Defining Temperature Temperature is a measure of the average kinetic energy of the particles in a...

Defining Temperature

• Temperature is a measure of the average kinetic energy of the particles in a substance.

• Adding or removing energy usually changes temperature.

• Internal energy is the energy of a substance due to both the random motions of its particles and to the potential energy that results from the distances and alignments between the particles.

Temperature and Thermal Equilibrium

Thermal Equilibrium

• Thermal equilibrium is the state in which two bodies in physical contact with each other have identical temperatures.– By placing a thermometer in contact with an object and waiting

until the column of liquid in the thermometer stops rising or falling, you can find the temperature of the object.

– The reason is that the thermometer is in thermal equilibrium with the object.

• The temperature of any two objects in thermal equilibrium always lies between their initial temperatures.


Thermal Expansion

• In general, if the temperature of a substance increases, so does its volume. This phenomenon is known as thermal expansion.

• Different substances undergo different amounts of expansion for a given temperature change.

• The thermal expansion characteristics of a material are indicated by a quantity called the coefficient of volume expansion.

• Gases have the largest values for this coefficient. Solids typically have the smallest values.


Measuring Temperature

• The most common thermometers use a glass tube containing a thin column of mercury, colored alcohol, or colored mineral spirits.

• When the thermometer is heated, the volume of the liquid expands.

• The change in length of the liquid column is proportional to the temperature.


Measuring Temperature, continued

• When a thermometer is in thermal equilibrium with a mixture of water and ice at one atmosphere of pressure, the temperature is called the ice point or melting point of water. This is defined as zero degrees Celsius, or 0°C.

• When the thermometer is in thermal equilibrium with a mixture of steam and water at one atmosphere of pressure, the temperature is called the steam point or boiling point of water. This is defined as 100°C.


Measuring Temperature, continued• The temperature scales most widely used today are the Fahrenheit, Celsius, and Kelvin scales. • Celsius and Fahrenheit temperature measurements can be converted to each other using this equation:


9 32.05

9 Fahrenheit temperatu

re

Celsius temperature 32.05

F CT T

• The number 32.0 indicates the difference between the ice point value in each scale: 0.0ºC and 32.0ºF.



• Temperature values in the Celsius and Fahrenheit scales can have positive, negative, or zero values.

• But because the kinetic energy of the atoms in a substance must be positive, the absolute temperature that is proportional to that energy should be positive also.

• A temperature scale with only positive values is suggested by the graph on the next slide. This scale is called the Kelvin scale.



• The graph suggests that if the temperature could be lowered to –273.15°C, the pressure would be zero.

• This temperature is designated in the Kelvin scale as 0.00 K, where K represents the temperature unit called the kelvin.

• Temperatures in the Kelvin scale are indicated by the symbol T.



• A temperature difference of one degree is the same on the Celsius and Kelvin scales. The two scales differ only in the choice of zero point. • Thus, the ice point (0.00°C) equals 273.15 K, and the steam point (100.00°C) equals 373.15 K. • The Celsius temperature can therefore be converted to the Kelvin temperature by adding 273.15:

T TC 273.15 Kelvin temperature Celsius temperature 273.15

Temperature Scales and Their Uses


Defining Heat

Heat and Energy

• Heat is the energy transferred between objects because of a difference in their temperatures.

• From a macroscopic viewpoint, energy transferred as heat tends to move from an object at higher temperature to an object at lower temperature.

• The direction in which energy travels as heat can be explained at the atomic level, as shown on the next slide.

Transfer of Particles’ Kinetic Energy as Heat

Defining Heat

Energy is transferred as heat from the higher-energy particles to the lower-energy particles, as shown on the left. The net energy transferred is zero when thermal equilibrium is reached, as shown on the right.

Defining Heat

Heat and Energy, continued

• The atoms of all objects are in continuous motion, so all objects have some internal energy. – Because temperature is a measure of that energy,

all objects have some temperature.

• Heat, on the other hand, is the energy transferred from one object to another because of the temperature difference between them. – When there is no temperature difference between

a substance and its surroundings, no net energy is transferred as heat.

