chapter 17 work, heat, and the first law of thermodynamics · phys 212 s’14 chapter 17 –work,...

PHYS 212 S’14 CHAPTER 17 – WORK, HEAT, & FIRST LAW OF THERMODYNAMICS

CHAPTER 17WORK, HEAT, and the

FIRST LAW OF THERMODYNAMICS

In this chapter, we will examine various thermal properties of matter, as well as several mechanisms by which energy

can be transferred to and from a system without the use of mechanical work.

Slide 1

CHAPTER 17 – WORK, HEAT, & FIRST LAW OF THERMODYNAMICSPHYS 212 S’14

17.1 It’s All About Energy

Slide 2

Most of chapter 17 is concerned with tracking changes in the total energy of a system. From PHYS 211, we know that

∆𝐸sys = ∆𝐾 + ∆𝑈 + ∆𝐸th = 𝑊ext

(in chapter 15, we omitted the ∆𝐸th term…why?)

We also know that ∆𝐾 + ∆𝑈 defines the change in mechanical energy∆𝐸mecof the system, and thus

∆𝐸sys = ∆𝐸mec + ∆𝐸th = 𝑊ext

In most thermodynamic processes of interest,𝑊ext is not equal to zero – that is, the system is not isolated. We perform external work on it by heating, cooling, pressurizing, etc. We wish to determine how energy is transferred between the system and its environment.



Slide 3

Two examples from PHYS 211 are shown here. We will discuss them in class.



Slide 4

Now let’s examine a situation that was not covered in PHYS 211. A pot of water is placed on the stove, and the burner is turned on. The water’s temperature increases (∆𝐸th > 0), but its mechanical energy remains the same (∆𝐸mec = 0). Furthermore, no external work is done on the water (since there’s neither a force nor a displacement). We must conclude that there’s a problem with the last equation on slide 10.As it turns out, there is more than one way for a system to exchange

energy with its environment (other than through the mechanical interaction 𝑊ext). There can also be a thermal interaction. The energy transferred by a thermal interaction is called heat, and has the symbol 𝑄:

∆𝐸sys = ∆𝐸mec + ∆𝐸th = 𝑊ext + 𝑄


17.2 Work in Ideal-Gas Processes

Slide 5

Again, let’s step back to PHYS 211 to remind ourselves how to calculate mechanical work.If a force 𝐹 acts on an object through a displacement ∆𝑠 = 𝑠𝑓 − 𝑠𝑖,

the work done on the object is

𝑊 = 𝑠𝑖

𝑠𝑓

𝐹 𝑑𝑠

(in most cases in PHYS 211, 𝐹 was constant or a trivial function of position, so you didn’t actually have to perform an integral).



Slide 6

Consider a gas as it is compressed or expanded by a moveable piston of area A.In the top figure, the piston is held in placed

by 𝐹ext, which must balance the force 𝐹gasarising due to the gas pressure:

𝐹ext = −𝐹gas = −𝑝𝐴

If the piston moves outward by a small distance 𝑑𝑥, the external force will do an amount of work equal to

𝑑𝑊 = 𝐹ext𝑑𝑥 = −𝑝𝐴 𝑑𝑥

If 𝑑𝑥 is positive (i.e. the gas is expanding), then 𝑑𝑊 is negative, since the force and the displacement are in opposite directions.



Slide 7

Now, as the piston moves a distance 𝑑𝑥, the volume of the gas changes by 𝑑𝑉 = 𝐴 𝑑𝑥. From the last equation on the previous slide, we then have

𝑑𝑊 = −𝑝𝑑𝑉

Finally, as the volume changes from 𝑉𝑖 to 𝑉𝑓,

the amount of work done on the gas is

𝑊 = − 𝑉𝑖

𝑉𝑓

𝑝 𝑑𝑉

Note that as the volume changes, 𝑝 generally changes as well. It is not a constant, and can not be brought out of the integral.



Slide 8

If we have already drawn a 𝑝𝑉 diagram for a particular process, we can find the work done by examining the area under the curve. This will be discussed in class.



