Oxford Revise AQA GCSE Physics | Chapter P6 answers

P6: Energy of matter

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Answers

Extra information

Mark

AO / Specification reference

01.1

volume before = 72 cm3

volume after = 97 cm3

difference in volume = volume of clay

= 97 cm3 – 72 cm3

= 15 cm3

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AO1

AO2

4.3.1.1

01.2

\(
\begin{array}{lr}
\rm{resolution} = \frac{1}{2}\textrm{smallest division}\\
= \frac{1}{2}\times 5 = 2.5\ \rm{cm}^3
\end{array}
\)

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AO1

A02

01.3

digital balance

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AO2

01.4

\(
\rm{density} = \frac{\textrm{mass}} {\textrm{volume}}
\)
\(
\rm{allow} \rho =
\frac{\rm{m}}{\rm{v}}
\)

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AO1

01.5

\(
\begin{array}{lr}
\rm{density }= \frac{23.41}{15}\\
= 1.56\ \rm{g/cm}^3
\end{array}
\)

accept 1.56 or 1.6 with no working shown for two marks

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AO2

4.3.1.1

01.6

measure the length of each side (in cm)/measure the length, breadth and height

cube the answer/multiply the length, breadth and height

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AO1

4.3.1.1

02.1

vibrating

potential

moving fast

kinetic

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AO1

4.3.2.1

02.2

the internal energy changes from mainly potential

to mainly kinetic/more kinetic

do not accept answers involving solids/liquids/gases

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AO1

4.3.2.1

02.3

the particles in a gas are in random motion

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AO1

4.3.2.1

03.1

20 °C

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AO2

03.2

70 °C

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AO2

03.3

energy transferred = power × time

allow E = P × t

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AO1

4.3.2.3

03.4

energy = 1000 × 2 × 60

= 120 000 J

energy transferred = mass × specific latent heat of vaporisation

\(
\begin{array}{lr}
120\ 000 = \rm{mass} \times 365\ 000\\
\rm{mass}= \frac{120\ 000}{365\ 000}\\
= 0.33\ \rm{kg}
\end{array}
\)

allow E = mL

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AO2

4.3.2.3

03.5

an overestimate

(the energy transferred to vaporise the water is lower because) some energy is transferred to the thermal energy store of the surroundings

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AO2

4.3.2.3

03.6

a line that starts at 20 C

steeper than the original line

becomes horizontal at 70 °C at about 2 minutes

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AO2

04.1

energy = power × time

allow E = P × t

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AO1

04.2

2 kW = 2000 W

2 minutes = 120 seconds

energy = 2000 × 120

= 240 000 J

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AO1

AO2

4.1.1.4

4.3.2.3

04.3

change in mass = 1.276 – 1.180 = 0.096 kg

energy transferred = specific latent heat × change in mass

240 000 J = specific latent heat of vaporisation × 0.096 kg

\(
\begin{array}{lr}
\textrm{specific latent heat of vaporisation} =
\frac{240\ 000}{0.096}\\
= 2\ 500\ 000\ \rm{J/kg}\\
= 2500\ \rm{kJ/kg}\\
\end{array}
\)

allow 2500 with no substitution for two marks

allow E = mL

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AO1

AO2

04.4

the energy transferred was heating the material of the kettle/air around it as well as vaporising the water

a lower mass of water vaporised than should have been the case

so the value calculated was bigger than the textbook value

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AO2

4.3.2.3

05.1

the gas has been compressed/is exerting a big pressure on the inside of the container

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AO1

4.3.3.2

05.2

the student has done work on the gas (so has transferred energy)

so has increased the internal energy of the system (which produces an increase in temperature)

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AO1/2

4.3.3.3

05.3

when the student does work, the particles move more quickly/energy is transferred to the particles

the average speed/kinetic energy of the particles increases, so the temperature increases

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AO1

4.3.3.3

06.1

(when the student is in the shower) water evaporates to make water vapour

the water vapour condenses when it comes in contact with the colder mirror

and energy is transferred to the mirror

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AO1

AO2

4.3.1.2

06.2

for converting all kJ to J, cm to m

\(
\begin{array}{lr}
\textrm{the mass of water that condenses} =
\frac{\rm{energy}}{\rm{specific\ latent\ heat\ of\ vaporisation}}\\
= \frac{\rm{730\ 000}}{\rm{2\ 265\ 000}}\\
= 0.322\ \rm{kg}\\
\textrm{mass = density} \times \rm{volume}\\
\textrm{volume of water} = \frac{0.322}{1 \times 10^-8}= 0.000/ 322\\
\rm{volume} = \rm{area} \times \rm{thickness}\\
\textrm{thickness of water} = \frac{0.000322}{0.6 \times 0.6}= 8.95 \times\ 10^-4\ \rm{m}
\end{array}
\)

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AO1

AO2

4.3.1.1

4.3.2.3

07.1

the particles in a solid are arranges in a regular pattern/array

the particles in a gas are moving in all/random direction

the particles in a solid are vibrating about a fixed position

the particles in a gas are moving quickly

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AO2

07.2

the particles collide with the walls of the container

each particle exerts a force on the wall

pressure is force per unit area

the total force of all the collisions of the particles per unit area is the pressure

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AO3

07.3

pressure on y-axis, temperature on x-axis

straight line with positive gradient

that intercepts y-axis above zero (does not need to be extrapolated)

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AO1

AO2

4.1.2.1

08.1

the steam is at a high temperature (and in the gas state)

steam transfers latent heat energy to the milk (steam cools/changes to a liquid state and the temperature of the milk rises)

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AO1

AO2

4.3.2.2

4.3.2.3

08.2

convert 242 g to 0.242 kg, and 3.93 kJ/kg °C to 3930 J/kg °C

temperature difference = 70 – 20 = 50 °C

energy required = mass × specific heat capacity × change in temperature

= 0.242 × 3930 × 50

= 47 553 J (\( \approx \) 48 000 J so about 48 kJ)

allow E = m c Δθ

allow 47533 with no substitution for three marks

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AO1

AO2

4.3.2.2

08.3

convert 2260 kJ/kg to 2260 000 J/kg

energy = mass × specific latent heat of vaporisation

\(
\begin{array}{lr}
47\ 553 = \rm{mass}\ \times\ 2\ 260\ 000\\
\rm{mass} = \frac{47\ 553}{2\ 260\ 000}\\
= 0.02\ \rm{kg}
\end{array}
\)

allow E = mL

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AO2

4.3.2.3

08.4

Assume that all the energy transferred to the milk to heat it comes from the change of state of the steam.