Defining Heat

Heat and Energy, continued

• Just as other forms of energy have a symbol that identifies them (PE for potential energy, KE for kinetic energy, U for internal energy, W for work), heat is indicated by the symbol Q.

• Because heat, like work, is energy in transit, all heat units can be converted to joules, the SI unit for energy.

Thermal Units and Their Values in Joules

Defining Heat

Defining Heat

Thermal Conduction

• The type of energy transfer that is due to atoms transferring vibrations to neighboring atoms is called thermal conduction.

• The rate of thermal conduction depends on the substance.

• Two other mechanisms for transferring energy as heat are convection and electromagnetic radiation.

When this burner is turned on, the skillet’s handle heats up because of conduction.

Defining Heat

Conservation of Energy

• If changes in internal energy are taken into account along with changes in mechanical energy, the total energy is a universally conserved property.

• In other words, the sum of the changes in potential, kinetic, and internal energy is equal to zero.

CONSERVATION OF ENERGYPE + KE + U = 0

the change in potential energy + the change in kinetic energy + the change in internal energy = 0

Defining Heat

Sample Problem

Conservation of Energy An arrangement similar to the one

used to demonstrate energy conservation is shown in the figure. A vessel contains water. Paddles that are propelled by falling masses turn in the water. This agitation warms the water and increases its internal energy. The temperature of the water is then measured, giving an indication of the water’s internal energy increase.

Defining Heat

Sample Problem, continued

Conservation of Energy, continued If a total mass of 11.5 kg falls 1.3 m

and all of the mechanical energy is converted to internal energy, by how much will the internal energy of the water increase? (Assume no energy is transferred as heat out of the vessel to the surroundings or from the surroundings to the vessel’s interior.)

Defining Heat


1. DefineGiven:

m = 11.5 kgh = 1.3 mg = 9.81 m/s2

Unknown:U = ?

Defining Heat


2. PlanChoose an equation or situation: Use the conservation of energy, and solve for U.

PE + KE + U = 0(PEf – PEi) + (KEf – KEi) + U = 0

U = –PEf + PEi – KEf + KEi

Tip: Don’t forget that a change in any quantity, indicated by the symbol ∆, equals the final value minus the initial value.

Defining Heat

Sample Problem, continuedBecause the masses begin at rest, KEi equals zero. If we assume that KEf is small compared to the loss of PE, we can set KEf equal to zero also.

KEf = 0 KEi = 0Because all of the potential energy is assumed to be converted to internal energy, PEi can be set equal to mgh if PEf is set equal to zero.

PEi = mgh PEf = 0Substitute each quantity into the equation for ∆U:

∆U = –PEf + PEi – KEf + KEi

∆U = 0 + mgh + 0 + 0 = mgh

Defining Heat


4. EvaluateThe answer can be estimated using rounded values. If m ≈ 10 kg and g ≈ 10 m/s2, then ∆U ≈ 130 J, which is close to the actual value calculated.

3. CalculateSubstitute the values into the equation and solve:

U = mghU = (11.5 kg)(9.81 m/s2)(1.3 m)

U = 1.5 102 J

Changes in Temperature and Phase

Specific Heat Capacity

• The specific heat capacity of a substance is defined as the energy required to change the temperature of 1 kg of that substance by 1°C.

• Every substance has a unique specific heat capacity.

• This value tells you how much the temperature of a given mass of that substance will increase or decrease, based on how much energy is added or removed as heat.


Specific Heat Capacity, continued

• Specific heat capacity is expressed mathematically as follows:

cp QmT

specific heat capacity = energy transferred as heat

mass change in temperature

• The subscript p indicates that the specific heat capacity is measured at constant pressure.

• In this equation, T can be in degrees Celsius or in degrees Kelvin.

Specific Heat Capacities



Calorimetry

• Calorimetry is used to determine specific heat capacity.

• Calorimetry is an experimental procedure used to measure the energy transferred from one substance to another as heat.

A simple calorimeter allows the specific heat capacity of a substance to be determined.