Slide 9

For an isochoric process, the volume doesn’t change, so

𝑊 = 0

For an isobaric process, the pressure doesn’t change…this makes for an easy integral!

𝑊 = −𝑝∆𝑉



Slide 10

For an isothermal process, we have to do a bit of work (pun intended). The ideal-gas law helps here. We know that

𝑝 =𝑛𝑅𝑇

𝑉, where 𝑛 and 𝑅 are constants.

Also, in this case, 𝑇 is a constant. Thus,

𝑊 = − 𝑉𝑖

𝑣𝑓

𝑝 𝑑𝑉 = − 𝑉𝑖

𝑣𝑓 𝑛𝑅𝑇

𝑉𝑑𝑉 = −𝑛𝑅𝑇

𝑉𝑖

𝑣𝑓 𝑑𝑉

𝑉= −𝑛𝑅𝑇 ln

𝑉𝑓𝑉𝑖

Finally, since 𝑛𝑅𝑇 = 𝑝𝑖𝑉𝑖 = 𝑝𝑓𝑉𝑓 for an isothermal process, we can

write

𝑊 = −𝑝𝑖𝑉𝑖 ln𝑉𝑓𝑉𝑖

= −𝑝𝑓𝑉𝑓 ln𝑉𝑓𝑉𝑖



Slide 11

Path DependenceThe “area-under-a-curve” interpretation of the work done on a gas during an ideal-gas process should convince you that 𝑊 depends not just on the initial and final states, but also on how the gas transitioned between these states (i.e. on the path between 𝑖 and 𝑓).


Problem #1: Path Dependence

Slide 12

RDK STT 17.2

Two processes take an ideal gas fromstate 1 to state 3. Compare the workdone on the gas by process A to that done by process B.

A 𝑊𝐴 = 𝑊𝐵 = 0

B 𝑊𝐴 = 𝑊𝐵, but neither is zero

C 𝑊𝐴 > 𝑊𝐵

D 𝑊𝐴 < 𝑊𝐵


Problem #2: Isobaric Process

Slide 13

80 J of work are done on the gas in the process shown in the figure. What is 𝑉1 in units of cm3?

Solution: in class

RDK EX 17.3


17.3 Heat

Slide 14

“Heat” is not a substance. It’s energy. Specifically, it’s the energy transferred between a system and its environment as a consequence of a temperature difference between them. For example, we can increase the temperature of a container of water by placing it over a flame, or by rapidly spinning a paddle in it. The former achieves a transfer of energy via heat, in what is termed athermal interaction, while the latter achieves it by mechanical work (which increases the kinetic energy of the water molecules, bringing us back to PHYS 211).

The details of the heat transfer mechanisms will be discussed in chapter 18.


17.3 Heat

Slide 15

The sign of heat (𝑄) depends on the direction of energy flow, which is from hot to cold. If a system is colder than its environment, heat flows into the system, and 𝑄 > 0. If the system is hotter than its environment, heat flows out of the system, and 𝑄 < 0. In thermal equilibrium, the system and its environment are at the same temperature. Here, 𝑄 = 0 (there is no energy flow).

Since heat is energy, it must have units of Joules. In some branches of science, it is measured in calories (1 cal = 4.186 J). For added confusion, the “calorie” that we know of in relation to food is actually a kilocalorie (4186 J).


17.3 Heat

Slide 16

We must distinguish among heat, temperature, and thermal energy:• Thermal energy 𝐸th is a quantity of energy of a system due to the

motion of its atoms. It is a state variable, so we are sometimes interested in how𝐸thchanges during a process.

• Heat 𝑄 is energy transferred between a system and its environment as they interact. It is not a property of the system itself, and therefore is not a state variable. If a system does not interact thermally with its environment, then 𝑄 = 0. If it does interact with its environment, then it is possible that 𝑄 ≠ 0. Furthermore, it is possible that heat may cause𝐸thto change.

• Temperature is a state variable that quantifies the “hotness” of a system. It is related to the thermal energy per molecule. A temperature difference between system and environment causes 𝑄 ≠ 0.


