Charge and Discharge of Capacitors

2 a) 470 s
b) 564 uF
c) 3.8 uC
d) 0.081V

3 a) 600 uC
b) 1.8 mJ
c) 2 s
d) 10 s

4 a) 90 uC
b) 7.5 s
c) 6.0 V
d) 12 uA
e) 7.5 s

5 1.39 s

6 a) 0.90 uC
b) 0.50 uC
c) 5.0 V
d) 4.0 V
e) 1.8 uA

7 120 ohms

y12 mock

Boyle's Law

Note. Any convenient units can be used for pressure and volume, provided that the units for P₁ are the same as those for P₂, and that the units for V₁ are the same as those for V₂.

(Use g = 9.8 N kg^-1)     1 atm =  Atmospheric pressure = 100 000 Pa = 100 000 Nm^-2

1. A fixed mass of gas at constant temperature has a volume of 500 cm² at a pressure of 1 atm. What will its volume be at a pressure of 4 atm?  125 cm³

2. A gas has a volume of 600cm³ at a pressure of 3.0 x 10⁵ Pa.  What is its volume at a pressure of 6.0 x 10⁵ Pa?  300 cm³

3. A gas has a volume of 1000 cm³ at a pressure of 1 atm. At what pressure will it occupy 100 cm³?   10 atm

4. The pressure exerted by a gas is due to the molecules of the gas bombarding the walls of the container. Why does the pressure increase when the volume is decreased at constant temperature? Less volume, molecules closer together, more collisions etc

5. A cylinder of compressed air has a volume of 51 cm³ at a pressure of 8 atm. When the cylinder is opened, what volume of air will escape?                                           

P₁V₁ = P₂V₂

rearranging V₂ = P₁V₁/ P₂           

V₂ = 8 x 51/ 1 = 408 cm³ minus the 51cm³ left in the cylinder = 357 cm³

 

6. A diver is working in sea-water of density 1025 kg m^-3 on a day when atmospheric pressure is 100 450 Pa. A bubble from his mask is three times the volume when it reaches the surface. At what depth is he working? 20m

7. A lake is 92.7 m deep. A bubble at the bottom of the lake has a volume of 0.5 cm³. What will its volume be at the surface if a water barometer would read 10.3 m (i.e. atmospheric pressure is the same as the pressure at this depth)? 5cm³

8. A uniform cylinder 0.60 m long is sealed at one end. It is lowered open end downwards into a lake so that the mouth just touches the bottom. When retrieved the inside is wet up to 0.50 m from the mouth. Find the depth of the lake if a water barometer would read 10.4 m of water. 52.5m

9. A cylindrical diving bell 8 m high is lowered into water so that the top of the bell is 0.75 m below the surface. If the atmospheric pressure is the same as that exerted by a column of water 10.25 m high, find how far the water rises inside the bell. 6m

A vessel of volume 500 cm³ containing air at a pressure of 8.0 x 10⁴ Pa is connected by a very narrow tube fitted with a tap, to a vessel of volume 700 cm³ containing air at a pressure of 2.0 x 10⁵ Pa. What is the resulting pressure in the vessels when the tap is opened?                     1.5 x10⁵ Pa

A compression pump is connected to a vessel of volume 500 cm³. For each stroke of the pump the piston sweeps 50 cm³ of air at atmospheric pressure into the vessel. If the initial pressure in the vessel is 1 atm, find the pressure in the vessel after 2 strokes.

1.2 atm

An exhaust pump is connected to a flask of volume 100 cm³. The volume swept out by the piston is 25 cm³ for each stroke. The pressure in the vessel is originally 1 atm. By considering one stroke at a time, find the pressure in the vessel after two strokes. 0.64atm

Conservation of energy answers

Conservation of Energy

(Use g = 9.8 N kg^-1)

1. A waterfall is 120 m high. If 75 % of the potential energy available is converted into heat energy, find the difference in temperature between the top and bottom of the fall. The specific heat capacity of water is 4200 J kg^-1 K^-1.                                       .21K

2. A lead bullet moving at 210 m s^-1 is brought to rest on hitting a target. If 60 % of its energy is converted into heat energy, find the rise in temperature of the bullet. The specific heat capacity of lead is