Calorimetry, continued

Because the specific heat capacity of water is well known (cp,w= 4.186 kJ/kg•°C), the energy transferred as heat between an object of unknown specific heat capacity and a known quantity of water can be measured.

energy absorbed by water = energy released by substance

Qw = –Qx

cp,wmw∆Tw = –cp,xmx∆Tx


Sample Problem

Calorimetry A 0.050 kg metal bolt is heated to an unknown initial

temperature. It is then dropped into a calorimeter containing 0.15 kg of water with an initial temperature of 21.0°C. The bolt and the water then reach a final temperature of 25.0°C. If the metal has a specific heat capacity of 899 J/kg•°C, find the initial temperature of the metal.



1. DefineGiven: mm = 0.050 kg cp,m = 899 J/kg•°C

mw = 0.15 kg cp,w = 4186 J/kg•°C

Tw = 21.0°C Tf = 25.0°CUnknown: Tm = ?

Diagram:



2. PlanChoose an equation or situation: The energy absorbed by the

water equals the energy removed from the bolt.

Qw –Qmcp,wmwTw –cp,mmmTmcp,wmw (T f Tw ) –cp,mmm (T f Tm )

Rearrange the equation to isolate the unknown:

Tm cp,wmw (T f Tw )

cp,mmmT f




4. EvaluateTm is greater than Tf, as expected.

,

,

(4186 J/kg• C)(0.15 kg)(25.0 C 21.0 C)25.0 C

(899 J/kg C)(0.050 kg)81 C

( )p w w f wm f

p m m

m

m

c m T TT T

c m

T

T

Tip: Because Tw is less than Tf, you know that Tm must be greater than Tf.


Latent Heat

• When substances melt, freeze, boil, condense, or sublime, the energy added or removed changes the internal energy of the substance without changing the substance’s temperature.

• These changes in matter are called phase changes.• The energy per unit mass that is added or removed

during a phase change is called latent heat, abbreviated as L.

Q = mL energy transferred as heat during phase change = mass latent heat


Latent Heat, continued

• During melting, the energy that is added to a substance equals the difference between the total potential energies for particles in the solid and the liquid phases. This type of latent heat is called the heat of fusion, abbreviated as Lf.

• During vaporization, the energy that is added to a substance equals the difference in the potential energy of attraction between the liquid particles and between the gas particles. In this case, the latent heat is called the heat of vaporization, abbreviated as Lv.

Multiple Choice

1. What must be true about two given objects for energy to be transferred as heat between them?

A. The objects must be large.B. The objects must be hot.C. The objects must contain a large amount of

energy.D. The objects must have different temperatures.

Multiple Choice, continued

2. A metal spoon is placed in one of two identical cups of hot coffee. Why does the cup with the spoon have a lower temperature after a few minutes?

F. Energy is removed from the coffee mostly by conduction through the spoon.G. Energy is removed from the coffee mostly by convection through the spoon.H. Energy is removed from the coffee mostly by radiation through the spoon.J. The metal in the spoon has an extremely large specific heat capacity.


Use the passage below to answer questions 3–4.

The boiling point of liquid hydrogen is –252.87°C.

3. What is the value of this temperature on the Fahrenheit scale?

A. 20.28°FB. –220.87°FC. –423.2°FD. 0°F



The boiling point of liquid hydrogen is –252.87°C.

4. What is the value of this temperature in kelvins?

F. 273 KG. 20.28 KH. –423.2 KJ. 0 K


5. A cup of hot chocolate with a temperature of 40°C is placed inside a refrigerator at 5°C. An identical cup of hot chocolate at 90°C is placed on a table in a room at 25°C. A third identical cup of hot chocolate at 80°C is placed on an outdoor table, where the surrounding air has a temperature of 0°C. For which of the three cups has the most energy been transferred as heat when equilibrium has been reached?

A. The first cup has the largest energy transfer.B. The second cup has the largest energy transfer.C. The third cup has the largest energy transfer.D. The same amount of energy is transferred as heat for all three cups.


6. What data are required in order to determine the specific heat capacity of an unknown substance by means of calorimetry?

