Nelson Thornes Electrcity and thermal physics Prac2

7/23/2019 Nelson Thornes Electrcity and thermal physics Prac2

1/26NAS Physics Teachers Guide 2005 Nelson Thornes Ltd.

Unit 2Electricity and Thermal Physics

Solutions to Practice Questions

Chapter 11 An electric current is a flow of electric charge

2 The cell does electrical work by applying a force to the charge carriers in the direction in whichthey move

Both electrical and mechanical work transfer energy

3 A cell is a single source of e.m.f. whereas a battery is a group of cells connected together

4 Circuit A see Figure 4.6 on page 9

Circuit B

Circuit C

Circuit D

The circuits will run out of fuel in order A, C, B

D need not run out of fuel at all, as the engines can be turned off and still produce the same effect!




Solutions to Practice Questions5 Direct current: the electrons travel in one direction only

Alternating current: the electrons move first one way and then the other; they keep movingbackwards and forwards

Chapter 2

1 +4 nC on A and 4 nC on B

2 The duster takes electrons from the surface of the rod

The duster now has a negative charge and the rod an equal positive one

3 Current is a base quantity

Charge is a derived quantity

Charge is measured in coulombs and current in amperes

1 ampere = 1 coulomb per second

4 (a) Q =It= 3 A 4 s = 12 C

(b)Q = 7 A (8 60) s = 3360 C

(c) Q = 0.25 A (2 60 60) s = 1800 C

5 (a) I=Q/t= 700 C/(35 s) = 20 A

(b) I=Q/t= 3600 C/(3 60 s) = 20 A

6 Q =It= 4.5 A (20 60) s = 5400 C

Number of electrons = 5400 C/(1.6 1019 C) = 3.4 1022

7 Charge = current time

C = A s

Chapter 3

1 Draw a large copy of the circuit diagram on a piece of paper

Place components on top of their circuit symbolsConnect components using wires of the correct length to keep the circuit looking like itscircuit diagram

2 Break the circuit at the required point

Insert the ammeter; an additional lead will be needed

Make sure the red terminal of the ammeter is connected nearest to the positive terminal of thepower supply

3 Torch bulb 0.3 A, LED 20 mA, small motor 1 A, buzzer 0.1A, mains lamp 0.25 A, electric kettle 10 A

4 A series circuit is one where the components are all connected in-line, one after the otherThe current passes through one component, then through the next, then the next, etc.

Current is not used up by any component (conservation of charge): what goes in comes out

so the current throughout a series circuit is the same


3/26

5 The current entering a house equals the current leaving, so company A makes no charge

The charge entering a house equals the charge leaving, so company B makes no charge

Electrical devices in a house remove energy from the supply, so company C charges its customers

Chapter 4

1 A parallel circuit is one where components are connected across each other

Current flowing into a parallel circuit splits at the junction so that a part of it goes througheach route

Current flowing out of each route of a parallel circuit join together at the junction

2 The sum of the currents entering a point is equal to the sum of the currents leaving that point

Conservation of charge

3

4 In series:

(a)

12 V

2.4 A 1.6 A

0.8 A 0.8 A 0.8 A

1.6 A2.4 A

0.8 A 0.8 A 0.8 A

NAS Physics Teachers Guide 2005 Nelson Thornes Ltd.






Solutions to Practice Questions(b)

5 Note that currents in series are the same and that currents add up to zero at junctions

Circuit A I1 =I2 = 1 A + 1 A = 2 A

Circuit B I3 =I4 = 1.5 A

Circuit C I5 =I6 = 1.2 A [current through parallel resistor = 0.4 A]

Circuit D I7 = 20 mA,I8 = 20 mA 0.1 mA = 19.9 mA, I9 = 0.1 mA

Chapter 5

1 Adding a resistor to a series circuit increases the total resistance

So that the current flowing through the whole circuit decreasesA circuit can be given additional resistance by:

making part of the wiring thinner

using longer wires

using a material through which electrons find it hard to move

using a material in which there are only a few electrons that can move

2

mA

3 V




Solutions to Practice Questions3

4

5 An LDR will not pass enough current to run a motor

See Figure 5.7 on page 11

As the level of illumination increases, the resistance of the LDR falls

Current flowing through LDR and relay coil increases sufficiently to close the relay switch

which allows current to flow directly through the motor from the supply

Chapter 6

1 3 1.5 V = 4.5 V

Three cells in parallel will supply energy for a longer time compared to a single cell

[Will also reduce the overall effect of any internal resistance as current splits between the three cells see Chapter 17]

2 The average resultant force on the electrons is zeroThe power supply pushes the electrons along in their direction of travel

The circuit resistances apply equal forces in the opposite direction

live

neutral

12.4 A

electricfire

12 A

12 A

0.4 A

0.4 A





4 V1 = 1.5 V

V2 = 1.5 V

V3 = 3.0 V

V4 = 12 V, V5 = 12 V

V6 = 9.0 V, V7 = 9.0 V

5 The two types of voltage difference are in opposite directions

Chapter 7

1 Two components that give energy to a circuit: cells (batteries), generators

Two components that take energy away from a circuit: lamps, motors

Voltage differences across components that give energy to a circuit are called e.m.f.s

