Calculate, with a suitable unit, the electrical power output of the power station.

Calculate the mass of CO₂ generated in a year assuming the power station operates continuously.

Explain, using your answer to (b), why countries are being asked to decrease their dependence on fossil fuels.

Describe, in terms of energy transfers, how thermal energy of the burning gas becomes electrical energy.

«55.5 × 14.6 × 0.59» = 4.78 × 108 W

A unit is required for this mark. Allow use of J s^–1.

No sf penalty.

«14.6 × 2.75 × 3.16 × 10⁷ =» 1.27 × 10⁹ «kg»

If no unit assume kg

CO₂ linked to greenhouse gas OR greenhouse effect

leading to «enhanced» global warming
OR
climate change
OR
other reasonable climatic effect

Internal energy of steam/particles OR KE of steam/particles

«transfers to» KE of turbine

«transfers to» KE of generator or dynamo «producing electrical energy»

Do not award mark for first and last energies as they are given in the question.

Do not allow “gas” for “steam”

Do not accept reference to moving OR turning generator

Show that the intensity of solar radiation at the orbit of Mars is about 600 W m^–2.

Determine, in K, the mean surface temperature of Mars. Assume that Mars acts as a black body.

The atmosphere of Mars is composed mainly of carbon dioxide and has a pressure less than 1 % of that on the Earth. Outline why the greenhouse effect is not significant on Mars.

use of $I\propto \frac{1}{r^2}$ «1.36 × 10³ × $\frac{1}{{1.5}^2}$» ✔

604 «W m^–2» ✔

use of $\frac{600}{4}$ for mean intensity ✔

temperature/K = «$\sqrt[4]{\frac{600}{4\times 5.67\times {10}^{-8}}}=$» 230 ✔

recognize the link between molecular density/concentration and pressure ✔

low pressure means too few molecules to produce a significant heating effect

OR

low pressure means too little radiation re-radiated back to Mars ✔

Outline, with reference to energy changes, the operation of a pumped storage hydroelectric system.

The hydroelectric system has four 250 MW generators. The specific energy available from the water is 2.7 kJ kg^–1. Determine the maximum time for which the hydroelectric system can maintain full output when a mass of 1.5 x 10¹⁰ kg of water passes through the turbines.

Not all the stored energy can be retrieved because of energy losses in the system. Explain one such loss.

At the location of the hydroelectric system, an average intensity of 180 W m^–2 arrives at the Earth’s surface from the Sun. Solar photovoltaic (PV) cells convert this solar energy with an efficiency of 22 %. The solar cells are to be arranged in a square array. Determine the length of one side of the array that would be required to replace the
hydroelectric system.

PE of water is converted to KE of moving water/turbine to electrical energy «in generator/turbine/dynamo»

idea of pumped storage, ie: pump water back during night/when energy cheap to buy/when energy not in demand/when there is a surplus of energy

total energy = «2.7 x 10³ x 1.5 x 10¹⁰ =» 4.05 x 10¹³ «J»

time = «$\frac{4.0\times {10}^{13}}{4\times 2.5\times {10}^8}$» 11.1h or 4.0 x 10⁴ s

For MP2 the unit must be present.

friction/resistive losses in walls of pipe/air resistance/turbulence/turbine and generator bearings

thermal energy losses, in electrical resistance of components

water requires kinetic energy to leave system so not all can be transferred

Must see “seat of friction” to award the mark.
Do not allow “friction” bald.

area required $=\frac{1\times {10}^9}{0.22\times 180}$ «= 2.5 x 10⁷ m²»

length of one side $=\sqrt{area}=5.0$ k«m»

Identify the laws of conservation that are represented by Kirchhoff’s circuit laws.

State the emf of the cell.

Deduce the internal resistance of the cell.

The voltmeter is used in another circuit that contains two secondary cells.

Cell A has an emf of 10 V and an internal resistance of 1.0 Ω. Cell B has an emf of 4.0 V and an internal resistance of 2.0 Ω.

Calculate the reading on the voltmeter.

Outline why electricity is a secondary energy source.

Some fuel sources are renewable. Outline what is meant by renewable.

A fully charged cell of emf 6.0 V delivers a constant current of 5.0 A for a time of 0.25 hour until it is completely discharged.

The cell is then re-charged by a rectangular solar panel of dimensions 0.40 m × 0.15 m at a place where the maximum intensity of sunlight is 380 W m⁻².

The overall efficiency of the re-charging process is 18 %.

Calculate the minimum time required to re-charge the cell fully.

Outline why research into solar cell technology is important to society.