140 J kg^-1 K^-1.                                   94.5K

3. A lump of lead falls to the ground and its temperature rises by 1.4 °C. If all the potential energy is converted into heat energy, find the height fallen, if the specific heat capacity of lead is 140 J kg^-1 K^-1.                20m

4. If 60 % of the potential energy available in a waterfall is converted into heat energy, find the height of the waterfall when the temperature difference between the top and the bottom of the fall is 0.21 °C. (The specific heat capacity of water = 4200 J kg^-1 K^-1.)                  150m

5. A paddle wheel is made to rotate inside a calorimeter containing water by a mass of 12 kg falling through a height of 2 m. The total heat capacity of the calorimeter and its contents is 800 J K^-1. Find the rise in temperature of the water.                        .294K

6. A car of mass 800 kg is moving at 108 km h^-1. What heat energy is produced in the brake drums when it is brought to rest by the brakes?             360 kJ

7. On entering the earth's atmosphere the velocity of a meteorite of mass 6 kg is reduced rapidly from 300 m s^-1 to 100 m s^-1. How much heat energy is produced?                  240 kJ

8. A lead bullet moving at 70 m s^-1 is brought to rest on hitting a target. If 80% of its energy is converted into heat energy, find the rise in temperature. (The specific heat capacity of lead = 140 J kg^-1 K^-1.)                  14K

9. A lead bullet at 40.5 °C before impact just melts when it hits a target without rebounding. If all the energy of the bullet is converted into heat energy, find the velocity of the bullet. (The specific heat capacity of lead = 140 J kg^-1 K^-1. The melting point of lead = 328 °C. The specific latent heat of fusion of lead = 21 000 J kg^-1.)                  350ms^-1

Latent Heat Answers

Latent Heat

Latent Heat of fusion of ice = 3.36 x 10⁵ J kg^-1

Latent Heat of vaporization of water = 2.25 x 10⁶ J kg^-1

Specific heat capacity of water = 4200 J kg^-1 K^-1

1. How much heat energy is needed to change 0.010 kg of water at 20^oC to steam at 100^oC?                    25860J
2. How much heat energy is needed to change 0.020 kg of ice at 0°C into steam at 100°C?           60 120J  
3. How much heat energy is needed to melt 0.010 kg of ice at 0^o C and then heat the water formed to 20^o C? 4200j
4. What mass of ice at 0 °C is needed to cool 0.086 kg of water from      54 °C to 6 °C?                     .048kg
5. When a kettle boils a space can be seen between the spout and the so called `cloud of steam'. If your finger were placed in the space a much more serious scald would result than if it were placed in the `cloud of steam'. Explain why, and criticize the expression `cloud of steam'.
6. A heating coil takes 14 min to heat 0.20 kg of water from 20 °C to boiling point. How much longer will it take to boil away 0.02 kg of the boiling water?     562.5s
7. A 1 kW immersion heater takes 225 s to boil away 0.1 kg of water at 100 °C. Find the specific latent heat of vaporization of water.                     2.25x 10⁶ J kg^-1
8. How long will it take an immersion heater of power 500 W to boil away a mixture of 0.05 kg of water and 0.02 kg of ice initially at 0 °C? 387 s
9. A mixture of ice and water at 0 °C contains 0.060 kg of ice. A 750 W immersion heater takes 26.8 s to melt all the ice. Deduce a value for the specific latent heat of fusion of ice.        336000J kg^-1
10. A mixture of ice and water at 0 °C has a total mass of 1 kg, and con­tains 0.5 kg of ice. An immersion heater melts all the ice in 168 s, and then needs a further 84 s to heat the mixture to 20 ^°C. Find the specific latent heat of fusion of ice.            336000 Jkg^-1
11. A vessel of heat capacity 115 J K^-1 contains 0.30 kg of water and 0.20 kg of ice at 0 °C. An immersion heater takes 134.4 s to melt the ice and a further 88.6 s to heat the vessel and its contents to 20 °C. Find the specific latent heat of fusion of ice. 336000 Jkg^-1
12. A heating coil takes 115 s to heat 0.05 kg of a liquid from 20 °C to its boiling-point of 80 °C. It needs a further 340 s to boil away 0.02 kg of the liquid. If the specific heat capacity of the liquid is 2300 J kg^-1 K^-1, find its specific latent heat of vaporization. 1.02 x 10⁶ Jkg^-1