Voltage differences across components that take energy away from a circuit are calledpotential differences

2 (a) W= VQ = 9 V 15 C = 135 J

(b)W= VQ = VIt= 9 V 0.5 A (2 60) s = 540 J

3 (a) Energy per second (power) = 36 000 J/(10 60 s) = 60 J s1

(b) 3 A = 3 C s1

(c) Energy per coulomb (voltage) = 60 J s1/(3 C s1) = 20 J C1

4 P = VI

V=P/I= 0.75 W/(0.3 A) = 2.5 V

5 Q =It= 2.5 A (10 60) s = 1500 C

e.m.f. = W/Q = 300 000 J/(1500 C) = 200 V

V

+

V

+

V

+

V

+


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Chapter 81

2

3 V1 = 12 V 3 V = 9 V

V2 = V3 = 6 V

V4 = Vlamp = 4 V

V5 = 9 V 4 V = 5 V

4 See Figure 8.6 on page 17

Source of power in the water circuit is the Sun

Source of power in the electric circuit is the battery

Assuming no water either evaporates from the streams or is absorbed into the ground then

the amount of water flowing down mountain equals amount of water evaporating from sea

5 When at A, VX> potential of right-hand terminal of ammeter, so current through ammeter flowsfrom left to right

When at E, VX= potential of right-hand terminal of ammeter, so no current flows through ammeter

When at B, VX< potential of right-hand terminal of ammeter, so current through ammeter flowsfrom right to left

12 V 12 V1 k 12 V1 k 12 V1 k

2 V

4V

4V

4V

1 k

1 k

1 k








Chapter 91 The voltage at a point in a circuit is the voltage difference between zero and that point

The voltage across a component is the difference in the voltages at its two ends

2 (a) Vlamp = 1.8 V 0.4 V = 1.4 V

(b)Vresistor= 5.7 V 2.1 V = 3.6 V

(c) Vwire = 2.1 V 1.8 V = 0.3 V

3

4 V1 = 1.5 V

V2 = 3.0 V, V3 = 1.5 V

V4 = 3.0 V, V5 = 1.5 V

V6 = 2.0 V, V7 = 4.0 V, V8 = 6.0 V

V9 = 3.0 V, V10 = 6.0 V

V11 = 6.0 V, V12 = 3.0 V, V13 = 3.0 V, V14 = 6.0 V

5 Around any closed loop, the sum of the e.m.f.s is equal to the sum of the p.d.s

6 For energy conservation:

total energy gained by a coulomb going round a complete circuit equals the total energy lost

and as voltage is a measure of the energy transferred per unit charge

total e.m.f. (energy providers) = total p.d. (energy takers)

Chapter 10

1 Eithervolt = J C1 = N m C1 = kg m s2 m C1 = kg m s2 m (A s)1 = kg m s2 m A1 s1

= kg m2 s3A1

orvolt = W A1 = J s1A1 = N m s1A1 = kg m s2 m s1A1 = kg m2 s3A1

ohm = V A1 = kg m2 s3A1A1 = kg m2 s3A2

2 See experiment on page 20

Voltage/V

6

5

4

3

2

1

0

Position




Solutions to Practice Questions3 Resistors are connected in parallel (combined resistance < individual resistance)

1/total = 1/R1 + 1/R2

1/R2 = 1/total 1/R1 = 1/(120 ) 1/(180 ) = 0.002778 1

R2 = 360

4 The seven arrangements are

(i) 18 , (ii) 36 , (iii) 54 , (iv) 9 , (v) 6 , (vi) 12 , (vii) 27

5 Power =I2R

I2 =P/R = 0.810 W/(25 ) = 0.0324 A2

I= (0.0324 A2) = 0.18 AV=IR = 0.18 A 25 = 4.5 V

Number of cells = 4.5 V/(1.5 V) = 3

Chapter 11

1 (i) Behaviour of circuit will be unaffected by presence of voltmeter as resistance of parallel(voltmeter) combination will be as close as possible to resistance of component

(ii) Behaviour of circuit will be unaffected by presence of ammeter as it does not change the circuitresistance

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)




Solutions to Practice Questions2 In Figure 11.1 on page 22:

Voltmeter records the correct voltage for the resistor

Ammeter records the current through both the resistor and the voltmeter

Calculated resistance value = correct V/largerI= lowerR

In Figure 11.2 on page 22:

voltmeter records the voltage across both the resistor and the voltmeter

ammeter records the correct current for the resistor

Calculated resistance value = larger V/correctI= higherR

3 First connect the probes of the digital ohmmeter together and note the reading

Connect the probes to the ends of the component and note the new reading

Resistance of the component is the difference in these two readings

4

5 The resistance of a light-dependent resistor decreases as the light intensity falling on it increases

Chapter 12

1 Number of tennis balls along each 1 m side = 1000 mm/(66 mm) = 15

Total number in 1 m3 box = 15 15 15 = 3375

2 Volume of wire V=Al= 3 106

m

2

30 102

m = 9 107

m

3

Number of free electronsN= nV= 6.4 1028 m3 9 107 m3 = 5.76 1022

Amount of free chargeQ =Ne = 5.76 1022 1.6 1019 C = 9216 C = 9200 C

3 A = 3 C s1 (I=Q/t)

t=Q/I= 9216 C/(3 C s1) = 3072 s = 3100 s

Average speed = distance/time = 30 102 m/(3072 s) = 9.8 105 m s1 = 0.1 mm s1

3 When the switch is closed, electrons throughout the circuit start moving almost straight away

The electromagnetic wave that starts the electrons moving travels around the circuit very quickly (atthe speed of light 3 108 m s1)

