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Physics Factsheet January 2003

Number 44

Heat Engines and Heat Pumps These devices are direct applications of the principles of thermodynamics and you should refer to Factsheets 25 (the Molecular Kinetic Theory) and 31 (the First Law of Thermodynamics).

Work done When a gas expands it pushes back the surroundings and does work on the surroundings – an expansion means that the state moves in the direction AB on the curve.

The first law of thermodynamics can be written as: ∆U = W +Q This means that: increase in internal = work done + heat supplied energy of a system on the system to the system

When a gas is compressed work is done on the gas and the state moves in the direction BA. The word done when the gas changes between two states is given by the area under the curve between those two states: p A

for example between states X and Y work done = shaded area X

The ideal gas equation pV = nRT summaries Boyle’s, Charles’ and Avogadro’s laws for an ideal gas – i.e. one that obeys the rules of the kinetic theory of gases.

Y B V

State variables p, V and T are known as state variables because knowledge of two of these values fully gives the state of the ideal gas.

In an isothermal change there is no change in internal energy and so the work done on the gas must equal the heat supplied to the surroundings by the gas or else the heat supplied to the gas must cause the gas to do an equal amount of work on the surroundings. The first law of thermodynamics shows W = -Q (or Q = -W)

A graph of p against V will give a curve for which there can only be one value of T for an ideal gas. p A

State changes other than isothermal can occur under different changes of conditions. An isobaric change occurs when the volume changes under constant pressure. p

p

B

V V1

This curve is called an isothermal and shows that p ∝ V -1. Isothermal changes occur very slowly so that the gas is allowed to return to its original temperature before any further expansion or compression occurs. The container enclosing the gas needs to be a good conductor to allow heat to conduct through its walls.

V

2

V

The work done is the area of the shaded rectangle. An adiabatic change occurs when no heat enters or leaves the system. This must be done very quickly and the walls of the container must be very good insulators to prevent heat loss or gain. Adiabatic curves between two volumes are steeper than isothermals. The work done is still the area between the curve and the V-axis.

Typical Exam Question An ideal gas in a sealed, fixed volume container is at a temperature of 25oC. What will be its temperature when the pressure is doubled?

p

Answer: using pV = nRT, everything except p and T is constant so if the pressure doubles so must the absolute temperature.

adiabatic isothermal

The temperature in K = 25+ 273 = 298 K so the new temperature must equal 596 K or 596-273 = 323oC V

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Physics Factsheet

Heat Engines and Heat Pumps Applying the first law of thermodynamics to adiabatic changes: ∆U = W + Q but since Q = 0, ∆U = W (or -∆U = -W) so either there is an increase of internal energy equal to the work done on the gas or the work done by the gas on the surroundings will equal the decrease in internal energy.

Heat Engines These are devices which convert heat into mechanical energy (do work on the surroundings) and then exhaust the heat that isn’t used to do work. The operation of a heat engine can be described on a p-V diagram showing a cycle: p

Isometric changes occur at constant volume and so no work is done during these changes.

p1

A

B

p2

D

C

p p1

V

p2

V1 V2 Starting at A: • along AB the gas does work on the surroundings in expanding and since the temperature will also rise (pV/nR is larger) this means that the gas must be heated too. • Along BC no work is done – since there is no change in volume; this means that the gas loses heat to the surroundings in order to lower its temperature (and thus internal energy). • Along CD the gas has work done on it by the surroundings in being compressed and the fall in temperature (and internal energy) indicates that the gas heats the surroundings by cooling. • Finally, to complete the cycle, along DA again no work is done – this means that the gas is heated to raise its temperature (and thus internal energy). This shows that there is a net amount of work done on the surroundings.

V

V

Again applying the first law of thermodynamics to this change: ∆U = W + Q but since W = 0, ∆U = Q or -∆U = -Q meaning that an heat added to the gas raises its internal energy (and temperature) whilst heat given out by the gas lowers its internal energy (and temperature). Typical Exam Question A rigid steel pressure cooker is heated by the flames of a gas hob. Explain how the first law of thermodynamics applies to the contents of the pressure cooker. The pressure is held constant by a value and you may ignore any steam leaving through the valve. Answer: Since the pressure is constant no work is done on or by the contents of the pressure cooker. Heat conducting to the contents through the pressure cooker and its internal energy must increase. This means that since W = 0, ∆U = Q.

Operating the cycle the other way round, means that the “engine” is now behaving as a heat pump. Typical Exam Question A fixed mass of an ideal gas initially at 0 oC and a pressure of 1.1 × 105 Pa expands isothermally to twice its original volume before being compressed isobarically back to its original volume. (a) Sketch the changes on a p -V diagram for the gas.

