.14 for C :T

C max

H

( ) 1T eT

= −

C max H( ) (1 )T T e= −

Solve: With H 400 KT = and 0.20,e =

C max H( ) (1 ) (400K)(1 0 20) 320 KT T e= − = − . =

Assess: This answer is 80 K below H.T (We’d be sure we had done something wrong if it had come out above H.T )

P11.55. Prepare: The efficiency of an ideal engine (or Carnot engine) depends only on the temperatures of the hot and cold reservoirs. Solve: (a) The engine’s thermal efficiency is

out out

H C out

10 J 0.40 4015 J 10 J

W WeQ Q W

= = = = = %+ +

(b) The efficiency of a Carnot engine is Carnot C H1 / .e T T= − The minimum temperature in the hot reservoir is found as follows:

HH

293 K0.40 1 488 K 215 CTT

= − ⇒ = = °

This is the minimum possible temperature, which should be reported to two significant figures as 220°C. Assess: In a real engine, the hot-reservoir temperature would be higher than 215°C because no real engine can match the Carnot efficiency.

P11.56. Prepare: Efficiency is given by Equation 11.13. Solve: If QC = 2

3 QH, then Wout = QH – QC = 13 QC. Thus the efficiency is

1Hout 3

H H

13

QWeQ Q

= = =

The efficiency of a Carnot engine is e = 1 – TC/TH. Thus

C C

H H

1 213 3

T TT T

− = ⇒ =

P11.57. Prepare: The efficiency of a Carnot engine depends only on the temperatures of the hot and cold reservoirs. The Carnot engine has the maximum theoretical efficiency of any engine. Solve: The efficiency of a Carnot engine is Carnot C H1 .e T T= − This can be increased by either increasing the hot-reservoir temperature or decreasing the cold-reservoir temperature. For a 40% Carnot efficiency and a cold-reservoir temperature of 7°C,

CarnotH

280 K0.40 1eT

= = − H 470 KT⇒ =

A higher efficiency of 60% can be obtained by raising the hot-reservoir temperature to HT ′ . Thus,

Using Energy 11-19

CarnotH

280 K0.60 1eT

= = −′ H 700 KT ′⇒ =

The reservoir temperature must be increased by 230 K to cause the efficiency to increase from 40% to 60%. Assess: Note this is a large temperature change for a relatively small increase in efficiency.

P11.58. Prepare: Comfortable room temperature is about 73 F = 23 C = 296 K. Hand temperature is usually not as warm as internal body temperature. To be generous to the heat engine manufacturer let’s assume a warm 90 F = 32 C = 305 K hand temperature. Solve:

Cmax

H

296 K= 1 = 1 = 3.0305 K

TeT

− − %

Assess: Your answer will vary based on your assumptions for the hand and air temperatures, but it will be small.

P11.59. Prepare: Assume that the refrigerator is an ideal or Carnot refrigerator. Equation 11.15 gives the efficiency of a Carnot refrigerator. Solve: (a) For the refrigerator, the coefficient of performance is

( )( )CC in

in

( )( ) 5.0 10 J 50 JQCOP Q COP WW

= ⇒ = = =

The heat energy exhausted per cycle is

H C in 50 J 10 J 60 JQ Q W= + = + =

(b) If the hot-reservoir temperature is 27 C° = 300 K, the lowest possible temperature of the cold reservoir can be obtained as follows:

C CCarnot C

H C C

5.0 250 K 23 C300 K

T TK TT T T

= ⇒ = ⇒ = = − °− −

Assess: These results are reasonable. Compare to Example 11.13.

P11.60. Prepare: The efficiency is what you get (the useful work output in this case) divided by what you had to pay (the thermal energy input in this case).

out

H

what you get 50 J 1what you had to pay 100 J 2

WeQ

= = = =

Equation 11.14 also says about efficiency that

Cmax

H

1 TeT

= −

The maximum efficiency will occur when the ratio of the temperatures (in kelvin) of the hot and cold reservoirs is a minimum. This minimum ratio is what we want to know. Solve: Solve Equation 11.4 for C

H.T

T

Cmax

H min

1 11 12 2

T eT

= − = − =

Assess: So 1/2 is the smallest possible ratio of the two temperatures in kelvin. This is certainly doable (say C 250 K T = and H 500 K,T = although both temperatures won’t be in a normal everyday weather-like range), so

this engine is possible.