F. cp,water, Twater, Tsubstance, Tfinal, Vwater, Vsubstance

G. cp,substance, Twater, Tsubstance, Tfinal, mwater, msubstance

H. cp,water, Tsubstance, mwater, msubstance

J. cp,water, Twater, Tsubstance, Tfinal, mwater, msubstance


7. During a cold spell, Florida orange growers often spray a mist of water over their trees during the night. Why is this done?

A. The large latent heat of vaporization for water keeps the trees from freezing.B. The large latent heat of fusion for water prevents it and thus the trees from freezing.C. The small latent heat of fusion for water prevents the water and thus the trees from freezing.D. The small heat capacity of water makes the water a good insulator.


Use the heating curve to answer questions 8–10. The graph shows the change in temperature of a 23 g sample as energy is added to the sample as heat.

8. What is the specific heat capacity of the liquid? F. 4.4 105 J/kg•°CG. 4.0 102 J/kg•°CH. 5.0 102 J/kg•°CJ. 1.1 103 J/kg•°C



9. What is the latent heat of fusion? A. 4.4 105 J/kgB. 4.0 102 J/kg•°CC. 10.15 103 JD. 3.6 107 J/kg



10. What is the specific heat capacity of the solid? F. 1.85 103 J/kg•°CG. 4.0 102 J/kg•°CH. 5.0 102 J/kg•°CJ. 1.1 103 J/kg•°C

Short Response

Base your answers to questions 11–12 on the information below.

The largest of the Great Lakes, Lake Superior, contains 1.20 1016 kg of fresh water, which has a specific heat capacity of 4186 J/kg•°C and a latent heat of fusion of 3.33 105 J/kg.

11. How much energy would be needed to increase the temperature of Lake Superior by 1.0°C?

Short Response



11. How much energy would be needed to increase the temperature of Lake Superior by 1.0°C?

Answer: 5.0 1019 J

Short Response, continued



12. If Lake Superior were still liquid at 0°C, how much energy would need to be removed from the lake for it to become completely frozen?




12. If Lake Superior were still liquid at 0°C, how much energy would need to be removed from the lake for it to become completely frozen?Answer: 5.00 1021 J


13. Ethyl alcohol has about one-half the specific heat capacity of water. If equal masses of alcohol and water in separate beakers at the same temperature are supplied with the same amount of energy, which will have the higher final temperature?


13. Ethyl alcohol has about one-half the specific heat capacity of water. If equal masses of alcohol and water in separate beakers at the same temperature are supplied with the same amount of energy, which will have the higher final temperature?

Answer: the ethyl alcohol


14. A 0.200 kg glass holds 0.300 kg of hot water, as shown in the figure. The glass and water are set on a table to cool. After the temperature has decreased by 2.0°C, how much energy has been removed from the water and glass?

(The specific heat capacity of glass is 837 J/kg•°C, and that of water is 4186 J/kg•°C.)


14. A 0.200 kg glass holds 0.300 kg of hot water, as shown in the figure. The glass and water are set on a table to cool. After the temperature has decreased by 2.0°C, how much energy has been removed from the water and glass?

(The specific heat capacity of glass is 837 J/kg•°C, and that of water is 4186 J/kg•°C.)

Answer: 2900 J

Extended Response

15. How is thermal energy transferred by the process of convection?

Extended Response

15. How is thermal energy transferred by the process of convection?

Answer: The increasing temperature of a liquid or gas causes it to become less dense, so it rises above colder liquid or gas, transferring thermal energy with it.

Extended Response, continued

16. Show that the temperature –40.0° is unique in that it has the same numerical value on the Celsius and Fahrenheit scales. Show all of your work.


16. Show that the temperature –40.0° is unique in that it has the same numerical value on the Celsius and Fahrenheit scales. Show all of your work.

Answer:

TF 95

(–40.0C) 32.0 (–72.0 32.0)F –40.0F

Measuring Temperature


Determining Absolute Zero for an Ideal Gas



Calorimetry

Heat, Work, and Internal Energy

• Heat and work are energy transferred to or from a system. An object never has “heat” or “work” in it; it has only internal energy.