Electrons move in, and give energy to, the lamp almost instantaneously

Delay time t= length of cable/(3 108 m s1)

tungsten filament lamp

NTC thermistor

Temperature

essance




Solutions to Practice Questions4 I= nAvq

v =I/nAq = 25 A/(7 1028 m3 2.5 106 m2 1.6 1019 C) = 8.9 104 m s1 = 0.89 mm s1


Chapter 13

1 Resistivity = resistance area/length

Units of resistivity = m2/m = m

In base units:

resistance = = kg m2 s3A2 (see solution 10.1)

resistivity = m = kg m2 s3A2 m = kg m3 s3A2

2 Resistivity of tungsten is greater than that of copper

Cross-sectional area of the tungsten filament is smaller than that of the copper connecting wires

AsR =l/A, filament resistance is greater than that of the same length of copper connecting wire

Since they are in series, the current through both the filament and the connecting wires is the same

PowerP =I2R R, so filament dissipates more power and gets hotter

3 R =l/A

A = wt= 2.0 103

m 8.5 103

m = 1.7 105

m

2

R = 1.7 108 m 4.0 102 m/(1.7 105 m2) = 4 105

V=IR = 25 103A 4 105 = 1 106V = 1 V

4 In both components:

lattice vibrations increase with temperature

producing increased carrier obstruction and reduced drift speed v

In the tungsten filament lamp:

currentIv (nAq are constant), so producing a smaller current and a larger resistance

In the NTC thermistor:

charge carrier density n increases massively with temperature

n increases much more than v decreases

currentInv (Aq are constant), so producing a larger current and a smaller resistance

5 I= V/R = 12 V/(300 ) = 0.04 A = 40 mA

Power = V2/R = (12 V)2/(300 ) = 0.48 W = 480 mW

Thermistor gains 180 mW (i.e. 480 mW generated 300 mW dissipated)

(a) Temperature increases

(b) Resistance decreases

Both the current flowing and the power generated within the thermistor will increase

If process continues:

temperature rises more producing further reductions in resistance and increases in current andpower so that thermal runaway occurs




Solutions to Practice Questions6 Resistivity increases with temperature: copper and iron

Resistivity decreases with temperature: carbon, glass, paraffin wax and silicon

Chapter 14

1 A potential divider circuit consists of a chain of resistors, all connected in series

It divides the voltage from a source in proportion to their resistances

2 Total resistance = 7 k + 3 k = 10 k

CurrentI= e.m.f./(total resistance) = 9 V/(10 k) = 9 104A

VOUT =IR = 9 104A 3 103 = 2.7 V

3 VOUTwill decrease

Larger current in 7 k resistor so greater p.d. across top resistor

(oreffective resistance of bottom resistor is reduced)

4 (a) 1/R = (1/600 ) + (1/400 ) = 0.0042 1

R = 1/(0.0042 1) = 240

(b) TotalR = 360 + 240 = 600

(c) I= e.m.f./totalR = 12 V/(600 ) = 0.02 A = 20 mA

(d)V360 =IR = 0.02 A 360 = 7.2 V(e) Vparallel = Vsupply V360 = 12 V 7.2 V = 4.8 V

[or V240 =IR = 0.02 A 240 = 4.8 V]

(f) I600 = V600/R = 4.8 V/(600 ) = 0.008 A = 8 mA

(g) I400 =IsupplyI600 = 20 mA 8 mA = 12 mA

[or I400 = V400/R = 4.8 V/(400 ) = 0.012 A = 12 mA]

5 Parallel combination:

1/R = (1/12 ) + (1/36 ) = 0.111 1

R = 1/(0.111 1) = 9

Circuit:

totalR = 18 + 9 = 27

I= e.m.f./totalR = 9 V/(27 ) = 0.33 A

I18 = 0.33 A

V18 =IR = 0.33 A 18 = 6 V

Vparallel = Vsupply V18 = 9 V 6 V = 3 V

I12 = V12/R = 3 V/(12 ) = 0.25 A

I36 =IsupplyI12 = 0.33 A 0.25 A = 0.08 A





Chapter 151 A rheostat is a variable resistance in series with the component (see Figure 15.1 on page 30)

The rheostat controls the current through the lamp

A potentiometer is a variable potential divider (see Figure 15.2 on page 30)

The potentiometer controls the voltage across the lamp

An advantage of using a potentiometer is that the voltage across the lamp can be reduced to zero