 

Specific Heat Capacity Answers

Specific Heat Capacity

Data

Use the following for Specific Heat capacities

Substance	Specific Heat Capacity c	Units
Water	4200	J kg-1 K-1
Copper	380	J kg-1 K-1
Methylated Spirit	2500	J kg-1 K-1
Turpentine	1800	J kg-1 K-1

Notes

• The heat capacity of an object is the heat required to raise the all of the object’s temperature by 1K
• When heating a liquid you have to heat the vessel that it is in as well.
• When an object at high temperature is dropped into a liquid at low temperature there is a net heat flow from the high temperature object to the low temperature object until they are both at the same temperature.
• Unless you state “Ignoring heat losses” in each question you will lose a mark!

1 Find the quantity of heat needed to heat 0.50 kg of copper from 20^oC to 50 °C.              5700J

2 When 0.25 kg of ice are heated from -10 °C to -2 °C the heat supplied is 4200 J. Find the specific heat capacity of ice.                      2100 Jkg^-1K^-1

3 A vessel of heat capacity 30 J K^-1 contains 0.30 kg of water at 10 °C. When heated by a Bunsen burner for 5 min the temperature rises to 90 °C. At what rate does the Bunsen supply heat energy?     344W

4 A piece of iron of mass 0.20 kg is heated to 100 °C and dropped into 0.15 kg of water at 20 °C. If the temperature of the mixture is 30 °C, find a value for the specific heat capacity of iron.     450 Jkg^-1K^-1

5 A piece of iron of mass 0.05 kg is heated in a flame and then quickly transferred to a calorimeter of mass 0.05 kg and specific heat capacity 360 J kg^-1 K^-1 containing 0.18 kg of water at 18 °C. If the temperature of the water rises to 43 °C, find the temperature of the flame. The specific heat capacity of the iron is 450 J kg^-1 K^-1.                 903 ^oC

6 Which would result in the more serious burn if dropped onto your outstretched hand a white-hot spark from a firework, or a lump of red-hot iron? Explain your answer.

The white hot spark has a higher temperature than the lump of iron

The lump of iron has a greater mass than the white hot spark

They both have the same specific heat capacity.

Lump of iron has greater store of heat energy

The rate of heat transfer from the white hot spark will be greater for the spark than the lump of iron.

A greater quantity of heat will be transferred from the iron

Causing more serious skin damage.

Power work sheet answers

Work, Energy and Power

(Use g = 9.8 Nkg^-1)

Examples

A Find the work done in lifting a mass of 6 kg through a height of 10 m.

W = Fx W work done in J

F force in N

x distance moved in the direction of the force in m

W = 6 x 9.8 x 10

= 588 J

B) An escalator carries 100 people of average mass 70 kg to a height of 6 m in one minute.

Find the power necessary to do this.

Work done = F x = 100 x (70 x 9.8) x 6 J

Power = rate of doing work = work done per second

= (100 x 70 x 9.8 x 6) / 60 s = 6860 W or 6.86 kW

1. Find the work done to lift a mass of 10 kg through a height of 5 m. If this takes 15s what is the power?

W= Fx = (10kg)(9.8Nkg^-1) (5m) = 490J

P = W/t = 490 J / 15s = 32.67 W

2         A mass of 5 kg is pulled along a surface at a constant speed. Frictional force between the mass and the surface is 4N what work is done in moving the mass 2 m along the surface? Find the power if this takes 2 s.

W = Fx = (4N) (2m) = 8J

P = W/t =  8J/ 2s = 4 W

3        A girl of mass 50 kg runs up a flight of steps 4.5 m high in 5 s. What power does she develop?

W = Fx             F = 50kg x 9.8 = 490.5 N                     x = 4.5m

W = 490.5 x 4.5 = 2207.25 J

P = W/t = 2207.25 J / 5s = 442w

4        2 x 10⁶ kg of water per second flow over a large waterfall of height 50 m.

W = Fx                   F = 2 x 10⁶ kg x 9.8 = 1.96 x10⁺⁷N      x=50m

W= (1.96 x10⁺⁷N) (50m) = 9.8 x10⁺⁸ J

What power is available?