Cycles A cycle is a series of changes which returns a gas (or “working substance”) to its original state. When this is sketched or plotted on a p – V graph the area enclosed by the shape that results from the cycle amounts to the net (overall) amount of work done either on or by the gas. Whether it is “on” or “by” depends on the direction of the cycle.

(b) (i) On your sketch shade the net work done. (ii) State whether the net work was done by or on the gas. (c) Calculate the final temperature of the gas.

p

p1

A

B

Answer (a) (b) (i)

p2

D

1.1

C V

V1

p/Pa × 105

0.55

V2

This cycle consists of two isometric changes and two isobaric changes. In going round the cycle in the direction ABCDA the work done on expansion (area under AB) is greater than the work done under compression (area under CD) so this means that the area enclosed gives the net work done by the gas on the surroundings. This makes the overall value of W negative in the first law of thermodynamics.

Vi

2Vi

V

(b) (ii) since the area under the expansion is greater than the area under the compression, work is being done by the gas overall.

The cycle ADCBA would result in a positive value for W since net work is being done on the gas.

(c) in expansion T remains at 273 K; in the compression pV=nRT V/T = constant 273 so the temperature must fall to 2 = 136(.5)K

The direction around a cycle determines whether work is done on the gas or by the gas. If the higher pressure volume change is an expansion then the net work is done by the gas.

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Physics Factsheet

Heat Engines and Heat Pumps Heat engines can also be described by “reservoir” models:

This series of changes produces the following cycle: p

Hot reservoir QH

isothermal adiabatic

W adiabatic isothermal

QC V

Cold reservoir Heat Pumps Although heat cannot spontaneously flow from a region of lower temperature to one at a higher temperature, it can be made to do so if enough work is done on the system. Refrigerators and other heat pumps operate in a reverse cycle to heat engines, meaning that net work is done on the system . In the refrigerator this causes the internal energy to fall and is brought about by the refrigerant liquid undergoing a phase change by evaporating thereby extracting energy from a cold area. The refrigerant is then compressed and forced to condense in a hotter area of the refrigerator meaning that latent heat of vaporisation is given out there.

This shows that the engine takes heat QH from the hot reservoir and does work W on the surroundings and rejects heat QC to the cold reservoir. useful work obtained heat input from hot reservoir QH − Qc W = = QH QH

Thermal efficiency =

A reservoir model for a heat pump is of the form: Typical Exam Question An engine takes in 50 kJ from a hot reservoir and rejects 35 kJ to the cold reservoir. Calculate: (a) the work done by the motor (b) the thermal efficiency of the engine.

Hot reservoir

QH

Answer (a) difference between input and out put heat = work done = 15kJ (b) thermal efficiency = 15/50 = 30% W QC

The Second Law of Thermodynamics and Heat Engines There are several ways of stating this law but the most appropriate to a heat engine for the purposes of A level can be stated as: “it is impossible to do an amount of work equal to the heat extracted from the hot reservoir. Some of the heat must be exhausted to the cold reservoir. This means that it is impossible to build a perfect heat engine.

Cold reservoir The effectiveness of a heat pump is given by its coefficient of performance:

The Carnot Cycle This is an idealised cycle and represents the maximum possible efficiency for a heat engine such that the efficiency is given by TH − QC efficiency = TH where T’s are absolute temperatures and the Hs and Cs refer to the hot and cold reservoirs.

coefficient of performance =

and in the ideal case of the Carnot cycle this is

cylinder

QH

Entropy The second law of thermodynamics is often stated in terms of the entropy of the system: in any cyclic process the entropy will either increase or remain the same. Entropy is a measure of the disorder of the system and is a state variable defined for a reversible process at absolute temperature T when amount of heat Q is absorbed by the Q where ∆S is the change in entropy in units of JK-1. relationship:∆S = T

piston

QC

Hot reservoir TH

Insulated stand

Cold reservoir TC

Insulated stand

isothermal expansion

adiabatic expansion

isothermal compression

adiabatic compression

TH − QC

. TH For a refrigerator it is the amount of heat that is extracted from the cold reservoir that matters more than the amount of heat rejected to the hot reservoir and so in this case the coefficient of performance is given by: Q Qc coefficient of performance = c = W QH − Qc

This cycle consists of an isothermal expansion followed by an adiabatic expansion, an isothermal compression and finally an adiabatic compression. We can visualise this as being applied to a gas trapped inside a cylinder by a moveable piston: gas

QH − Qc W = QH QH

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Physics Factsheet

Heat Engines and Heat Pumps Exam Workshop (a) State what is meant by an adiabatic change. no energy is lost to the surroundings!

(iii) The temperature of the isothermal change A is 450 K. Calculate the temperature of the second isothermal change C. [2]

[1] 0/1

P1V1 P2V2 6 × 10 −5 0.11×106 × 2 ×10 −4 = = = 450 T1 T2 T1

No it’s heat not energy i.e.doesn’t include work done – think of the first law of thermodynamics

T1 = 1.227 × 10-3 K "!