P11.61. Prepare: Given the efficiency of the power plant we can calculate the amount of waste heat. Solve: The amount of heat discharged per second is calculated as follows:

11-20 Chapter 11

( ) 9out outC out

H C out

1 11 900 MW 1 1.9 10 W0.32

W We Q WQ Q W e

= = ⇒ = − = − = × +

That is, each second the electric power plant discharges 91.9 10 J× of energy into the ocean. Since a typical American house needs 42.0 10 J× of energy per second for heating, the number of houses that could be heated with the waste heat is 9 4(1.9 10 J)/(2.0 10 J) 95,000.× × = Assess: This is a relatively efficient electric power plant. Even so, over two times more energy is lost as waste heat than is delivered.

P11.62. Prepare: The power plant is to be treated as a heat engine. Solve: (a) Every hour 300 metric tons or 3 × 105 kg of coal is burnt. The volume of coal is

5 34 33.00 10 kg 1.5 m 24 hour 1.08 10 m

1 hour 1000 kg×

× × = ×

The height of the room will be 110 m. (b) The thermal efficiency of the power plant is

12out

65H

2.7 10 J 3228 10 J(3.00 10 kg)

kg

WeQ

×= = = %

×× ×

Assess: An efficiency of 32% is typical of power plants.

P11.63. Prepare: The power plant is treated as a heat engine. Solve: (a) The maximum possible thermal efficiency of the power plant is

Cmax

H

303 K1 1 0.471 47573 K

TeT

= − = − = ≈ %

(b) The plant’s actual efficiency is 6

out6

H

700 10 J/s = 0.35 352000 10 J/s

WeQ

×= = = %

×

Assess: If the plant were operating at the maximum possible thermal efficiency, the output would be (2000 MJ)(0.471) = 942 MJ. There is a loss of 242 MJ due to inefficiencies in the plant.

P11.64. Prepare: The amount of energy that needs to be pumped to the cold reservoir is C = (250 students)(125 J) = 31250 JQ every second.

Solve:

C Ci

i

31250 JCOP = = = = 6250 JCOP 5n

n

Q QWW

⇒

That is the work needed per second. We round to two significant figures, so the power needed is 6.3 kW. Assess: The cost for this cooling isn’t too much for one hour, but it adds up over many hours.

P11.65. Prepare: The needed force to keep a constant velocity is just the opposite of the backward rolling friction force. The energy obtained in burning 1 g of gasoline is 44 kJ. The density of gasoline is

3= 719.7 kg/m .ρ Solve: (a) 2

r r= = (0.06)(1500 kg)(9.8 m/s ) = 882 N 880 NF nm ≈ (b) = = (882 N)(5.0 m/s) = 4410 W 4400 W = 4.4 kWP Fv ≈ (c) Use =W Fd and lots of unit conversions.

Using Energy 11-21

= = = =W EP Fv v vd d

3

6 3

1 gal 1 mi 3785 ml m 719.7 kg 1000 g 44 kJ (5.0 m/s) = 24.8 kW15 mi 1609 m 1 gal 1 10 ml 1 m 1 kg 1 g

× ×

what you get 4.4 kW= = = 18what you had to pay 24.8 kW

e %

Assess: 18% seems to be in the typical range.

P11.66. Prepare: Say the outside temperature is 95 F = 35 C = 308 K, and the desired indoor temperature is

75 F = 24 C = 297 K. Solve:

Cmax

H C

297 KCOP = = = 27308 K 297 K

TT T− −

max maxSEER = 3.4 COP = (3.4)(27) = 92×

Assess: Your answer may vary if you assumed different inside and outside temperatures.

P11.67. Prepare: The maximum possible efficiency of a heat engine is given by Equation 11.14:

Cmax

H

1 TeT

= −

In this problem H 27 C 300 KT = ° = and C 3 C 276 KT = ° = Solve:

Cmax

H

276 K1 1 0 080 8 0300 K

TeT

= − = − = . = . %

Assess: An efficiency of 8% does not sound wonderful, and 1200 m is reasonably deep. But the hot and cold reservoirs are huge, and since nature will maintain the temperatures of the reservoirs for us, if the construction of the heat engine is safe, and cheap, it might be worth it.