As 1W = I J per second the water falling over the fall transfers 9.8 x10⁺⁸ J of GPE per second the available power is 9.8 x10⁺⁸ W (ignoring losses due to friction)

5        The first-stage rocket motors of Saturn 5 burnt fuel at the rate of 13600 kg per second. The work done by the pumps to drive the fuel into the combustion chambers is the same as that necessary to lift the fuel through 1680 m.

W= Fx                    F = 13600kg x 9.8= 1.3328 x 10⁺⁵ N    x = 1680m

W = (1.3328x 10⁺⁵ ) (1680) = 2.24 x 10⁺⁸ J

What power is needed to do this? (Your answer is about twice the engine power of the largest ocean liner.)

As this mass of fuel is burnt in 1 second then 2.24 x 10⁺⁸ J of work is done in 1 second so the power is 2.24 x 10⁺⁸ W

6        A gas turbine locomotive developing 6330 kW pulls a train at a steady velocity of 108 kmh^{- 1}. What is the total force in Newtons resisting the motion?

At a steady velocity all work done by the engine is against friction or gravity.

The engine does 6.33 x10⁶ J of work every second

As W = Fx the force exerted by the engine is equal Work done per second divided by the distance travelled per second

F = W/x            108 kmh^-1 = (108000m)/(60 min x 60 sec) = 30 ms^-1

F = 6.33 x10⁶ J/ 30m = 211 000N = 2.11 x 10⁵ N

7        A mechanical loader driven by an engine developing 3 kW lifts a total load of 200000 kg to a height of 3 m in one hour. What is the efficiency of the system? 54.4%

Work done lifting 2 x 10⁵kg through 3m

W= Fx              F = (2 x 10⁵kg) (9.8) = 1.96 x 10⁶ N     x = 3m

W = (1.96 x 10⁶ N) (3m) = 5.88 x 10⁷ J

Time = 1hr = 3600 seconds

Power out = (5.88 x 10⁷ J) / (3600) = 1630 W

(Efficiency = Power out / Power in = 1630 / 3000 = 0.54 = 54%)

Power

(Use    g = 9.8 Nkg^-1)

Examples

A         Find the work done in lifting a mass of 6 kg through a height of 10 m.

            W = Fx            W        work done in J

 F         force in N

x          distance moved in the direction of the force in m

W = 6 x 9.8 x 10

    = 588 J

B) An escalator carries 100 people of average mass 70 kg to a height of 6 m in one minute.

Find the power necessary to do this.

Work done = F x = 100 x (70 x 9.8) x 6 J

Power = rate of doing work = work done per second

           = (100 x 70 x 9.8 x 6) / 60 s = 6860 W or 6.86 kW

1. Find the work done to lift a mass of 10 kg through a height of 5 m. If this takes 15 s what is the power?

2. A mass of 5 kg is pulled along a surface at a constant speed. Frictional force between the mass and the surface is 4N what work is done in moving the mass 2 m along the surface? Find the power if this takes 2 s.

3. A girl of mass 50 kg runs up a flight of steps 4.5 m high in 5 s. What power does she develop?

4. 2000000 kg of water per second flow over a large waterfall of height 50 m. What power is available?

5. The first-stage rocket motors of the Saturn 5 moon rocket burnt fuel at the rate of 13600 kg per second. The work done by the pumps to drive the fuel into the combustion chambers was the same as that necessary to lift the fuel through 1680 m. What power is needed to do this? (Your answer is about twice the engine power of the largest ocean liner.)

6. A gas turbine locomotive developing 6330 kW pulls a train at a steady velocity of 108 kmh^{- 1}. What is the total force in Newtons resisting the motion?