• A system is a set of particles or interacting components considered to be a distinct physical entity for the purpose of study.

• The environment the combination of conditions and influences outside a system that affect the behavior of the system.

Relationships Between Heat and Work

Heat, Work, and Internal Energy, continued

• In thermodynamic systems, work is defined in terms of pressure and volume change.


( )

work = pressure volume change

A FW Fd Fd Ad P VA A

W P V

• This definition assumes that P is constant.


• If the gas expands, as shown in the figure, V is positive, and the work done by the gas on the piston is positive.

• If the gas is compressed, V is negative, and the work done by the gas on the piston is negative. (In other words, the piston does work on the gas.)



• When the gas volume remains constant, there is no displacement and no work is done on or by the system.

• Although the pressure can change during a process, work is done only if the volume changes.

• A situation in which pressure increases and volume remains constant is comparable to one in which a force does not displace a mass even as the force is increased. Work is not done in either situation.


Thermodynamic Processes

• An isovolumetric process is a thermodynamic process that takes place at constant volume so that no work is done on or by the system.

• An isothermal process is a thermodynamic process that takes place at constant temperature.

• An adiabatic process is a thermodynamic process during which no energy is transferred to or from the system as heat.


Energy Conservation

• If friction is taken into account, mechanical energy is not conserved.

• Consider the example of a roller coaster:– A steady decrease in the car’s total mechanical energy

occurs because of work being done against the friction between the car’s axles and its bearings and between the car’s wheels and the coaster track.

– If the internal energy for the roller coaster (the system) and the energy dissipated to the surrounding air (the environment) are taken into account, then the total energy will be constant.

The First Law of Thermodynamics

Energy Conservation


Energy Conservation, continued

• The principle of energy conservation that takes into account a system’s internal energy as well as work and heat is called the first law of thermodynamics.

• The first law of thermodynamics can be expressed mathematically as follows:

U = Q – WChange in system’s internal energy = energy transferred to or from system as heat – energy

transferred to or from system as work


Signs of Q and W for a system


Sample Problem

The First Law of Thermodynamics A total of 135 J of work is done on a gaseous

refrigerant as it undergoes compression. If the internal energy of the gas increases by 114 J during the process, what is the total amount of energy transferred as heat? Has energy been added to or removed from the refrigerant as heat?



1. DefineGiven:

W = –135 JU = 114 J


Tip: Work is done on the gas, so work (W) has a negative value. The internal energy increases during the process, so the change in internal energy (U) has a positive value.

Diagram:

Unknown:Q = ?


2. PlanChoose an equation or situation: Apply the first law of thermodynamics using the values for U and W in order to find the value for Q.U = Q – W


Rearrange the equation to isolate the unknown:Q = U + W



Q = 114 J + (–135 J)

Q = –21 J


Tip: The sign for the value of Q is negative. This indicates that energy is transferred as heat from the refrigerant.


4. EvaluateAlthough the internal energy of the refrigerant increases under compression, more energy is added as work than can be accounted for by the increase in the internal energy. This energy is removed from the gas as heat, as indicated by the minus sign preceding the value for Q.


Cyclic Processes

• A cyclic process is a thermodynamic process in which a system returns to the same conditions under which it started.

• Examples include heat engines and refrigerators.

• In a cyclic process, the final and initial values of internal energy are the same, and the change in internal energy is zero.

Unet = 0 and Qnet = Wnet


Cyclic Processes, continued

• A heat engine uses heat to do mechanical work.

• A heat engine is able to do work (b) by transferring energy from a high-temperature substance (the boiler) at Th (a) to a substance at a lower temperature (the air around the engine) at Tc (c).


• The internal-combustion engine found in most vehicles is an example of a heat engine.

The Steps of a Gasoline Engine Cycle


The Steps of a Refrigeration Cycle


Thermodynamics of a Refrigerator


Efficiency of Heat Engines

• The second law of thermodynamics can be stated as follows:

No cyclic process that converts heat entirely into work is possible.