« conservation of » charge ✓

« conservation of » energy ✓

Allow [1] max if they explicitly refer to Kirchhoff’ laws linking them to the conservation laws incorrectly.

12 V ✓

I = 2.0 A OR 12 = I (r +4) OR 4 = Ir OR 8 = 4I ✓

«Correct working to get » r = 2.0 «Ω» ✓

Allow ECF from (b)(i)

Loop equation showing EITHER correct voltages, i.e., 10 – 4 on one side or both emfs positive on different sides of the equation OR correct resistances, i.e. I (1 + 2) ✓

10−4 = I (1 + 2) OR I = 2.0 «A» seen✓

V = 8.0 «V» ✓

Allow any valid method

is generated from primary/other sources ✓

«a fuel » that can be replenished/replaced within a reasonable time span

OR

«a fuel» that can be replaced faster than the rate at which it is consumed

OR

renewables are limitless/never run out

OR

«a fuel» produced from renewable sources

OR

gives an example of a renewable (biofuel, hydrogen, wood, wind, solar, tidal, hydro etc..) ✓

OWTTE

ALTERNATIVE 1

«energy output of the panel =» VIt OR 6 x 5 x 0.25 x 3600 OR 27000 «J» ✓

«available power =» 380 x 0.4 x 0.15 x 0.18 OR 4.1 «W» ✓

$t=$ «$\frac{27000}{4.1}$=» 6600 «s» ✓

ALTERNATIVE 2

«energy needed from Sun =» $\frac{Vlt}{eff}$ OR $\frac{6\times 5\times 0.25\times 3600}{0.18}$ OR 150000 «J» ✓

« incident power=» 380 x 0.4 x 0.15 OR 22.8 «W» ✓

$t=$ «$\frac{150000}{22.8}$=» 6600 «s» ✓

Allow ECF for MP3

Accept final answer in minutes (110) or hours (1.8).

coherent reason ✓

e.g., to improve efficiency, is non-polluting, is renewable, does not produce greenhouse gases, reduce use of fossil fuels

Do not allow economic reasons

a) Most just stated Kirchhoff's laws rather than the underlying laws of conservation of energy and charge, basically describing the equations from the data booklet. When it came to guesses, energy and momentum were often the two, although even a baryon and lepton number conservation was found. It cannot be emphasised enough the importance of correctly identifying the command verb used to introduce the question. In this case, identify, with the specific reference to conservation laws, seem to have been explicit tips not picked up by some candidates.

bi) This was probably the easiest question on the paper and almost everybody got it right. 12V. Some calculations were seen, though, that contradict the command verb used. State a value somehow implies that the value is right in front to be read or interpreted suitably.

bii) In the end a lot of the answers here were correct. Some obtained 2 ohms and were able to provide an explanation that worked. A very few negative answers were found, suggesting that some candidates work mechanically without properly reflecting in the nature of the value obtained.

ci) A lot of candidates figured out they had to do some sort of loop here but most had large currents in the voltmeter. Currents of 2 A and 10 A simultaneously were common. Some very good and concise work was seen though, leading to correct steps to show a reading of 8V.

cii) This question was cancelled due to an internal reference error. The paper total was adjusted in grade award. This is corrected for publication and future teaching use.

di) The vast majority of candidates could explain why electricity was a secondary energy source.

dii) An ideal answer was that renewable fuels can be replenished faster than they are consumed. However, many imaginative alternatives were accepted.

ei) This question was often very difficult to mark. Working was often scattered all over the answer box. Full marks were not that common, most candidates achieved partial marks. The commonest problem was determining the energy required to charge the battery. It was also common to see a final calculation involving a power divided by a power to calculate the time.

eii) Almost everybody could give a valid reason why research into solar cells was important. Most answers stated that solar is renewable. There were very few that didn't get a mark due to discussing economic reasons.

Describe the difference between photovoltaic cells and solar heating panels.

A solar farm is made up of photovoltaic cells of area 25 000 m². The average solar intensity falling on the farm is 240 W m^–2 and the average power output of the farm is 1.6 MW. Calculate the efficiency of the photovoltaic cells.

Determine the minimum number of turbines needed to generate the same power as the solar farm.

Explain two reasons why the number of turbines required is likely to be greater than your answer to (c)(i).

solar heating panel converts solar/radiation/photon/light energy into thermal energy AND photovoltaic cell converts solar/radiation/photon/light energy into electrical energy

Accept internal energy of water.

power received = 240 × 25000 = «6.0 MW»

efficiency «$=\frac{1.6}{6.0}$ = 0.27 / 27%

area = $\pi$ × 17² «= 908m²»

power = $\frac{1}{2}\times 908\times 1.3\times {7.5}^3$ «= 0.249 MW»

number of turbines «$=\frac{1.6}{0.249}=6.4$» = 7

Only allow integer value for MP3.

