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A2 Physics Knowledge > §11 THERMAL PHYSICS Knowledge > Flashcards

§11 THERMAL PHYSICS Knowledge Flashcards

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1
Q

What two processes can cause energy transfer between two objects.

A
  1. Heat transfer to/from object
  2. Work done on/by object.
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2
Q

What is ‘heat transfer’, ΔQ

A
  • Heat transfer is energy transfer due to a temperature difference.
  • Heat energy moves from a higher T region to a lower T region.
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3
Q

What is mechanical ‘work’; can you give an example of how it produces a change in U?, ΔW

A
  • Is an energy transfer process where there is no transfer of heat, i.e. internal energy is changed by doing mechanical work,
  • Work is the name given to the energy transferred when a force is applied across a distance • Work is done and there is a rise in temperature due to an increase in internal energy (remember however that no heat transfer has taken place).
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4
Q

What is the ‘direction’ of heat transfer under ordinary circumstances.

A

• Heat transfer takes place from the hotter object to the colder object.

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5
Q

What are the 3 types of heat transfer process?

A

• (1 )Radiation • (2) Convection • (3) Conduction • Can occur either to/from an object.

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6
Q

What is ‘internal energy’?

A

The internal energy of an object, is the sum of the random distribution of the kinetic and potential energies of its molecules.

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7
Q

How is internal energy related to the microscopic structure of a substance?

A

U [internal energy] is the energy of an object due to the constituent particles’: • Individual positions (potential energy) • Motion (kinetic energy)

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8
Q

What is ‘thermal energy’?

A

• Thermal energy: the internal energy of an object due to its temperature.

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9
Q

What’s the difference between ‘internal energy’ and ‘thermal energy’

A

• Internal energy U may not equal the object’s thermal energy, • ∵ Some of the U may be from sources other than thermal ones.

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10
Q

Can you give an example of part of an object’s internal energy not coming from a ‘thermal’ source?

A

• A magnetic Fe bar has greater U when magnetised then when demagnetised. • This is U but not due to an object’s temperature.

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11
Q

Can you give an example of work done ON an object?

A

• Stirring a teacup very vigorously can increase its internal energy (• This is more than offset by convection resulting in heat transfer out of the cup)

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12
Q

What do you know if ΔU = 0 (so U is constant)?

A

(1LT) ΔU = ΔQ + ΔW • If ΔU = 0, ΔQ = - ΔW • So (1) no heat transfer/work done to/from/on/by the object. OR (2) Thermal equilibrium (3) Work done on object = heat energy lost by object. (Isothermal)

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13
Q

What is thermal equilibrium

A

• No overall transfer of heat… • … between two objects of the same temperature.

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14
Q

Why do the brake pads of a braking car gain U?

A

• Moving vehicle has KE • Friction does work on BP when pressed. • KE → ΔW → U increase • Higher U manifest as higher temperature of the brake pad

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15
Q

How could ΔU = 0 for a filament in a running filament bulb?

A

• FB turned on → Elec. supply does ΔW in pushing electrons through high resistance filament. • ΔW on filament ΔU = ΔW, U increases. • ∴ Filament temperature increases to operating temp. • At O.T. negative ΔQ: heat transferred to surroundings (temp-diff) and light is radiated. • … Until ΔW = – ΔQ → ΔU = 0 • ∴Constant internal energy of lightbulb filament.

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16
Q

Give 4 examples of how doing mechanical work on an object changes its internal energy

A

• Work done by electricity e.g. on a filament bulb • banging your head on a table • hitting a golf ball with a club etc. • An important example of changing internal energy by doing work is the compression or expansion of a gas (compression - work done ON gas , expansion - work done BY gas)

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17
Q

What is the definition of ‘temperature’

A

• A measure of the average internal energy an object has. • In simple terms, ‘the degree of hotness of an object’.

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18
Q

What is the ‘specific latent heat of fusion’

A

The specific latent heat of fusion is: • the energy needed to change 1 kg of the substance from solid to liquid… • … when the solid is at its melting point (at constant temperature). • Unit of SLHF = J/kg

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19
Q

What’s an expression for the energy required to change (m) kg of solid to liquid at constant temperature (m.p.)