Problem #3: Heat or Not Heat?

Slide 17

RDK STT 17.3

Which of the following processes involves heat?

A The brakes in your car get hot when you stop

B A steel block is held over a candle

C You push a rigid cylinder of gas across a frictionless surface

D You push a piston into a cylinder of gas, increasing the temperature of the gas

EYou place a cylinder of gas in hot water. The gas expands, causing a piston to rise and lift a weight. The temperature of the gas does not change


17.4 The First Law of Thermodynamics

Slide 18

As we learned earlier in this chapter, a complete statement of the law of conservation of energy requires a heat term:

∆𝐸sys = ∆𝐸mech + ∆𝐸th = 𝑊 +𝑄

For our study of thermodynamics, we are not interested in systems that have any macroscopic degree of motion. That is, we will assume ∆𝐸mech = 0, and therefore claim that

∆𝐸th = 𝑊 + 𝑄

This is a statement of the first law of thermodynamics. It’s illustrated by the figure to the right, which we will discuss in class.



Slide 19

Three Special Ideal-Gas ProcessesIt is possible to define processes in which one of the three terms in the first law (∆𝐸th,𝑊, or 𝑄) is zero. The setup shown here has three properties:1. A locking pin can hold the piston in

place, keeping volume constant.2. Adding or removing masses on top

of the piston can change the pressure. Work is done on the masses as they move.

3. The gas can be warmed or cooled through the bottom surface. Other surfaces are insulated.



Slide 20

When the piston is unlocked but stationary, the gas pressure is related to the mass 𝑀 as

𝑝gas = 𝑝0 +𝑀𝑔

𝐴where 𝐴 is the area of the piston. When the piston is locked, there is no such relation.

We are now ready to examine the three aforementioned processes…



Slide 21

Isochoric Cooling As we learned previously, no work is done in an isochoric process, since the volume doesn’t change. This is achieved as follows:• the locking pin is inserted• the cylinder is placed on a block of

ice…this produces a 𝑄 < 0…the gas temperature and pressure fall

• the cylinder is removed from the ice when the desired pressure is reached

• masses are removed from the piston until the desired pressure is reached

• the locking pin is removed

Overall: 𝑊 = 0, 𝑄 < 0, ∆𝐸th < 0.



Slide 22

Isothermal Expansion This is achieved as follows:• the cylinder is placed over the

flame…this produces a 𝑄 > 0…the gas will begin to expand

• since 𝑝𝑉 must be constant for an isothermal process, remove masses as the gas expands in order to reduce the gas pressure…the temperature remains constant, since heat energy balances the negative work done on the expanding gas

• when the desired volume is reached, remove the cylinder from the flame.

Overall: 𝑊 < 0, 𝑄 > 0, ∆𝐸th = 0.



Slide 23

Adiabatic Compression An adiabatic process is one in which no heat energy is transferred between the system and its environment. • add insulation beneath the cylinder• add masses to the piston…this will

increase the pressure…the piston will descend, decreasing the volume

• stop adding masses when the desired volume is reached

Overall: 𝑊 > 0, 𝑄 = 0, ∆𝐸th > 0.


Problem #4: First-Law Processes

Slide 24

RDK STT 17.4

Which first-law bar chart describesthe process shown in this 𝑝𝑉 diagram?

A B

C D


Problem #5: Thermal Processes

Slide 25

A gas is compressed from 600 cm3 to 200 cm3 at a constant pressure of 400 kPa. At the same time, 100 J of heat energy is transferred out of the gas. What is the change in thermal energy of the gas during this process?

Solution: in class

RDK EX 17.9


17.5 Thermal Properties of Matter

Slide 26

We’re now aware that the thermal energy of a system can increase or decrease. This can have two effects – it can change the temperature of the system, or it can change its phase (i.e. the system can melt, boil, etc.)In the first case, the natural question is: what quantity of thermal energy increase corresponds to a given temperature increase? It’s not hard to predict that the temperature change depends on the amount of the substance – that is, if a certain ∆𝐸th produces a 1 K temperature increase in 1 liter of water, then it should produce a 2 K temperature increase in ½ liter of water (since the thermal energy per molecule) is twice as great.However, it turns out that the relation between temperature change and thermal energy also depends entirely on which material is being heated or cooled. This leads to the concept of specific heat.