A disadvantage of using a potentiometer is that circuit current still flows even when there is zerocurrent in the lamp

2 Digital voltmeter shows that there is 1.5 V across bottom resistor of potential divider

Torch bulb is connected in parallel with the bottom resistor

Their combined resistance is less so p.d. across them is less than 1.5 V

3 Using a digital voltmeter (infinite resistance):

Total resistance = 2000 + 3000 + 1000 = 6000

I= e.m.f./totalR = 12 V/(6000 ) = 0.002 A

VBC =IRBC = 0.002 A 3000 = 6 V

Using analogue voltmeter (resistance = 6000 ):

Combined resistanceRBC = 1/[(1/3000 ) + (1/6000 )] = 2000

Total resistance = 2000 + 2000 + 1000 = 5000 I= e.m.f./totalR = 12 V/(5000 ) = 0.0024 A

VBC =IRBC = 0.0024 A 2000 = 4.8 V

Using both voltmeters:

combined resistanceRBC = 1/[(1/3000 ) + (1/6000 ) + (1/ )] = 2000

so both voltmeters read 4.8 V

4 The terminals must be joined together by a wire of negligible resistance [short-circuited]

R1 = Vsupply/Imax= 4 V/(0.5 A) = 8

Note that the current of 0.1 A flows between the output terminals and notthroughR2VR1 = Vsupply Vout = 4 V 3 V = 1 V

IR1 = VR1/R1 = 1 V/(8 ) = 0.125 A

IR2 =IR1Iout = 0.125 A 0.100 A = 0.025 A

R2 = VR2/IR2 = Vout/IR2 = 3 V/(0.025 A) = 120

5 (a) At 1000 lux,Rldr= 150 so total circuit resistance = 1050

Current flowing = 5 V/(1050 ) = 0.0048 A

V900 =IR = 0.0048 A 900 = 4.29 V = 4.3 V

(b) At 90 lux,Rldr= 1500 so total circuit resistance = 2400

Current flowing = 5 V/(2400 ) = 0.0021 A

V900 =IR = 0.0021 A 900 = 1.875 V = 1.9 V





Chapter 161 The current through a component is directly proportional to the voltage across it

See experiment on page 32

2 The resistance of the tungsten filament of a lamp increases as it gets hotter

its current-voltage graph is curved

If the filament is maintained at a constant temperature (e.g. by keeping it immersed in water)

its current-voltage graph is a straight line through the origin

showing that the material from which the filament is made is ohmic

3 With switch S closed:

Vsupply= VR1 + Vdiode

Graph shows that at 50 mA, Vdiode = 0.77 V

So VR1 = 2.5 V 0.77 V = 1.73 V

R1 = VR1/I= 1.73 V/(0.05 A) = 34.6 = 35

With switch open:

Vsupply= VR1 + VR2 + Vdiode

Graph shows that at 10 mA, Vdiode = 0.65 V

VR1 =IR1 = 0.01 A 34.6 = 0.346 V

So VR2 = 2.5 V (0.65 V + 0.346 V) = 1.504 V

R2 = VR2/I= 1.504 V/(0.01 A) = 150.4 = 150

Diode power when switch S closed =IdiodeVdiode = 0.05 A 0.77 V = 0.0385 W = 39 mW

4 Voltage across series resistance VR= Vsupply VLED = 6.0 V 1.7 V = 4.3 V

R = VR/I= 4.3 V/(20 103A) = 215 = 220

5

Chapter 17

1 Voltage across resistance =IRexternal = 0.5 A 12 = 6 V

lost volts =E voltage across resistance = 9 V 6 V = 3 V

Internal resistance r= lost volts/I= 3 V/(0.5 A) = 6

Potential difference

Resistance


15/26


3 E.m.f. = terminal voltage + lost volts

E= V+Ir

V= (r)I+E

Graph of terminal voltage Vagainst currentIhas gradient equal torand intercept equal toE

E.m.f. = intercept = 1.64 V

Internal resistance = gradient = 1.62

4 Working resistanceR = V/I= 3.5 V/(0.35 A) = 10

Total circuit resistance =E/I= 12 V/(0.35 A) = 34

Required series resistance = 34 (4 + 10 ) = 20

lost volts = 4.5 V 3.5 V = 1.0 V

Internal resistance = lost volts/I= 1.0 V/(0.35 A) = 2.9

5 Digital voltmeter reads 12 V when switch S is open

CurrentI= e.m.f./total resistance = 12 V/(2 + 4 ) = 12 V/(6 ) = 2 A

Terminal voltage =IRexternal = 2 A 4 = 8 V

Power dissipated in internal resistancePr=I2r= (2 A)2 2 = 8 W

Power dissipated in external circuitPR=I2R= (2 A)2 4 =16 W

6 A car battery has a very small internal resistance (10 m)

so that it can supply a large current (e.g. 200 A) to the starter motor

A school e.h.t. supply has a very large internal resistance (5 M)

to prevent it from supplying dangerously large currents

0.30 0.1

Current A

1.8

1.6

1.4

1.2

1

0.8

0.6

Terminalvoltage/V

0.2 0.4 0.5 0.6

equation of line: terminal voltage=(1.62 current)+1.64V = r I + E

+








Chapter 181 Pressure is the force acting per unit area

Pressure = force/area

Pa = N m2 = kg m s2 m2 = kg m1 s2

2 (a) Pressure = force/area = 150 N/[(0.06 m)2] = 1.3 104 N m2

Force on piston =pA = 1.33 104 N m2 (0.96 m)2 = 38 000 N

(b) For volume of fluid to remain the same

(Distance moved area)piston = (distance moved area)plungerDistance moved by piston = 0.64 m (0.06 m)2/[(0.96 m)2] = 0.0025 m = 2.5 mm