A

• Q = m x SPECIFIC latent heat of fusion

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20
Q

What is the ‘specific latent heat of vaporisation’

A

The specific latent heat of vaporisation is: • the energy required to turn unit mass (1 kg) of the substance from liquid state to vapour, • without causing a change of temperature (at its b.p.) Unit : J / kg - Once again, when boiling point is reached, the energy supplied does not change temperature but is used to free molecules from the liquid to become gas (or vapour).

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21
Q

What’s an expression for the energy required to change (m) g of gas to a liquid at constant temperature?

A

• Q = m x Specific latent heat of Vaporisation

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22
Q

When the latent heat of vaporisation is supplied to a liquid at its boiling point, where does the energy go?

A

• When boiling point is reached….(or when the liquid evaporates in general) • … the energy supplied does not change temperature…. • … but is used to free molecules from the liquid, to change the phase to gas (or vapour).

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23
Q

What are the 4 phases of matter

A
  1. Solid, 2. Liquid. 3. Gas 4. Plasma - but only at v.high temperatures (e.g. fusion reactors or stars) so not in spec.
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24
Q

What’s the difference between the terms ‘state of matter’ and ‘phase of matter’?

A
  • Phase of matter ≠ state of matter necessarily • Phase : Solid / liquid / gas
  • State of a system can be changed by changing temperature, pressure or volume
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25
Q

What does this show?

A
  • The gradients of the solid, liquid and gas are not necessarily the same.
  • The time axis is directly proportional to an energy axis because W = Pt, the constant of proportionality is the power.
  • The time D to E is greater than that of B to C because the latent heat of vaporisation is greater than the latent heat of fusion.
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26
Q

What assumptions are made in the experiment used to draw this graph?

A

Assuming:

  1. No heat loss occurs during the heating phase
  2. That energy is supplied at a constant rate P to that substance
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27
Q

Look between A and B

  • What is the physical meaning of the gradient?
  • How is the gradient related to the power supplied to the solid?
A
  • The gradient (ΔT/Δt)[solid] is the rate of increase of the temperature of the solid.
  • Related to power supplied by:
    • P = mc[solid] (ΔT/Δt) [solid]
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28
Q

Look between C and D:

  • What is represented by the gradient?
  • How is the gradient related to the power supplied?
A
  • Gradient, (ΔT/Δt) [liquid] = rate of change of temperature.
  • Gradient is related to the power supply by:
    • P = mc[liquid] (ΔT/Δt) [liquid]
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29
Q

Why are the gradients of this graph for solid, liquid and gas phases not necessarily the same?

A
  • Gradient = Rate of change of temperature , (ΔT/Δt)phase
  • Power = mcphase (ΔT/Δt)phase
  • So, assuming constant power,
    • Gradient ∝ 1/(specific heat capacity)Phase
  • ∴ If the liquid phase has Cliquid < Csolid then the liquid heats up faster than the solid → steeper gradient.
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30
Q

What is an expression for the

  • melting time
  • boiling time

in this scenario?

A
  • For a change of phase at constant temperature,
    • ΔQ = m l fusion or vaporisation = P Δt
  • ∴ melting or boiling time = (m l fusion or vaporisation )/ Power
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31
Q

Which ‘heat equation’ is true in this situation:

  • When a substance’s temperature changes when heat is transferred
  • Phase does not change
A
  • ΔQ = mc ΔT ‘holds’/equals the energy transferred
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32
Q

Which ‘heat equation’ holds in this situation:

  • When a substance’s phase changes when heat is transferred,
  • … and its temperature stays constant:
A
  • ΔQ = ml
    • where l is either the latent heat of fusion (lf) or the latent heat of vaporisation (lv).
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33
Q

(Exam tech) If in any situation the volume of

the substance is given __________________

A

convert this into the mass by using V𝜌 = m

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34
Q

What is the temperature of steam (specifically referring to water in the gas phase, formed when water boils).