Slide 27

The specific heat of a substance is the amount of energy required to raise the temperature of 1 kg of the substance by 1 K. It has the symbol c, and SI units of J/kg·K.

Based on this description, we can write the relation between thermal energy change and temperature change as

∆𝐸th = 𝑀𝑐∆𝑇

We will examine the specific heat of gases in section 17.7.

1 calorie / g·K



Slide 28

From the First Law, we can write𝑊 +𝑄 = 𝑀𝑐∆𝑇

For liquids and solids, we’re almost always changing the thermal energy by heating, in which case 𝑊 = 0 (since liquids and solids are “incompressible”, we can’t really perform work on them by changing their volume). In this case, we can write

𝑄 = 𝑀𝑐∆𝑇

The molar specific heat is the thermal energy required to change the temperature of 1 mole of a substance (rather than 1 kg) by 1 K. It has a symbol C and units of J/mol·K. The relevant relation is

𝑄 = 𝑛𝐶∆𝑇According to the table, for elemental solids, C is always very close to 25 J/mol·K. We’ll find out why this is in the next chapter.



Slide 29

Phase Change and Heat of TransformationFrom the previous slide, it should be clear that if we plot 𝑇 vs. 𝑄 for any substance, the graph should appear as a straight line with slope 1/𝑀𝑐. However, plotting actual experimental data results in the figure shown here.

The first interesting property of this plot is that the solid, liquid, and gas phases of a substance have a different specific heat(notice that the slopes are not identical). Even more interesting is that there are regions of zero slope, corresponding to the melting and boiling temperatures, 𝑻𝒎 and 𝑻𝒄.



Slide 30

Phase Change and Heat of Transformation cont’Physically, what is happening in these latter regions is that the additional heat energy does not raise the temperature of the material; rather, it is used to break molecular bonds in order to produce the phase change.The heat of transformation (𝑳, units of J/kg) is the amount of heat energy required to cause a phase change in 1 kg of material. A subscript indicates if the phase change is freezing/melting (𝑓, “heat of fusion”) or boiling/condensing (𝑣, “heat of vaporization”). The heat needed to cause a phase change in a mass 𝑀 of a substance is therefore 𝑄 = ±𝑀𝐿𝑓 (melt / freeze)

𝑄 = ±𝑀𝐿𝑣 (boil / condense)(the negative sign indicates that heat must be removed from the system in order to freeze or condense the substance).



Slide 31

Phase Change and Heat of Transformation cont’A few heats of transformation are shown here. Notice that 𝐿𝑣 is always a few times larger than 𝐿𝑓.


Problem #6: Boiling Booze

Slide 32

What is the maximum mass of ethyl alcohol you could boil with 1000 J of heat, starting from 20°C?

Solution: in class

RDK EX 17.15


17.6 Calorimetry

Slide 33

Calorimetry is a fancy term that refers to the science of measuring the heat of physical or chemical reactions. The latter is done in most introductory (or high school) chemistry labs; we’ll be concerned with the former.Experiments in calorimetry involve systems or collections of systems that are thermally isolated. This means that the net heat 𝑄net

is zero. There’s really not much more to say…it’s best explained by working through a problem.


Problem #7: Hot Pan!

Slide 34

A 750 g aluminum pan is removed from the stove and plunged into a sink filled with 10.0 L of water at 20.0°C. The water temperature quickly rises to 24.0°C. What was the initial temperature of the pan, in °C?

Useful data: 𝑐Al = 900 J/kg·K, 𝑐H2O = 4190 J/kg·K

Solution: in class

RDK EX 17.17


17.7 The Specific Heats of Gases

Slide 35

Our previous discussion of specific heat was concerned only with solids and liquids. This is because the relation between heat and temperature change of a gas is more complicated – as we have seen, there are many mechanisms by which the temperature of a gas can change.