3 See parts 2 and 3 of experiment Blowing up balloons on page 38

Use a pressure gauge to measure the pressure of the gas and a thermometer to measureits temperature

4 E.m.f. generated = 35 V C1 160 C = 5600 V = 5.6 mV

CurrentI= e.m.f./total resistance = 5.6 103V/(16 + 4 ) = 2.8 104A

Power producedP = VI= 5.6 103V 2.8 104A = 1.57 106W

5 Mark liquid level at 0 C (ice/water) and 100 C (steam) using elastic bands

divide interval between these two marks into 100 equal divisions

For this thermometer:

100 C (18.0 2.0) cm = 16.0 cm

So scale is 0.16 cm C1

(a) 35 C 35 C 0.16 cm C1 = 5.6 cm

Alcohol level is (5.6 + 2.0) cm = 7.6 cm above the bulb

(b) 8 C 8 C 0.16 cm C1 = 1.28 cm

Alcohol level is (1.28 + 2.0) cm = 0.72 cm above the bulb

Chapter 19

1 Macroscopic means large scale

Volume m3

Pressure Pa (orN m2)

Temperature K

Amount of gas present

2 Boyles law: for a fixed mass of gas at constant temperature, the product of the pressure andvolume is constant


Precaution: after each compression, wait for gas to cool before taking readings


17/26

3

4 Pressure volume = 1.8 106 Pa 0.06 m3 = 1.08 105 Pa m3

When tap opened:

new pressure new volume = 1.08 105 Pa m3

new volume = 1.08 105 Pa m3/(100 103 Pa) = 1.08 m3

air leaving cylinder = (1.08 0.06) m3 = 1.02 m3 = 1.0 m3

5 As the bubble rises, there is less water pushing down on it from above

so pressure on the fixed mass of gas in the bubble decreases as the bubble rises

and its volume increases so thatpVremains constant

Volume of bubble has increased by a factor of 5 (20 mm3/4 mm3)

so pressure at surface must be 1/5 of pressure at bottom of lake

Pressure at bottom of lake = 5 pressure at surface = 5 atmospheric pressure

Pressure at bottom is equivalent to 5 10 m of water = 50 m of water

Depth of lake = (50 10) m = 40 m

Chapter 201 Pressure law: for a fixed mass of gas at constant volume, the pressure is directly proportional to the

Kelvin temperature


Precautions:

submerge as much of the flask as possible in the water

use a short length of tubing to connect flask to pressure gauge

allow time for the gas in the flask to reach the temperature of the water before taking readings

Pressure

Volume

Higher temperature

Lower temperature







Solutions to Practice Questions2 As p Tthenp/T= constant

For the given results:

34 C is the incorrect temperature

Correct temperature = 111 kPa/(average of otherp/Tvalues) = 111 kPa/(0.3506 kPa K1) = 317 K= 44 C

3 Pressure/temperature = 100 kPa/[(273 + 15) K] = 0.347 kPa K1

New pressure/new temperature = 0.347 kPa K1

New temperature = new pressure/(0.347 kPa K1) = (100 + 20) kPa/(0.347 kPa K1) = 345.6 K= 72.6 C

4 Vreduces to 1

4V0whenp increases to 4p0 (constant T)

1

4V0 increases to V0when Tincreases to 4T(constantp)

4T= 4 (273 + 27) K = 1200 K = 927 C

To return to original conditions: temperature must be reduced back to 27 C at constant volume

5 The critical temperature of a gas is the temperature above which it cannot be liquefied

An ideal gas is one that would obey the gas laws at all temperatures and pressures and wouldnever liquefy

p is the pressure of the gas in Pa

Vis the volume of the gas in m3

n is the number of moles of gas present in mol

R is the molar gas constant in J K1 mol1

Tis the Kelvin temperature of the gas in K

Pressure

0

Volume

V014 V0

12 V0

34 V0

4p0

3p0

2p0

p0

0

C

A

B

Temperature/C 1 12 29 34 58 78

Pressure / kPa 96 100 106 111 116 123

Temperature / K 274 285 302 307 331 351

p/T/ kPa K1 0.350 0.351 0.351 0.362 0.350 0.350





Chapter 211 See experiment on page 44

When viewed through the microscope, the smoke particles appear as very small bright dots whichdance around randomly, moving first one way and then immediately another