• As seen in the last section, Wnet = Qnet = Qh – Qc.– According to the second law of thermodynamics, W

can never be equal to Qh in a cyclic process. – In other words, some energy must always be

transferred as heat to the system’s surroundings (Qc > 0).

The Second Law of Thermodynamics

Efficiency of Heat Engines, continued

• A measure of how well an engine operates is given by the engine’s efficiency (eff ).

• In general, efficiency is a measure of the useful energy taken out of a process relative to the total energy that is put into the process.


• Note that efficiency is a unitless quantity. • Because of the second law of thermodynamics, the

efficiency of a real engine is always less than 1.

eff Wnet

QhQh –QcQh

1 QcQh

Sample Problem

Heat-Engine Efficiency Find the efficiency of a gasoline engine that, during

one cycle, receives 204 J of energy from combustion and loses 153 J as heat to the exhaust.


1. DefineGiven: Diagram: Qh = 204 J

Qc = 153 J

Unknown eff = ?


2. PlanChoose an equation or situation: The efficiency of a heat engine is the ratio of the work done by the engine to the energy transferred to it as heat.


eff Wnet

Qh1

QcQh




eff 1 QcQh

1 153 J204 J

eff 0.250

4. EvaluateOnly 25 percent of the energy added as heat is used by the engine to do work. As expected, the efficiency is less than 1.0.

Entropy

• In thermodynamics, a system left to itself tends to go from a state with a very ordered set of energies to one in which there is less order.

• The measure of a system’s disorder or randomness is called the entropy of the system. The greater the entropy of a system is, the greater the system’s disorder.

• The greater probability of a disordered arrangement indicates that an ordered system is likely to become disordered. Put another way, the entropy of a system tends to increase.


Entropy, continued

• If all gas particles moved toward the piston, all of the internal energy could be used to do work. This extremely well ordered system is highly improbable.


• Greater disorder means there is less energy to do work.

Entropy, continued

• Because of the connection between a system’s entropy, its ability to do work, and the direction of energy transfer, the second law of thermodynamics can also be expressed in terms of entropy change:

The entropy of the universe increases in all natural processes.

• Entropy can decrease for parts of systems, provided this decrease is offset by a greater increase in entropy elsewhere in the universe.


Energy Changes Produced by a Refrigerator Freezing Water


Because of the refrigerator’s less-than-perfect efficiency, the entropy of the outside air molecules increases more than the entropy of the freezing water decreases.

Multiple Choice

1. If there is no change in the internal energy of a gas, even though energy is transferred to the gas as heat and work, what is the thermodynamic process that the gas undergoes called?

A. adiabatic B. isothermalC. isovolumetricD. isobaric


2. To calculate the efficiency of a heat engine, which thermodynamic property does not need to be known?

F. the energy transferred as heat to the engine G. the energy transferred as heat from the engine

H. the change in the internal energy of the engine

J. the work done by the engine

3. In which of the following processes is no work done?

A. Water is boiled in a pressure cooker. B. A refrigerator is used to freeze water.

C. An automobile engine operates for several minutes.

D. A tire is inflated with an air pump.


4. A thermodynamic process occurs in which the entropy of a system decreases. From the second law of thermodynamics, what can you conclude about the entropy change of the environment?

F. The entropy of the environment decreases. G. The entropy of the environment increases. H. The entropy of the environment remains unchanged. J. There is not enough information to state what happens to the environment’s entropy.


Use the passage and diagrams to answer questions 5–8.A system consists of steam within the confines of a steam engine, whose cylinder and piston are shown in the figures below.

5. Which of the figures describes a situation in which U < 0, Q < 0, and W = 0?

A. (a)B. (b)C. (c)D. (d)



6. Which of the figures describes a situation in which U > 0, Q = 0, and W < 0?

F. (a)G. (b)H. (c)J. (d)



7. Which of the figures describes a situation in which U < 0, Q = 0, and W > 0?

A. (a)B. (b)C. (c)D. (d)



8. Which of the figures describes a situation in which U > 0, Q > 0, and W = 0?

F. (a)G. (b)H. (c)J. (d)


9. A power plant has a power output of 1055 MW and operates with an efficiency of 0.330. Excess energy is carried away as heat from the plant to a nearby river. How much energy is transferred away from the power plant as heat?