As an example, the figure shows an isochoric process (A) and an isobaric process (B), both of which have the same initial and final temperatures. Since work is done in process B but not in process A, a different amount of heat is required in the two cases. Thus, we require a different “version” of specific heat for the two cases.



Slide 36

Since we usually deal with moles rather than mass when performing gas calculations, we will only be concerned with the molar specific heats.

As before, the quantity of heat needed to raise the temperature of 𝑛moles of gas by ∆𝑇 is

𝑄 = 𝑛𝐶V∆𝑇 (constant volume)𝑄 = 𝑛𝐶P∆𝑇 (constant pressure)

For a general gas process, in which neither volume nor pressure is constant, there is no way to define specific heat.

isobaric isochoric R



Slide 37

Path Dependence of HeatWe have already seen that the work done in changing the state of a gas depends on the path taken on the 𝑝𝑉 diagram. In fact, the heat required for the state change depends on the path as well.

This can be shown using the First Law, ∆𝐸th = 𝑊 +𝑄. Recall that

∆𝐸th is a state variable, and thus it depends on the state of the gas, and not the process by which the gas arrived at that state. Therefore, 𝑊𝐴 + 𝑄𝐴 = 𝑊𝐵 + 𝑄𝐵. But we already know that W is path-dependent (𝑊𝐴 ≠ 𝑊𝐵) Therefore, 𝑄𝐴 ≠ 𝑄𝐵. The heat added or removed during an ideal-gas process depends on the path followed through the 𝒑𝑽 diagram.



Slide 38

Adiabatic ProcessesRecall that an adiabatic process is one for which 𝑄 = 0 (no heat energy is transferred). This can be achieved by thoroughly insulating a system from its environment. Alternatively, it can come about by expanding or compressing a gas rapidly enough that heat transfer –while allowed – simply doesn’t have enough time to occur.

The relationship of isothermal, isochoric, and adiabatic processes to the First Law is shown here:



Slide 39

Adiabatic Processes cont’For any ideal-gas process, it can be shown (see p. 487) that

∆𝐸th = 𝑛𝐶V∆𝑇Since ∆𝐸th = 𝑊 for an adiabatic process, it follows that

𝑊 = 𝑛𝐶V∆𝑇 (adiabatic process)

This provides a fourth description of work, along with𝑊 = 0 (isochoric process)𝑊 = −𝑝∆𝑉 (isobaric process)

𝑊 = −𝑝𝑖𝑉𝑖 ln𝑉𝑓𝑉𝑖

= −𝑝𝑓𝑉𝑓 ln𝑉𝑓𝑉𝑖

(isothermal process)



Slide 40

Adiabatic Processes cont’Finally, we must examine the trajectory of an adiabatic process on a 𝑝𝑉 diagram. In order to do this, we start by defining the specific heat ratio:

𝛾 =𝐶𝑝

𝐶𝑉=

1.67 monatomic gas1.40 diatomic gas

(these numbers represent the fractions 5/3 and 7/5, respectively…there’s some slick physics in there, but it’s well beyond the scope of PHYS 212).

Then, an adiabatic process is one for which 𝑝𝑉𝛾 is constant; i.e.

𝑝𝑖𝑉𝑖𝛾= 𝑝𝑓𝑉𝑓

𝛾. The resulting curve is called an adiabat. There’s a

proof of this result on p. 490 of the text. Don’t sweat it.


Problem #8: Adiabatic Expansion

Slide 41

A gas cylinder holds 0.10 mol of O2 at a temperature of 150°C and a pressure of 3.0 atm. The gas expands adiabatically until the pressure is halved.a) What is the final volume of the gas?b) What is the final temperature of the gas?c) How would these answers change if the gas was N2?

Solution: in class

RDK EX 17.25


17.8 Heat-Transfer Mechanisms

Slide 42

Much of this chapter has involved the transfer of heat. In this last section of the notes, we will investigate the various physical mechanisms by which this transfer occurs – conduction, convection, and radiation. A fourth mechanism – evaporation – was already described earlier in these notes (heat of vaporization).