A

At atmospheric pressure, steam always has a temperature of 100℃.

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35
Q

When a substance changes phase from a solid to a liquid, when is this latent heat of fusion released?

A
  • LHF (supplied during phase transition from solid → liquid) is released when a liquid solidifies.
  • During solidification (liquid → solid)
    • Liquid’s molecules slow down
    • mean KE ↓
    • Molecules unable to overcome ‘force bonds’
      • ∴ start to lock molecules into fixed positions of a solid
  • Some of the latent heat released during solidification keeps temperature at melting point, until all the liquid has solidified.
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36
Q

Why is it called the “latent” heat of fusion/vaporisation?

A
  • “Latent” = “Hidden”
  • Latent heat supplied to melt any solid can be considered ‘hidden’
    • ∵ no temperature change takes place,
    • Even though the solid is being heated
    • (and its internal energy is increasing therefore).
  • Fusion = the term referring to the melting of a solid.
  • Vaporisation = term referring to boiling of a liquid.
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37
Q

Explain what happens when a liquid is heated near its boiling point.

A
  • The molecules gain enough kinetic energy to overcome intermolecular forces between them (which previously held them close together as a liquid), and to break away from from being ‘in contact with each other’.
  • Liquid molecules gain enough KE
    • .. Able to overcome intermolecular forces between them
      • (Which previously held them close together - liquid).
      • Now able to become separated by rel. large distances
  • Energy required to vaporise any mass of liquid → gas at constant temperature is called ‘latent heat of vaporisation’
  • This latent heat of vaporisation (= m l V) released again when a vapour condenses.
38
Q

Explain in molecular terms the process of a vapour condensing.

A
  • As the vapour’s temperature↓ to boiling point.
    • Vapour molecules <ke>↓</ke>
  • Molecules eventually have low enough <ke>
    </ke><ul>
    <li>Intermolecular forces able to pull molecules together to form a liquid.</li>
    </ul></ke>
39
Q

What is sublimation

A

Sublimation is the process whereby a solid vaporises directly into a solid when heated.

40
Q

Compare the size of:

  • Specific latent heat of fusion (lF)
  • Specific latent heat of vaporisation (lV)
A

In general:

  • Specific Latent Heat of Fusion << Specific Latent Heat of Vaporisation
  • “The specific latent heat of vaporisation of a substance is always much greater than the specific latent heat of fusion”
  • Example:
    • For 1 kg of water (S)LHV = 2.25 MJ, much larger than its (S)LHF = 0.336 MJ.
41
Q

What is the specific latent heat of fusion: lF

A
  • The energy needed…
  • …. to change unit mass of the substance …
  • …from solid to liquid…
  • … without changing its temperature.
42
Q

How is the energy of a molecule in a solid substance affected by:

  • Heat transfer to the solid which results in increased temperature.
A
  • Heat energy transferred to ↑temperature of the solid:
    • Causes ↑<ke> of solid's molecules.</ke>
43
Q

How is the energy of a molecule in a solid substance affected by:

  • Heat transfer to the solid which results in the solid melting (‘fusing’).
A
  • Energy supplied during the melting of a solid
    • goes towards increasing the potential energy of the solid’s molecules
    • because they break free from each other
    • — potential energy increases (since work has to be done against forces of attraction between the molecules in the solid’s 3D structure).
  • When a substance is melting (or fusing),
    • the energy supplied does not raise the temperature.
    • All the energy supplied is used to break intermolecular force bonds and change the phase.
    • The energy input becomes internal potential energy. <v></v>
44
Q

What are the molecular properties of liquids?

A
  • Liquids have higher internal energy than solids.
    • Relatively weak Intermolecular forces between liquid’s molecules not strong enough to hold molecules in fixed positions (as in solid phase).
  • Molecules in random motion, able to change relative positions.
    • But unable to become separated from each other.
  • Heat/Work into liquid → Increased Liquid Temperature (TE↑) → increased (∵ molecular speed↑).
  • Liquid vaporises if sufficient energy is supplied because eventually the molecules will have sufficient KE + PE to break free ad move away from other molecules.
    • Therefore lower KE and PE than the gas phase.
45
Q

What are the molecular properties of a gas?