Award [2 max] for 25 turbines (ECF from incorrect power)

Award [2 max] for 26 turbines (ECF from incorrect radius)

«efficiency is less than 100% as»

not all KE of air can be converted to KE of blades

OR

air needs to retain KE to escape

thermal energy is lost due to friction in turbine/dynamo/generator

Allow velocity of air after turbine is not zero.

State what is meant by binding energy of a nucleus.

Outline why quantities such as atomic mass and nuclear binding energy are often expressed in non-SI units.

Show that the energy released in the reaction is about $180 \mathrm{MeV}$.

Estimate, in $\mathrm{J} {\mathrm{kg}}^{-1}$, the specific energy of U-235.

The power station has a useful power output of $1.2 \mathrm{GW}$ and an efficiency of $36 \%$. Determine the mass of U-235 that undergoes fission in one day.

Write down the proton number of nuclide X.

State the half-life of Sr-94.

Calculate the mass of Sr-94 remaining in the sample after $10$ minutes.

energy required to «completely» separate the nucleons
OR
energy released when a nucleus is formed from its constituent nucleons ✓

Allow protons AND neutrons.

the values «in SI units» would be very small ✓

$140\times 8.29+94\times 8.59-235\times 7.59$ OR $184 «\mathrm{MeV}»$ ✓

see $«\mathrm{energy}=» 180\times {10}^6\times 1.60\times {10}^{-19}$ AND $«\mathrm{mass}=» 235\times 1.66\times {10}^{-27}$ ✓

$7.4\times {10}^{13} «\mathrm{J} {\mathrm{kg}}^{-1}»$ ✓

energy produced in one day$=\frac{1.2\times {10}^9\times 24\times 3600}{0.36}=2.9\times {10}^{14} «\mathrm{J}»$ ✓

mass$=\frac{2.9\times {10}^{14}}{7.4\times {10}^{13}}=3.9 «\mathrm{kg}»$ ✓

$39$ ✓

Do not allow ${}_{39}{}^{94}\mathrm{X}$ unless the proton number is indicated.

$75 «\mathrm{s}»$ ✓

ALTERNATIVE 1

$10 \min$$=8 t_{1/2}$ ✓

mass remaining$=1.0\times {\left(\frac{1}{2}\right)}^8=3.9\times {10}^{-3}«\mathrm{kg}»$ ✓

ALTERNATIVE 2

decay constant$=«\frac{\ln 2}{75}=»9.24\times {10}^{-3} «{\mathrm{s}}^{-1}»$ ✓

mass remaining$=1.0\times e^{-9.24\times {10}^{-3}\times 600}=3.9\times {10}^{-3} «\mathrm{kg}»$ ✓

Determine the orbital period for the satellite.

Mass of Earth = 6.0 x 10²⁴ kg

Determine the mean temperature of the Earth.

Suggest how the difference between λ_S and λ_E helps to account for the greenhouse effect.

Not all scientists agree that global warming is caused by the activities of man.

Outline how scientists try to ensure agreement on a scientific issue.

$\frac{m{v}^2}{r}=G\frac{Mm}{r^2}$

leading to T² = $\frac{4{\pi }^2{r}^3}{GM}$

T = 5320 «s»

Alternative 2

«$v=\sqrt{\frac{G{M}_E}{r}}$» = $\sqrt{\frac{6.67\times {10}^{-11}\times 6.0\times {10}^{24}}{6600\times {10}^3}}$ or 7800 «ms^–1»

distance = 2$\pi$r = 2$\pi$ x 6600 x 10³ «m» or 4.15 x 10⁷ «m»

«T = $\frac{d}{v}=\frac{4.15\times {10}^7}{7800}$» = 5300 «s»

Accept use of ω instead of v

T = «$\frac{2.90\times {10}^{-3}}{{\lambda }_{\text{max}}}=$» $\frac{2.90\times {10}^{-3}}{10.1\times {10}^{-6}}$

= 287 «K» or 14 «°C»

Award [0] for any use of wavelength from Sun

Do not accept 287 °C

wavelength of radiation from the Sun is shorter than that emitted from Earth «and is not absorbed by the atmosphere»

infrared radiation emitted from Earth is absorbed by greenhouse gases in the atmosphere

this radiation is re-emitted in all directions «including back to Earth»

peer review

international collaboration

full details of experiments published so that experiments can repeated

[Max 1 Mark]

Explain why the power incident on the planet is

                                                                $\frac{P}{4\pi d^2}\times \frac{A}{4}.$

The albedo of the planet is ${\alpha }_{\text{p}}$. The equilibrium surface temperature of the planet is T. Derive the expression

$T=\sqrt[4]{\frac{P(1-{\alpha }_{\text{p}})}{16\pi d^{2}e\sigma }}$

where e is the emissivity of the planet.