7. A mechanical loader driven by an engine developing 3 kW lifts a total load of 200000 kg to a height of 3 m in one hour. What is the work done on the load? How much power is developed lifting the load? How does this compare with the work done by the motor in the same time?

mark scheme for cosmology questions

Answers to questions on Stellar Distances

Questions on Stellar distances

y13 assessment mark scheme

1. (a) (i) Cp = 2 + 4 = 6 μF A1


(ii) 1/C = 1/2 + ¼ C1

Cs = 4/3 =1.33 μF A1

(b) (i) 6.0 V A1

(ii) Q = CpV C1

= 6 × 6 = 36 μC A1

(c) E = ½ CsV2 C1

= 24 × 10–6 A1

(d) (i) The capacitors discharge through the voltmeter. B1

(ii) V = V0e–t/CR

1/4 =e–t/(6×12) C1

ln 4 = t / 72 C1

t = 72 ln 4 ≈ 100 s A1

[12]



2. (a) (i) Q = VC; W = ½ VC.V ( = ½ CV2) (2)

(ii) parabolic shape passing through origin (1)

plotted accurately as W = 1.1 V2 (1) 4

(b) (i) T = RC; = 6.8 × 103 × 2.2 = 1.5 × 104 s = 4.16 h 2

(ii) ΔW = ½ C(V12 –V22) = 1.1(25 – 16) ; = 9.9 (J) 2

(iii) 4 = 5 exp(–t/1.5 × 104) ; giving t = 1.5 × 104 × ln 1.25 = 3.3 × 103 (s) 2

(iv) P = ΔW/Δt = 9.9/3.3 × 103 = 3.0 mW ecf b(ii) and (iii) 1

allow P = Vav2 /R = 4.52/6.8 × 103 = 2.98 mW

[11]



3 (a) (i)

capacitor capacitance / µF charge / µC p.d. / V energy / µJ

X 5 30 = Q/C

= 6 (V) (1) = ½ CV 2(1)

= ½ × 5 × 62

= 90 (1)

Y 25 = CV

= 25 × 6

=150 (µC) (1) = 6 (V) (1) = 450 (1)

Z 10 30 + 150 =

180 (µC) (1) = Q/C

= 180/10

= 18 (V) (1) = 1620 (1)

Each box correctly calculated scores (1) + (1) for ½ CV2 9

(ii) 1 18 V + 6 V = 24 (V) (1)

2 180 (µC) (1)

3 180 / 24 = 7.5 (1)

4 90 + 450 + 1620 = 2160 (µJ) (1) 4

(b) (i) Kirchhoff’s second law OR conservation of energy (1) 1

(ii) Kirchhoff’s first law OR conservation of charge (1) 1

(c) (i) time constant = CR (1)

= 7.5 × 10–6 × 200 000 = 1.5 (s) (1) 2

(ii) (1)

Q/Qo = e–4 = 0.0183 (1) 2

[19]



4. (a) Positive as E-field is downwards/top plate is positive/like charges repel/AW (1) 1

(b) (i) k.e. = QV; = 300 × 1.6 × 10–19 = (4.8 × 10–17 J) (2) 2

(ii) 1/2mv2 = 4.8 × 10–17; = 0.5 × 2.3 × 10–26 × v2 so v2 = 4.17 × 109;

(giving v = 6.46 × 104 m s–1) (2) 2

(c) E = V/d; so d = V/E = 600/4 × 104 = 0.015 m (2) 2

Y12 assessment mark scheme

Note Some symbols have not been copied across accurately

1. (a) either (If in parallel) when one bulb fails, other bulbs stay on


or (If in parallel) can identify which bulb has failed; (1) 1

(b) (i) P = VI (1)

0.5 = 240 I

I = 2.1 × 10–3 A 1 s.f. in answer (–1) once only (1) 2

(ii) R = V/I (1)

= 240/(2.1 × 10–3)

= 1.14 × 105 Ω or 1.15 × 105 Ω ans

accept (1.1 to 1.2) × 105 Ω. (1) 2

(iii) A = ρ l / R (1)

= 1.1 × 10–6 × 6.0 × 10–3 / (1.14 × 105) (= 5.79 × 10–14 m2)

A = πr2 (1)

5.79 × 10–14 = πr2 so r = 1.4 × 10–7 m (1) 3

(iv) filament too thin / fragile to be manufactured / used without damage;

allow ecf from (iii). (1) 1



4

[20]



2. (a) current  p.d / voltage (for a metallic conductor) M1

as long as temperature is constant / physical conditions remain constant A1

(b) (i) (R =) (= 0.0349) B1

(ii) R = (Allow any subject) C1

ρ = = C1

resistivity = 5.6  10–5 A1

unit: ohm metre /  m (Allow V m A–1) A1

(5.6  10–n without unit or incorrect unit and n  5 or 3 – can score 2/4)