2 The smoke particles are being bombarded by the gas particles

Although very light, the gas particles are moving very fast and undergo a significant changein momentum

At any instant, more gas particles will hit one side of the smoke particle than another

creating a resultant force that momentarily pushes the smoke particle in that direction

then the collision imbalance and the direction of the resultant force changesso the smoke particle continually gets pushed in different directions

3 (a) Use more ball bearings in the Perspex tube

(b) Increase the speed of the motor to make the molecules move faster

(c) Add mass to the cardboard disc

Place a small polystyrene sphere, representing a smoke particle, in with the ball bearings

4 Collisions of gas particles with the container walls exert forces and create pressures on them

Increasing the temperature of a gas causes its particles to move faster; the gas particles collide with

the walls harder and more often, producing a greater pressureIncreasing the volume of a container decreases the packing density of the gas particles within; fewercollisions per unit wall area occur per unit time and a lower pressure is produced

5 If collisions were inelastic, the average kinetic energy of the gas molecules would decrease

The molecules would slow down and stop moving, just like the ball bearings in the model

The temperature of the gas would fall and it would change into a liquid and then a solid

Chapter 22

1 For one collision:

Change of momentum = mv mu = 0.2 kg (15 15) m s1 = 0.2 kg 30 m s1 = 6 kg m s1

For 600 collisions:

Total change of momentum = 600 6 kg m s1 = 3600 kg m s1

Average force = total change of momentum/time taken = 3600 kg m s1/(12 s) = 300 N

600 collisions in 12 s = an average of 1 collision every 20 ms = 5 collisions in 100 ms

Average force acting on wall = total area under the graph/(100 ms)

Force

0

Time/ms

20 40 60 80 100




Solutions to Practice Questions2 p = pressure of the gas

= density of the gas

= mean square speed of the gas molecules

Left-hand side:

pressure = N m2 = kg m s2 m2 = kg m1 s2

Right-hand side:

= kg m3 (m s1)2 = kg m3 m2 s2 = kg m1 s2

E.g. (any four)

gas consists of a large number of particles in rapid, random motion

all collisions are elastic orparticles assumed to be hard elastic spheres

molecular size is negligible compared with the volume occupied by the gas

intermolecular forces are negligible except during collisions

time of collision is negligible compared with time between collisions

3 Sincep = /3

= 3p/= 3 101 103 Pa/(0.09 kg m3) = 3.37 106 m2 s2

Root mean square speed = (3.37 106 m2 s2) = 1835 m s1


4 (a) Left-hand side:

1

2mv2 = kg (m s1)2 = kg m2 s2

Right-hand side:

3kT/2 = J K1 K = J = N m = kg m s2 m = kg m2 s2

(b) Molar gas constant = Avogadro constant Boltzmann constant

(c) Since 1

2 mv2 = 3kT/2

v2 = 3kT/m so v2 T

T2/T1 = (847 + 273) K/[(7 + 273) K] = 1120 K/(280 K) = 4

v2 increases by a factor of 4

R.m.s. speed increases by a factor of 2 (i.e. 4)R.m.s. speed = 2 380 m s1 = 760 m s1

5 (a) Mean speed = (350 + 420 + 280 + 610 + 680 + 540 + 590 + 490) m s1/8 = 3960 m s1/8= 495 m s1 = 500 m s1

(b) Mean velocity = (350 + 420 280 + 610 680 540 + 590 490) m s1/8 = 20 m s1/8= 2.5 m s1

(c) Mean square speed = (3502 + 4202 + 2802 + 6102 + 6802 + 5402 + 5902 + 4902) m2 s2/8 =2 091 600 m2 s2/8 = 261 450 m2 s2 = 260 000 m2 s2

(d) Mean square velocity = [3502 + 4202 + (280)2 + 6102 + (680)2 + (540)2 + 5902 +(490)2] m2 s2/8 = 2 091 600 m2 s2/8 = 261 450 m2 s2 = 260 000 m2 s2

(e) Root mean square speed = (261 450 m2 s2) = 511 m s1 = 510 m s1

(f) Root mean square velocity = (261 450 m2 s2) = 511 m s1 = 510 m s1





Chapter 231 Internal energy: the total of the random kinetic and potential energies of all the molecules of that

body

Examples: working on it [e.g. hammering or passing an electric current through it (electricalworking)] or heating it (e.g. placing body in a hot fire)

2 Solid: molecules constantly vibrate about fixed positions

Liquid: molecular vibrations sufficient to allow molecules to interchange positions

Gas: molecules move throughout their container at high speeds

Internal energy is stored as both molecular kinetic energy (due to all kinds of motion including

vibration and spin) and potential energy (in the repulsive fields between the molecules)

3 A spoonful of hot water has more energy per degree of freedom than a bucketful of cold water as itis at a higher temperature, but it has less total internal energy as it has a much smaller mass

4 Random shuffling of energy quanta between the two bodies will favour movement from hot to coldas a hot body has more energy per degree of freedom than a cold body