A
  • Very weak intermolecular attraction between gas molecules.
  • Particles separated by relatively large distances (relatively high PE therefore) ∴ Vol1 kg liquid much less than Vol1 kg gas of a substance.
  • High <ke> , molecules moving at high speeds.</ke>
  • Internal energy = KE + PE,
    • So ↑U →
      • higher temperature
      • higher pressure (assuming constant vol. and mass).
46
Q

How could ↑U of a gas result in higher:

  • Pressure
  • Temperature

assuming constant volume and mass.

A
  • ↑U means ↑KE or ↑PE of molecules.
    • ↑KE → higher temperature
    • ↑PE → greater separation between molecules → higher pressure
47
Q

Why does the temperature of a gas change when

  1. It does work e.g. by pushing out a piston by increasing in volume.
  2. Work is done on it, e.g. compression by a piston
A
  • Piston may compress or expand a constant mass of gas.
  • Compression:
    • Piston does mechanical work on the gas, ΔW = P ΔV for small ΔV.
    • ↑U → ↑Temperature
  • Expansion
    • Gas does mechanical work on piston as it expands in volume.
    • ↓U causing decrease in temperature.
  • Effectively this is a way of converthing the gas’s internal energy into mechanical work.
48
Q

What is thermal equilibrium?

A
  • When there is no overall heat transfer…
  • … between objects at the same temperature.
49
Q

What’s an example of a system omving towards thermal equilibrium.

A
  • Placing an object in higher/lower temperature water → heat transfer to/from the object → reduces temperature difference
    • Moves towards thermal equilibrium
  • If water is at same temperature as the object
    • No temperature difference
    • ∴ no overall heat transfer occurs.
50
Q

What is a

  • Molecule
  • Atom
A
  • Molecule: smallest particle of a substance that is characteristic of the substance.
  • Atom: the smallest particle of an element that is characteristic of the element.
51
Q

Compare the volume occupied by 1 kg of a substance

  • In gas phase,
  • In liquid phase
A
  • The 1kg of gas occupies much more volume than the same matter in liquid phase.
  • ∵ Gas molecules have much larger distances between them.
52
Q

How does a gas exert pressure on its container?

A
  • Gas molecules
    • Constantly moving at high speed
  • Make frequent elastic collisions with container walls
    • i.e. without KE loss.
  • During rebound, momentum of particles changes
    • Change in momentum over a period of time occurs,
    • ∴ A force acts on the rebounding molecule (N2L)
    • ∴ The molecule exerts a force on the container wall (N3L)
  • Numerous molecular collisions over the area of container surface produce macroscopic gas-pressure.
53
Q

Describe the molecular structure and properties of a solid

A

Molecules :

  • Organised in a regular 3D structure.
  • Have ‘strong’ intermolecular forces attraction between them.
  • Vibrate randomly about fixed points, but cannot move out of position in structure.
54
Q

Explain in molecular terms the process of a solid melting.

A
  • Heating a solid increases KE of molecules.
    • Molecules have higher speed/vibrate faster
  • With enough energy… molecules can be made to :
    • Break out of positions
    • Move around freely.
  • … solid melts into a liquid.
55
Q

What holds molecules attracted to each other in a solid?

A
  • Intermolecular forces
    • Arise from the electrical charges of the protons/electrons in constitutent molecules.
56
Q

What relates the temperature of a solid, and the rate at which its molecules vibrate?

A
  • ↑Temperature → ↑ ∴ faster vibration.
57
Q

What happens when heat is continuously added to a substance in solid phase?

A
  • Temp.↑ up to melting point.
  • Temp. stays at MP until latent heat of fusion transferred.
    • Meanwhile phase changes to liquid.
  • Liquid temp↑ up to boiling point.
  • Temp. stays at BP until latent heat of vaporsiation transferred
    • Meanwhile phase change from liquid to gas.

Note: BP and MP only well defined temperatures for pure substances.