A. 0.348 109 J/sB. 0.520 109 J/s C. 0.707 109 J/s D. 2.14 109 J/s


10. How much work must be done by air pumped into a tire if the tire’s volume increases from 0.031 m3 to 0.041 m3 and the net, constant pressure of the air is 300.0 kPa?

F. 3.0 102 JG. 3.0 103 JH. 3.0 104 JJ. 3.0 105 J



An air conditioner is left running on a table in the middle of the room, so none of the air that passes through the air conditioner is transferred to outside the room.

11. Does passing air through the air conditioner affectthe temperature of the room? (Ignore the thermaleffects of the motor running the compressor.)

Short Response



11. Does passing air through the air conditioner affectthe temperature of the room? (Ignore the thermaleffects of the motor running the compressor.)

Answer: No, because the energy removed from the cooled air is returned to the room.

Short Response



12. Taking the compressor motor into account, whatwould happen to the temperature of the room?




12. Taking the compressor motor into account, whatwould happen to the temperature of the room?

Answer: The temperature increases.


13. If 1600 J of energy are transferred as heat to anengine and 1200 J are transferred as heat awayfrom the engine to the surrounding air, what is theefficiency of the engine?


13. If 1600 J of energy are transferred as heat to anengine and 1200 J are transferred as heat awayfrom the engine to the surrounding air, what is theefficiency of the engine?

Answer: 0.25


14. How do the temperature of combustion and thetemperatures of coolant and exhaust affect theefficiency of automobile engines?

Extended Response

14. How do the temperature of combustion and thetemperatures of coolant and exhaust affect theefficiency of automobile engines?

Answer: The greater the temperature difference is, the greater is the amount of energy transferred as heat. For efficiency to increase, the heat transferred between the combustion reaction and the engine (Qh) should be made to increase, whereas the energy given up as waste heat to the coolant and exhaust (Qc) should be made to decrease.

Extended Response

Use the information below to answer questions 15–18. In each problem, show all of your work.

A steam shovel raises 450.0 kg of dirt a vertical distance of 8.6 m. The steam shovel’s engine provides 2.00 105 J of energy as heat for the steam shovel to lift the dirt.


15. How much work is done by the steam shovel in lifting the dirt?




15. How much work is done by the steam shovel in lifting the dirt?

Answer: 3.8 104 J




16. What is the efficiency of the steam shovel?




16. What is the efficiency of the steam shovel?

Answer: 0.19




17. Assuming there is no change in the internal energy of the steam shovel’s engine, how much energy is given up by the shovel as waste heat?




17. Assuming there is no change in the internal energy of the

steam shovel’s engine, how much energy is given up by the shovel as waste heat?

Answer: 1.62 105 J




18. Suppose the internal energy of the steam shovel’s engine increases by 5.0 103 J.

How much energy is given up now as waste heat?




18. Suppose the internal energy of the steam shovel’s engine increases by 5.0 103 J. How much energy is given up now as waste heat?

Answer: 1.57 105 J

19. One way to look at heat and work is to think of energy transferred as heat as a “disorganized” form

of energy and energy transferred as work as an“organized” form. Use this interpretation to showthat the increased order obtained by freezingwater is less than the total disorder that resultsfrom the freezer used to form the ice.


19. One way to look at heat and work is to think of energy transferred as heat as a “disorganized” formof energy and energy transferred as work as an“organized” form. Use this interpretation to showthat the increased order obtained by freezingwater is less than the total disorder that resultsfrom the freezer used to form the ice.


Answer: Disorganized energy is removed from water to form ice, but a greater amount of organized energy must become disorganized to operate the freezer.

Entropy


Energy Changes Produced by a Refrigerator Freezing Water


Defining Temperature Temperature is a measure of the average kinetic energy of the particles in a...

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