5 The internal energy of an ideal monatomic gas is only kinetic, as there are no forces betweenits atoms

The internal energy of a real monatomic gas is both kinetic and potential, as there are forces

between its atoms

Chapter 24

1 Copper has conduction (free) electrons in its structure

The conduction electrons in the hotter part of the material gain energy and pass this on to thecolder part as they diffuse (move) into it

Quartz has very few conduction electrons in its structure but its atoms are strongly linked together

Large vibrational energy at the hot end is transferred along the atoms to atoms at the cold end

2 From flame to base of pan: direct contact of hot and cold bodies, convection within the flame andradiation from it

Through base of pan: conduction

Into water: adjacent to the base by conduction and throughout the water by convection

3 Energy conducts from the hot water to the cooler metal of the radiator

Energy mainly conducts from the warm radiator to the cooler air in contact with it

(Radiation from the radiator will be limited as its temperature is fairly low)

The heated air expands and rises, carrying energy into the room by convection




Solutions to Practice Questions4 Conduction: helpful in allowing energy to reach the contents of a pan through its base; unhelpful

in allowing energy to escape through the walls of a warm house

Convection: helpful in heating a whole room from a single source of heat e.g. one radiator;unhelpful when convection currents in the atmosphere result in a bumpy flight

Radiation: helpful in allowing the Sun to heat the Earth; unhelpful in radiating heat from a cookingpot even when it has a lid to prevent evaporation and convection

5 The two bodies in thermal equilibrium are at the same temperature

It is a dynamic situation as energy flows between the two bodies, although at the same rate inboth directions

In thermal equilibrium, all the temperatures are the same and there is no net flow of

internal energyIn steady state, the temperatures are constant but different and there is a steady flow ofinternal energy

Chapter 25

1 Specific heat capacity: energy needed to raise the temperature of 1 kg of that substance by 1 Kwithout a change of state

Unit: J kg1 K1

In base units: J kg1 K1 = N m kg1 K1 = kg m s2 m kg1 K1 = m2 s2 K1

2 See second experiment on page 52

Precautions: lagging, good thermal contact of block with heater and thermometer, measuremaximum temperature reached after heater turned off

Calculation: energy absorbed by block = mass specific heat capacity temperature rise

Energy supplied = potential difference current time = VIt

Specific heat capacity c = VIt/(mass temperature rise)

3 Energy supplied = VIt= V2t/R = (5.0 V)2 60 s/(25 ) = 60 J

mcT= 60 J

T= 60 J/(mc) = 60 J/(0.030 kg 380 J kg1 K1) = 5.3 K

4 Kinetic energy dissipated = 1

2 mv2 =

1

2 900 kg (30 m s1)2 = 405 000 J

Energy absorbed by each disc = 1

4 405 000 J = 101 250 J

mcT= 101 250 J

T= 101 250 J/(mc) = 101 250 J/(2.8 kg 460 J kg1 K1) = 78.6 K = 79 K

Either perform second experiment on page 52, or

Place a known massMof water at a measured temperature 1 in a calorimeter of known mass m

Increase temperature of disc to a known temperature (e.g. 100 C using boiling water)

Quicklydry and transfer hot disc to water in calorimeter

Measure maximum final temperature 2 of liquid

Energy lost by disc = energy gained by water and calorimeter

Specific heat capacity of disc = [Mcwater (2 1)] + [mccalorimeter (2 2)]/[mass of disc (100 2)]




Solutions to Practice Questions5 Atomic masses m

A:

Aluminium = 4.5 1026 kg

Copper = 1.1 1025 kg

Iron = 9.3 1026 kg

Lead = 3.5 1025 kg

If c 1/mAthen cmA= constant:

For aluminium, 880 J kg1 K1 4.5 1026 kg = 4.0 1023J K1

For copper, 380 J kg1 K1 1.1 1025 kg = 4.2 1023J K1

For iron, 450 J kg1 K1 9.3 1026 kg = 4.2 1023J K1

For lead, 130 J kg1 K1 3.5 1025 kg = 4.6 1023J K1

Since the products are all approximately the same, c 1/mANumber of atoms determines energy required

Larger atomic masses result in fewer atoms in each kilogram of substance

Less energy required to raise temperature of fewer atoms by 1 K

Concrete can be raised through a much higher temperature than water and so store more energyper kilogram

Concrete is cheap and safe to use (best notto use water with electric storage heaters!)

Chapter 26


Precautions: lag the cylinder; make sure heater is fully immersed

Calculation: energy used to vaporise water = mass of water vaporised specific latent heatof vaporisation

Energy supplied = potential difference current time = VIt

Specific latent heat of vaporisationL = VIt/(mass of water vaporised)

2 Energy required = mL = 0.016 kg 2.25 106J kg1 = 36 000 J

Minimum power = energy/time = 36 000 J/(60 s) = 600 W

Actual power will be greater to compensate for energy transferred to the surroundings by radiation,convection and conduction

3 Power of kettle = VI= 230 V 8 A = 1840 W

Energy required to heat water = mcT= 0.6 kg 4200 J kg1 K1 (100 15) K = 214 200 J

Energy required to heat kettle = 350 J C1 (100 15) C = 29 750 J

Time taken = total energy required/power = (214 200 + 29 750) J/(1840 W) = 132.6 s = 130 s