(b) The diagram shows an engine cycle in which there are two isothermal changes and two changes at constant volume.

1/2

Right method but wrong answer – largely because of the wrong value for p1V1. Since the power is wrong it means that the temperature is wrong. (correct answer is 1.23 × 103 K) The student could also be penalised for using too many significant figures – use 2 or 3 normally.

p/MPa

0.80

(iv) Estimate the net energy output to the surroundings.

B

C

0.30

This is a good answer but where is the working? [On other occasions you would be given a grid and be able to work out the energy equivalent of one square and, by counting the squares estimate the total energy].

D

0.11

A 75

[4] 2/4

area enclosed = 37.5 J""!!

200

V/10−6m3

Examiner’s Answers (a) a change in which no heat enters or leaves the system " (b) (i) A" Compression takes place, which requires an external force" (ii) Initially p = 0.8 × 106 Pa, V = 75 × 10-6 m3 ⇒ pV = 60 Pa m" Finally p = 0.3 × 106 Pa, V = 200 × 10-6 m3 ⇒ pV = 60 Pa m" Since pV unchanged, C is isothermal (" for either calculation, " for both correct answers) max 2 marks

(i) State and explain which change in the cycle corresponds to work having been done by the working substance. [2] A because there is a decrease in volume. "!

1/2

Yes but continue to explain this by saying that this means that an external force (from the surroundings)is doing work in compressing the working substance

(iii) (ii) Using the data on the graph show that change C is approximately isothermal [2] pV = constant = 0.8 × 75 × 10-6 = 6 × 10-5!!

P1V1 P2V2 = T1 T2

0.11× 10 6 × 200 × 10 −5 " 0.8 × 10 6 ×75 × 10 −4 = 450 T2

0/2

0.8 × 10 6 ×75 × 10 −4 × 450 0.11× 10 6 × 200 × 10 − 5 = 1230 K (3 SF) "

T2 =

Nearly but how is this showing that it is constant? The student needs to show that for the second point the answer is the same – so showing that this is constant andis therefore isothermal. NB the student has missed out the mega in the unit (but this doesnot really effect the isothermal idea – but it will in the next part)

" (of side (iv) – approximating the enclose area to a trapezium 6 6 " lengths 0.3 × 10 Pa and 0.19 × 10 Pa and the separation to be 125 × 10-6 " m3) gives a value of about 43 J and so the approximation will be about 40J"

Questions

Answers

1. Explain the difference between heat and internal energy. 2. Write down the first law of thermodynamics and explain what each term represents, when it is negative. 3. Explain what the area represents when a gas is taken around a cycle: a) in a clockwise direction b)in an anticlockwise direction. 4. Explain the difference between a heat engine and a heat pump. Draw a block diagram, representing the heat and energy transfers, and write down an equation for the thermal efficiency of each type of device. 5. A power station uses a heat engine operating between two heat reservoirs, the hot one consisting of steam at 100oC and the cold one consisting of water at 20oC. What is the maximum amount of electrical energy, which could be produced for every joule of heat extracted from the steam? 6. A fridge is working in a room at temperature 22oC. Its motor to extracts heat at a rate of 300 W from the inside in order to ensure that the temperature of its contents does not rise above 4oC. Calculate how much power must be supplied to the motor if its efficiency is 75%?

Answers to 1 –4 can be found in the text. 5. The maximum efficiency comes from using a Carnot (reversible) engine, for which

 T QH TH = , and so W = QH − QC = QH 1 − C QC TC  TH

   

converting temperatures into kelvin: TH = 373 K and TC = 293 K. For every joule of heat extracted from the hot reservoir (taking Q H = 1J), Wmax = 0.21 J. 6. In this case work is done on the contents of the fridge which means heat QC is extracted from the cold reservoir (at a rate of 300 W). For maximum efficiency a Carnot cycle operates – where  T W = QH − QC = QC 1 − C  TH

 .  

In a time of 1s W = 300 J(1- 277/295) = 18.3 J but since the fridge’s motor only works at 75% efficiency total heat needed in 1s = 18.3/0.75 = 24.4 J or a power of 24.4W. Acknowledgements: This Physics Factsheet was researched and written by Mike Bowen Jones The Curriculum Press,Unit 305B, The Big Peg,120 Vyse Street, Birmingham, B18 6NF. Physics Factsheets may be copied free of charge by teaching staff or students, provided that their school is a registered subscriber. No part of these Factsheets may be reproduced, stored in a retrieval system, or transmitted, in any other form or by any other means, without the prior permission of the publisher. ISSN 1351-5136

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