58
Q

What happens when heat is transferred to steam in a piston

  • When the volume is held constant.
  • When the pressure is held constant (atmospheric pressure)
A
  • Constant volume:
    • ΔU = ΔW + ΔQsupplied
    • Vol constant, ∴ΔW = 0
    • ΔU = ΔQsupplied<strong> </strong>→ temperature of the gas increases
      • In turn implies pressure of gas must have increased.
  • Constant atmospheric pressure → temp. of steam ≡ 100℃ ∴ isothermal conditions,
    • (Assuming no heat loss to surroundings is allowed
      • ΔU = ΔW + ΔQsupplied
      • Isothermal → ΔU = 0 → ΔW = - ΔQsupplied , so gas does work on surroundings.
      • Piston is pushed out by the expanding gas, doing mechanical work equal to ΔQsupplied
59
Q

Compare the 𝜌 of a gas with the 𝜌 of a liquid / solid.

A
  • 𝜌[gas] <<< 𝜌[liquid] < 𝜌[solid]
  • Molecules of Liquid/Solid packed ‘in contact’ together without much distance separating them.
  • Gas molecules typically spaced far apart on average by relatively large distances.
60
Q

Why are liquids / gases fluid while solids are not?

A
  • Solid:
    • molecules prevented from moving relative to another by strong force bonds.
  • Liquids/Gases:
    • molecules have too much KE/PE allowing them to overcome (to varying extents) the attraction of the forces between molecules,
    • allowing the molecules to move relative to one another and exhibit fluid behaviour on a macroscopic scale.
    • The force bonds are not strong enough to keep the molecules fixed to each other.
61
Q

Explain in terms of molecules what happens when a solid or liquid is heated (when not near its melting/boiling points).

A
  • Solid → ΔQ → molecules vibrate more quickly about fixed positions (mean positions)
  • Liquid → ΔQ → molecules move more quickly relative to each other, still keep in contact with each other but free to move about.
62
Q

What happens when a solid is heated near its melting point?

A
  • Energy is needed to break bonds when a substance changes phase.
  • This energy is called latent heat,
    • and temperature of a substance remains constant during a change of phase.
  • The energy needed for a change of state is calculated as follows
    • ΔQ = mlFUSION
63
Q

How can you adapt the equation for heat capacity to the number of moles in a substance (Rather than the mass).

A
  • It is sometimes more convenient to relate heat capacity to the number of moles of a substance rather than its mass in kg.
  • The specific heat capacity equation then becomes
    • ΔQ = nCΔT
  • and the latent heat capacity equation becomes
    • ΔQ = nL

Where

  • ΔQ Heat energy transferred, J
  • n Number of moles, mol
  • C Molar heat capacity, J mol-1 K-1
  • Lf or v Molar latent heat of fusion or vaporisation, J mol-1
  • ΔT Temperature change, K
64
Q

Give an example of how different substances can heat up at different rates

A
  • On a beach during a warm day, the sand may become extremely warm wile the water remains refreshingly cool.
  • Sand and water have different specific heat capacities.
65
Q

What are the 3 factors which affect the temperature rise (Δt) of an object when it is heated with a certain amount of heat energy transferred to the object (ΔQ).

A
  • The mass of the object.
  • The amount of energy supplied to it.
  • The specific heat capacity of the substance (c) from which the object is made.
66
Q

Define ‘specific heat capacity’, c, of a substance.

A
  • Specific heat capacity of a substance is the energy needed to raise the temperature of unit mass (1 kg) by 1 Kelvin without change of state.
  • Unit: J/kg/K
67
Q

Give the equation and derivation of the relationship between m, ΔE, c, ΔT.

A

To raise the temperature of mass m of a substance from temperature T1 to temperature T2:
Energy needed, ΔQ = mc(T2 — T1) = mc ΔT where ΔT = T2 — T1
2. *Derivation?