Slide 43

ConductionThe figure shows a material sandwiched between a higher temperature 𝑇𝐻 and a lower temperature 𝑇𝐿, resulting in a transfer of thermal energy from the hot side to the cold side. The microscopic mechanism of conduction will be described in class.An experimental observation of conduction reveals two properties:

i. More heat is transferred if ∆𝑇 = 𝑇𝐻 − 𝑇𝐿 is larger

ii. Materials with a larger cross-section 𝐴transfer more heat, while those with a larger length 𝐿 transfer less heat



Slide 44

Conduction cont’If a heat 𝑄 is transferred in a time interval ∆𝑡, then the rate of heat transfer is 𝑄/∆𝑡 (units of J/s…recall from PHYS 211 that a rate of energy transfer is a power, with units of Watts). Based on our experimental observations, we can write

𝑄

∆𝑡= 𝑘

𝐴

𝐿∆𝑇

Here, 𝑘 is the thermal conductivity of the material (SI units: W/m·K); a large value of 𝑘means that the material is a good conductor of heat.


Problem #9: Leaky Window

Slide 45

A glass window of area 2.0 m2 has a thickness of 4.0 mm. If the indoor temperature is 25°C and the outdoor temperature is 0°C, how much heat energy transfers through the window in 1 hour?

Solution: in class



Slide 46

ConvectionIn this process, thermal energy is transferred by the physical motion of a fluid. For example, when water is heated in a pot, the water that contacts the bottom surface is heated by conduction, but then it rises to the surface via convection.Mathematically, convection is too complex to describe in any detail in PHYS 212. Just be aware that it exists!

Aerogels (shown here) are extraordinarily good insulators, since they almost entirely prevent both conduction and convection.



Slide 47

RadiationThere’s ample proof that another mechanism of heat transfer must exist in addition to conduction and convection. Although there is essentially no material between the sun and the earth with which to conduct or convect thermal energy, we can still feel the sun’s heat.

Radiation here refers to any process by which electromagnetic energy is generated at one location and “beamed” to another. It requires no intervening medium – and in fact, it’s more efficient if there’s no medium at all.The properties of this electromagnetic energy depend on an object’s temperature. This is useful for satellite imaging of ocean temperature, for example.



Slide 48

Radiation cont’The rate of energy transfer by radiation (i.e. the radiated power) depends on an object’s temperature and on its surface area:

𝑄

∆𝑡= 𝑒𝜎𝐴𝑇4, where 𝜎 = 5.67 × 10−8 W/m2 ∙ K4

σ is known as the Stefan-Boltzmann constant. Here, 𝑒 is the emissivity of the surface, a unitless number between 0 and 1. It is a measure of how effectively the surface radiates. When 𝑒 = 1, the radiation is referred to as black-body radiation.



Slide 49

Radiation cont’All objects that emit radiation also absorb radiation from their surroundings. Therefore, we are more interested in the net rate of energy transfer,

𝑄net

∆𝑡= 𝑒𝜎𝐴(𝑇4 − 𝑇0

4)

where 𝑇 and 𝑇0 are the temperatures of the object and its environment, respectively. This equation indicates that an object in thermal equilibrium with its surroundings undergoes no net thermal radiation, while an object that is colder than its surroundings has a negative value of 𝑄net. Be careful about the sign convention here. Since radiation, by definition, refers to energy leaving an object, a positive 𝑄net refers to a net energy transfer out of the object.


Problem #10: Lead Sphere

Slide 50

What maximum power can be radiated by a 10-cm-diameter solid lead sphere? Assume that 𝑒 = 1.

Useful information: the surface area of a sphere of radius 𝑟 is 𝐴 = 4𝜋𝑟2

More useful information: you’ll need some data from a table elsewhere in these notes

Solution: in class

RDK EX 17.29

chapter 17 work, heat, and the first law of thermodynamics · phys 212 s’14 chapter 17 –work,...

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