(5.6  10–3  m – can score 3/4)

(5.6  10–3  cm – can score 4/4)



3. (a) Line crosses ‘y-axis’ at 1.4 (V) / V = E or 1.4(V) when I = 0

V = E – Ir; since I = 0 (Hence V = E or 1.4(V)) B1

(b) (i) (Graph extrapolated to give) current = 2.0 (A)

(Allow tolerance  0.1A) B1

(ii) E = I(max) r gradient = r (Ignore sign) C1

(r = ) (Attempt made to find gradient)

r = 0.7(0) () r = 0.7(0) () (Possible ecf) A1

(iii) (excessive) heating of cell / energy wasted internally /

cell might ‘explode’ / cell goes ‘flat’ (quickly) B1

[5]



4. (a) Energy (transformed by a device working) at 1 kW for 1 hour B1

(b) E = Pt / 5.8 = 0.12  time / (time =) 48.3 (hr) C1

(time =) 1.74  105  1.7  105 (s) A1

[3]



5. (i) v = I / nAe = 0.0025 / (8.5 × 1028 × 1.1 × 10–7 × 1.6 × 10–19) (1)

= 1.67 × 10–6 m s–1 (1) 2

(ii) Free electron concentration (or wtte) is much smaller in the thermistor than

in the wire. 1

[3]



6. (a) R = R1 + R2 / R = 200 + 120 / R = 320 C1

current = C1

current = 2.5  10–2 (A) A0

(b) V = 25  10–3  120 / V =

V = 3.0 (V) (Possible ecf) B1

(c) p.d. across the 360 () resistor = p.d. across the 120 () resistor /

There is no current between A and B / in the voltmeter B1

(Allow ‘A & B have same voltage’ - BOD)

The p.d. calculated across 360  resistor is shown to be 3.0 V /

The ratio of the resistances of the resistors is shown to be the same. B1

[5]





7. (i) evidence of knowledge of:

full/max transmission when the (transmission axis of) polarising sheet is

parallel to the light’s plane of polarisation/vibrations B1

no transmission when the (transmission axis of) polarising sheet is at right

angles to light’s plane of polarisation/vibrations B1

(ii) reflected light from surface is partially plane polarised B1

polarising sheet is placed at right angles to reflected light’s polarisation B1

plane/AW

[4]



8. (a) (i) amplitude = 1.2 (mm) B1

(ii) period = 2.4 (ms) B1

{allow 2.4 × 10–3 ms if 2.4 × 10–3 is correctly used in

substitution in (b)(i)}

(b) (i) frequency = 1/period C1

1/0.0024 = 417Hz (OR 420) A1

{1/2.4 = 0.417 OR 0.42 OR 0.4 scores 1 mark}

{allow ecf from cand’s period value}

(ii) recall of v = f  OR c = f  OR  = vT OR 1500 = 417  C1

 = 3.6 m A1

{ecf for cand’s f: e.g. λ= 1500/0.417 = 3600 m scores 2 marks

OR λ = 1500/0.4 = 3750 m scores 2 marks}

λ = 1500/0.42 = 3571 m scores 2 marks

(iii) valid scale for cand’s λ shown on position axis AND at least two full

‘sine’ waves drawn (waves can be very rough but not square waves } B1

amp. shown as 1.2 mm +/– ½ sq.: check first peak + trough only B1

first wavelength correct as 3.6 m +/– 1 sq.{allow ecf from (b)(ii)} B1

{NB If there is no scale on the position axis the 1st and 3rd marks cannot be scored}

[9]

y12 Homework on Resistivity

1 See note on resistivity, or check mail blog
2 3.1 ohm
3 17.8 ohm
4 0.64mm
5 2.78g

This homework is now closed

OCR questions Shm and Gravity

 

Y12 Homework on resistivity

The answers are below.

1. 1.4 ohm
2. 2.16 x 10 –7 m2
3. 1.1 x 10 –7 ohm m
4. 0.66 m
5. 2.5 ohm
6. 24 ohm

questions on graviational fields and SHM