68
Q

What are the issues of applying the specific heat capacity equation to a gas

A
  • Specific Heat Capacity of a Gas: This can take any value depending on how much the gas expands during heating.
  • Also, the specific heat capacity of a gas in state changes at constant pressure is greater than state changes occurring with constant volume.
69
Q

Describe the inversion tube expriment

A

The Inversion Tube Experiment

  • Setup: Vertical Tube of length L with facility of inserting a thermometer to measure temperature.
  • Inverting tube causes GPE of object within to be transformed: ΔGPE —> ΔKE —(work)—> ΔU
  • Purpose is to experimentally determine the specific heat constant of the solid object within.
70
Q

Derive an expression for the specific heat capacity in the context of the inversion tube experiment.

A
  • Mass of lead shot used := m
  • Tube Length := L
  • Inversions := n

Analysis

  • ΔGPE each inversion = mgL —> ΔGPE = n.m.g.L
  • ΔGPE lead shot —> ΔU lead shot [assumption]
  • = ΔQ
  • n.m.g.L = ΔQ = mc ΔT, ΔTof lead shot = T2 — T1
  • n.g.L = c ΔT
  • c = n.g.L/ΔT allows you to calculate the specific heat capacity of the lead shot.
71
Q

Describe the experimental setup for measuring the specific heat capacity using experimental methods.

A
  • Metal block, mass = m, contained within an insulated container.
    • 1st Hole drilled inside metal and a 12v electrical heater is inserted into it.
    • Supplies known amount of electrical energy
  • Circuit: cell, switch, veristor, ammeter, voltmeter (in parallel with heater)
  • 2nd hole — thermometer inserted into the metal
    • Measures temperature ΔT = T2 — T1
  • Small amount of water/oil inserted into the metal improves the thermal contact between the thermometer
72
Q

How can you calculate the SHC of a metal using the electrical method (knowing P, I, V of heater, and having measured the heating time?).

A

Heater —> {P, I, V}

{t = heating time]

  • P = IVt = ΔQ
  • = mc ΔT
  • c = (IVt)/(mΔT) is an alternate way of calculating the specific heat capacity.
73
Q

Describe the experiment for measuring the specific heat capacity of a liquid

A
  • Use same circuit as electrical method - with heater.
  • Liquid (known mass m L but unknown specific heat capacity cL) placed in a calorimeter (of known mass mc and specific heat capacity cc) assuming no heat loss of calorimeter.
  • Liquid is heated directly using 12v electrical heater.
  • Thermometer inserted into liquid to measure temperature change ΔT [stir liquid before measurement].
74
Q

Describe how to calculate the specific heat capacity CL of a liquid knowing

  • Mass of liquid,
  • Mass and C of calorimeter
  • Temperature change of Liquid and Calorimeter
A
  • E supplied = IVt = current ⨉ pd ⨉ heating time
  • E —> ΔU liquid + ΔU calorimeter (where U generally stands for internal energy).
  • = mLcLΔT + mcccΔT assuming both calorimeter and liquid undergo same ΔT, and there’s no heat loss to surroundings.

So: IVt = mLcLΔT + mcccΔT

  • All quantities except cL are known/ can be determined experimentally.
75
Q

Apply the equation relating temperature change and work done on the solid/liquid to a continuous flow heating process.

A

(ΔQ/Δt) = (Δm/Δt) c (T2 - T1)

Where ΔQ/Δt = Heat energy transferred to the fluid per second [J/s]
Δm/Δt = Mass flow rate of the liquid per second [kg/s]
c = Specific heat capacity of fluid, J kg-1K-1
T1 & T2 : Temperature of fluid before and after heating, K.

76
Q

Explain the concept of a temperature scale

A
  • Defined in terms of two ‘standard’ fixed point temperatures
  • …. which are standard degrees of hotness
  • …. which can easily and accurately be reproduced.
77
Q

Standard fixed points of the Celsius scale:

A
  • Ice point 0℃ — the temperature of pure melting ice.
  • Steam point 100℃ — the temperature of steam at standard atmospheric pressure.
78
Q

What are the fixed points of the Kelvin (or ‘Absolute’) temperature scale.