Energy supplied by kettle in two minutes = 1840 W 120 s = 220 800 J

Mass of water vaporised = energy supplied/latent heat = 220 800 J/(2.25 106J kg1) = 0.098 kg

Mass of water remaining in kettle = (600 98) g = 502 g = 500 g

4 (a) Energy given out by water = mcT= 0.284 kg 4200 J kg1 K1 22 K = 26 200 J

mL = 26 200 J

mass of ice that melts = 26 200 J/L = 26 200 J/(330 103J kg1) = 0.080 kg





The lead has solidified

(a) Energy lost by solid lead = mcT= 3 kg 130 J kg1 K1 6 K = 2340 J

Rate of energy loss = energy loss/time = 2340 J/(10 s) = 234 J s1 (or W)

(b) Energy loss in 5 minutes = 234 J s1 (5 60) s = 70 200 J

mL = 70 200 J

Specific latent heat of fusion of lead = 70 200 J/m = 70 200 J/(3 kg) = 23 400 J kg1

(c) Energy loss in 16 s = 234 J s1 16 s = 3744 J

mcT= 3744 J

Specific heat capacity of molten lead = 3744 J/mT= 3744 J/(3 kg 8 K) = 156 J kg1 K1

Chapter 27

1 Working involves a moving force

Heating involves a temperature difference

Mechanical working:

force squashes material and does mechanical workFx

Electrical working:

power supply forces electrons through material and does electrical work VIt

2 KE = 12 mv2 = 12

0.3 kg (15 m s1)2 = 33.75 J = 34 J

Lead is soft compared with the hammer head and yields under the hammer blows

Mechanical work is done on the lead as the force compresses it

The leads internal energy rises

(The lead is now able to heat the hammer head as it is at a higher temperature but the two are incontact only for a short period so little energy is transferred)

Total energy from 50 blows = 33.75 J 50 = 1690 J = 1700 J

mcT= 1690 J

Temperature rise = 1690 J/mc = 1690 J/(0.15 kg 130 J kg1 K1) = 87 K

3 Heating is the transfer of energy through a temperature difference from hot to cold

The hot lamp can heat the cold cell but not vice versa

The cold cell is forcing electrons to move through the hot lamp, an example of electrical working

1500 300

Time/s

610608

606

604

602

600

598

596

594

592

Temperature/K

50 100 200 250 350




Solutions to Practice Questions4 Electrical work done = VIt= 0.12 V 3.5 A 30 s = 12.6 J = 13 J

I= nAqv

v =I/nAq = 3.5 A/(1.7 1029 m3 1.0 106 m2 1.6 1019 C) = 1.29 104 m s1

= 0.13 mm s1

x= vt= 1.29 104 m s1 30 s = 3.9 103 m = 3.9 mm

Work done =Fx= 12.6 J

F= 12.6 J/x= 12.6 J/(3.9 103 m) = 3264 N = 3300 N

5 Electric mains supply forces electrons to move through the fire

The mains supply does electrical work on the electrons

The electrons then transfer their energy by collisions to the lattice of the heating elementThe mains supply cannot heat the electric fire since it is at a lower temperature than the fire

Some energy will radiate into the room from the glowing element but convection currents will carrymost of it around the room

Chapter 28

1 U: increase in the internal energy of the system

Q: energy transferred to the system by heating

W: energy transferred to the system by working

Conservation of energy

2 Q = 0 either when the system is thermally isolated or when the system is at the same temperatureas its surroundings

3 Thermos flasks do not completely isolate their contents from the surroundings

Some energy will always flow through either the walls or the lid

It is impossible to produce a completely isolated system

4 U= 0

The filament has reached a steady temperatureW=Pt= 24 W 5 s = 120 J

The power supply is working on the filament

Q = U W= 0 120 J = 120 J

The filament is heating the surroundings

5 U= Q + W

When a gas rapidly expands, Q = 0 as there is insufficient time for energy to enter or leave thesystem

so U= W

As the gas is doing work on its surroundings, Wis negativeso Uis also negative

The temperature of a rapidly expanding gas decreases


26/26



Chapter 291 Heat engine: a device that takes energy from a hot source, uses some of this to do mechanical work,

and gives the rest to a cold sink


2 Efficiency = useful output/input

Power of light emitted = (2/100) 60 W = 1.2 W

The other 58.8 W increases the internal energy of the surroundings

3 Make source temperature very high and sink temperature very low

Maximum efficiency = 1 T2/T1 = 1 300 K/(673 K) = 0.55

4 In a heat pump, work is done to force energy to flow from cold to hot

In a heat engine, energy flowing from hot to cold is used to do work

Heat pumps are used in refrigerators, freezers and air conditioners

5 Maximum efficiency = 1 3 K/(5 000 000 K) = 0.999 999 4

Eventually, the Suns temperature will decrease greatly (and that of the Universe willincrease slightly)

Documents

Nelson Thornes Electrcity and thermal physics Prac2