P11.68. Prepare: In part (a) we simply need to multiply the rate of energy consumption (i.e., power) by the time. In part (b) we use the efficiency equation:

what you getwhat you had topay

=e

“What you get” is the energy needed by the 250 000 houses. “What you had to pay” is the energy collected by an area A of solar cells (at 5% efficiency), where A is what we want to know.

Solve: (a) 73600 s(1 kW)(24 hr) 8.64 10 J 86.4 MJ 86 MJ1 hr

= ∆ = = × = ≈

E P t

(b) 2

(250 000 houses)(86.4 MJ/house)0.05(20 MJ/m )A

= % =e

Then solve for A:

22 272

(250 000 houses)(86.4 MJ/house) m mi mi2.16 10 8.34 8.3(0.05)(20 MJ/m )

= = × = ≈A

Assess: 86.4 megajoules sounds like a lot, but one joule isn’t a lot of energy. 86.4 megajoules is only ten times as much as an average person consumes in metabolic energy in a day. An area of 8.34 mi2 is a lot, but that’s for 250 000 houses; for one house it is 86.4 m2.

11-22 Chapter 11

Of course there are complicating factors such as clouds and storage of the energy for use at night, etc. And so we’ll need much more efficient solar cells (also known as photovoltaic cells) before they’ll become a viable factor in the mix of sources of electrical energy. This is an active area of research. Solar cells with efficiencies in the teens are now available, and there is some speculation that the efficiencies may get up to 40% eventually.

P11.69. Prepare: The graph of power used versus speed shows a straight line approximately through the origin for humans, which means that running twice as fast takes approximately twice the power. However, the time needed at a fast speed is less, and so to cover the same distance would require about the same amount of energy at either speed. Solve: C. The total energy is about the same for a fast speed and a slow speed. Assess: This result differs from that for kangaroos. Because fast kangaroo hops are more efficient the kangaroo could cover the distance using less total energy at a fast speed than at a slow one. See also Problem 11.69.

P11.70. Prepare: As mentioned, kangaroo hops are not as efficient at slow speeds as they are at high speeds, so it will take less energy to go 1 km at high speed. Looked at another way, since the necessary power is approximately the same for low and high speeds it would take approximately the same amount of energy for the kangaroo to hop for equal amounts of time, but it could cover the 1 km distance in a lot less time at a fast speed. Solve: A. A faster speed requires less total energy. Assess: This result may be somewhat surprising, at least compared to humans. For humans, running twice as fast requires about twice as much power, so the energy expended over the 1 km distance would be about the same either way. Not so for kangaroos.

P11.71. Prepare: Figure P11.69 shows oxygen uptake versus speed, and it mentions that oxygen uptake is a measure of energy use per second, or power. Solve: The graph shows that at 4 m/s the line for the running human is below the line for the hopping kangaroo, so the human uses less energy per second to go at a speed of 4 m/s and is therefore more efficient at that speed. So the answer is A. A running human is more efficient than an equal-mass hopping kangaroo. Assess: At slow speeds, running humans are more efficient than equal-mass hopping kangaroos, but the reverse is true at higher speeds. As the graph shows, the crossover is at about 7 m/s. Humans have trouble running much faster than 10 m/s as that is about the speed of an olympic sprinter who runs 100 m in about 10 s.

P11.72. Prepare: Figure P11.69 shows oxygen uptake versus speed, and it mentions that oxygen uptake is a measure of energy use per second, or power. So we are simply looking on the graph for the speed at which the value on the line for humans is half the value on the line for kangaroos; it appears that it happens at about 3 m/s. Solve: A. 3 m/s. Assess: That’s the speed at which a human would use half the power of an equal-mass kangaroo.

P11.73. Prepare: Figure P11.69 shows oxygen uptake versus speed, and it mentions that oxygen uptake is a measure of energy use per second, or power. We are simply looking for the speed where the lines cross and the kangaroo line is below the line for humans. When the line for kangaroos is below the line for humans the kangaroos are using less power to go that speed, and are therefore more efficient. It appears that it happens at about 7 m/s. Solve: C. 7 m/s. Assess: This agrees with the information in the text of the passage problems.