A
  • Absolute zero: 0K, the lowest possible temperature associated with minimum internal energy.
  • Triple point of water: 273.15 K the temperature at which water, ice and water vapour coexist in thermodynamic equilibrium (i.e. no heat flow).
  • Therefore Temperature in oC = Absolute temperature in K – 273.15
79
Q

Explain the concept of ‘absolute zero’

A
  • An object at absolute zero has minimum internal energy, regardless of the substances that the object consists of.
  • Therefore no object can have a temperature below absolute zero.
80
Q

Explain how absolute zero is determined experimentally.

A
  • Take a fixed mass of gas and a fixed volume sealed container.
  • Measure gas pressure as temperature changes
  • Line of best fit always cuts the temperature axis at -273℃, regardless of which gas is used or how much gas is used.
81
Q

State some of the ways in which the properties of matter change at very low temperatures:

A
  • Superconductivity - zero resistivity.
  • Superfluids - can empty themselves out of containers.
  • These properties become apparent when matter is cooled to a few micro-Kelvin of absolute zero, which can be achieved in laboratories using scientific approaches.
82
Q

Explain the Zeroth Law of Thermodynamics

A

(1) If two systems are at the same temperature, there is no resultant flow of heat energy between them.

83
Q

Give an example to illustrate the Zeroth Law of Thermodynamics

A

Example: is well illustrated by goldfish.

  • Fish are cold blooded so they are at the same temperature as the water.
  • Because there is no flow of heat energy between them we say that they are in thermal equilibrium.
  • This is the zeroth law… ‘if two systems are at the same temperature, there is no resultant flow of heat between them”
84
Q

Explain the 1st Law of Thermodynamics.

A

1st Law of Thermodynamics: Is a conservation of energy law, showing that if you add heat or work to a system its internal energy increases.

∆U of system = ∆Q into system + ∆W on system.

Be accurate with direction of work / heat energy transfers.

85
Q
  • Explain
    • (a) what an isothermal change is, and
    • (b) what this implies about the link between work done on/by an object and heat transfer in/out of an object.
A
  • An isothermal change is a change in state which occurs at constant temperature, i.e. no change in internal energy.
  • If ΔU = 0, and ΔU = ΔQ+ΔW,
    • then ΔW = -ΔQ,
  • i.e. In an isothermal situation if work is done on a system, an equal amount of heat must be lost from the system.
86
Q
  • Explain
    • (a) what is meant by an adiabatic situation and
    • (b) What this implies about the relationship between ΔW and ΔQ.
A
  • An adiabatic situation means no energy is lost to the surroundings.
  • This usually happens because adiabatic changes occur so quickly that there is no time for a system to lose heat to the surroundings.
  • Thus ΔQ = 0 &
    • ΔU = ΔW
87
Q

Illustrate an adiabatic situation with an example?

A
  • An adiabatic change in which work is done on the system (i.e. system is compressed) results in increased internal energy & thus increased temperature.
  • An adiabatic change in which work is done by the system (i.e. system expands) results in decreased internal energy & thus reduced temperature.
88
Q

Explain the relationship between ΔU and ΔQ, ΔW when Volume is held constant (and other state variables change). Apply this concept.

A
  • In a constant volume change,
    • no work is done because force (arising from gas pressure) does not move across a distance.
    • Thus ΔW = 0 and ΔU = ΔQ.
  • The heating of a solid or liquid is, more or less, a constant volume state change.
  • Although liquids do expand more than a solid under heating,
    • the work done in pushing back the surroundings is small compared with the energy needed to change the kinetic energy (& temperature) of the molecules.
89
Q

Derive an expression for the work done ΔW in compressing a gas at pressure P resulting in a volume change of ΔV in a cylindrical piston of cross sectional area A.

A

ΔW = F Δs
P = F/A
F = PA
ΔW = P (A Δs) = P ΔV
Hence ΔW = P ΔV
Work Done = Pressure ⨉ Change in Volume
[J] [N/m2] x [m3]

Note: only applies for small volume changes as pressure changes with volume.

90
Q
A
91
Q
A

A2 Physics Knowledge (5 decks)

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