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AO2

4.3.2.3

09.1

B

D

the temperature isn’t changing/doesn’t change

even though the substance is being heated

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AO1

4.3.1.2

09.2

solid

it changes state twice/goes from solid to liquid, then liquid to gas

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AO1

4.3.1.2

09.3

A

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AO1

4.3.1.2

09.4

C

E

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AO1

4.3.1.2

10.1

volume = (5)3

= 125 cm3

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AO2

4.3.1.1

10.2

\(
\begin{array}{lr}
\rm{density} = \frac{\rm{mass}}{\rm{volume}}\\
1.5 = \frac{\rm{mass}}{125}\\
\rm{mass} = 1.5\ \times\ 125\rm{cm}^3\\
= 187.5\rm{g}
\end{array}
\)

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AO1

AO

4.3.1.1

10.3

sublimation

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AO1

4.3.1.2

10.4

if the process is reversed, the material recovers its original properties

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A01

4.3.1.2

10.5

the internal energy of the gas is bigger than the internal energy of the solid

the particles have more kinetic energy/are moving faster

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AO2

4.3.1.2

11.1

one mark for correct symbol

one mark for correct label

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AO1

4.2.1.1

11.2

(when they put the test tube into iced water) the temperature of the water decreases

the resistance of the thermistor increases

the potential difference across the thermistor increases

Vout decreases

because the total potential difference across the thermistor and the resistor = 12 V at all times

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AO1

AO2

4.2.1.4

11.3

find the potential difference across the resistor by measuring Vout

find potential difference across thermistor by subtracting Vout from 12 V

use
\(
\begin{array}{lr}
\frac{\rm{potential\ difference\ across\ the\ resistor}}{\rm{potential\ difference\ across\ the\ thermistor}}\\
= \frac{\rm{resistance\ of\ resistor}}{\rm{resistance\ of\ thermistor}}
\end{array}
\)
to find the resistance of the thermistor

use a graph/table of the resistance of the thermistor at different temperatures to work out the temperature of the water.

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AO1

AO2

4.2.2

11.4

sensible suggestions e.g., the human body continually generates thermal energy, but the water in the test tube does not

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AO3

12

Level 3: Well organised answer with descriptions of reasons for calculations. Equations for power, efficiency and gravitational potential energy used

5-6

AO1/1

AO2/1

4.1.1.4

4.1.1.2

Level 2: Some relevant calculations, but descriptions lacking detail or missing, or parts of calculations/conversions incorrect.

3-4

Level 1: Some relevant calculations completed, but unit conversions may be missing, and no explanation of method.

1-2

No relevant comment.

0

Indicative content:

  • calculate energy needed to be transferred to bulb:
    • \(
      \begin{array}{l}
      \rm{energy} = \rm{power} \times \rm{time}\\
      = 0.24 \times 60\\
      = 14.4\ \rm{J}
      \)
  • energy transferred to motor must be greater than this because the generator is only 90% efficiency
    • \(\rm{efficiency} = \frac{\rm{energy\ out}}{\rm{energy\ in}} \times\ 100\)
    • \(90 = \frac{14.4}{\rm{energy\ in}} \times\ 100\)
    • \(\rm{energy in} = \frac{14.4}{90} \times\ 100 \)
    • = 16 J
  • this energy is transferred by the falling mass.
    • gravitational potential energy = mass × gravity × height
    • 16 = 0.3 × 9.8 × height
    • height = 5.4 m

13.1

voltmeter

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AO2

4.2.1.3

13.2

potential difference = current × resistance

allow V = IR

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AO1

13.3

potential difference = 15 × 10

= 1.5 V

allow 1.5 with no substitution for two marks

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AO1

4.2.1.3

13.4

charge = current × time

allow Q = It

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AO1

13.5

charge = 0.15 × 60

= 9

C/coulombs

allow nine with no substitution for two marks

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AO1

AO2

4.2.1.2

13.6

it would increase

the potential difference has increased (but the clock resistance is the same)

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AO1

AO2

4.2.1.3

14.1

before the air is pumped out the pressure of the air in the marshmallow pockets = the pressure of the air outside the marshmallow

the particles of the air (inside and outside the marshmallow) collide with the (surface/material of the) marshmallow,

producing a force, force = pressure × area

initially the forces are the same

when the teacher pumps the air out, the force/pressure exerted by the particles of the air on the outside is smaller than the force exerted by the particles inside, (so the marshmallow expands)

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AO2

AO3

4.3.3.1

4.3.3.2

14.2

initially: pressure × volume = 101 × 103 × 8 × 10–6

= 0.808 (Pa m3)

= final pressure × final volume

final pressure =
\(
\frac{0.808}{2.7 \times 10^{-5}}
\)

= 2.99 × 104 Pa

= 29.9 kPa.

change in pressure = 101 – 29.9

= 71.1 kPa (71 100 Pa)

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AO1

4.3.3.2

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