On average, the Moon is the same distance from the Sun as the Earth. The Moon can be assumed to have an emissivity e = 1 and an albedo ${\alpha }_{\text{M}}$ = 0.13. The solar constant is 1.36 × 103 W m⁻². Calculate the surface temperature of the Moon.

$\frac{P}{4\pi d^2}$ is the power received by the planet/at a distance d «from star» ✓

$\frac{A}{4}$ is the projected area/cross sectional area of the planet ✓

use of $e\sigma A{T}^4 \mathit{\text{OR}} \frac{P}{4\pi d^2}\times \frac{A}{4}\times (1-{\alpha }_{\text{p}})$ ✓

with correct manipulation to show the result ✓

$\sqrt[4]{\frac{1.36\times {10}^3\times 0.87}{4\times 5.67\times {10}^{-8}}}$ ✓

T = 268.75 «K» ≅ 270 «K» ✓

Explain why the output potential difference to the external circuit and the output emf of the photovoltaic cell are different.

Calculate the internal resistance of the photovoltaic cell for the maximum intensity condition using the model for the cell.

The maximum intensity of sunlight incident on the photovoltaic cell at the place on the Earth’s surface is 680 W m⁻².

A measure of the efficiency of a photovoltaic cell is the ratio

$\frac{\text{energy available every second to the external circuit}}{\text{energy arriving every second at the photovoltaic cell surface}}.$

Determine the efficiency of this photovoltaic cell when the intensity incident upon it is at a maximum.

State two reasons why future energy demands will be increasingly reliant on sources such as photovoltaic cells.

there is a potential difference across the internal resistance
OR
there is energy/power dissipated in the internal resistance ✓

when there is current «in the cell»/as charge flows «through the cell» ✓

Allow full credit for answer based on $V=\epsilon -Ir$

ALTERNATIVE 1
pd dropped across cell $=6.5 «\text{V}»$✓

internal resistance $\mathrm{=}\frac{6.5}{0.9}$ ✓

$7.2 «\text{Ω}»$✓

ALTERNATIVE 2

$\epsilon =I(R+r)$ so $\epsilon =V+Ir$ ✓

$21.0=14.5+0.9\times r$ ✓

$7.2 «\text{Ω}»$ ✓

Alternative solutions are possible

Award [3] marks for a bald correct answer

power arriving at cell = 680 x 0.35 x 0.45 = «107 W» ✓

power in external circuit = 14.5 x 0.9 = «13.1 W» ✓

efficiency = 0.12 OR 12 % ✓

Award [3] marks for a bald correct answer

Allow ECF for MP3

«energy from Sun/photovoltaic cells» is renewable
OR
non-renewable are running out ✓

non-polluting/clean ✓

no greenhouse gases
OR
does not contribute to global warming/climate change ✓

OWTTE

Do not allow economic aspects (e.g. free energy)

Identify the missing information for this decay.

On the graph, sketch how the number of boron nuclei in the sample varies with time.

After 4.3 × 10⁶ years,

$$\frac{\text{number of produced boron nuclei}}{\text{number of remaining beryllium nuclei}}=7.$$

Show that the half-life of beryllium-10 is 1.4 × 10⁶ years.

Beryllium-10 is used to investigate ice samples from Antarctica. A sample of ice initially contains 7.6 × 10¹¹ atoms of beryllium-10. State the number of remaining beryllium-10 nuclei in the sample after 2.8 × 10⁶ years.

State what is meant by thermal radiation.

Discuss how the frequency of the radiation emitted by a black body can be used to estimate the temperature of the body.

Calculate the peak wavelength in the intensity of the radiation emitted by the ice sample.

Derive the units of intensity in terms of fundamental SI units.

${}_4^{10}\text{Be}{\to }_5^{10}\text{B}+\beta +{\overline{\text{V}}}_{\text{e}}$

conservation of mass number AND charge ${}_5^{10}\text{B}$, ${}_4^{10}\text{Be}$

Correct identification of both missing values required for [1].

[1 mark]

correct shape ie increasing from 0 to about 0.80 N₀

crosses given line at 0.50 N₀

[2 marks]

ALTERNATIVE 1

fraction of Be = $\frac{1}{8}$, 12.5%, or 0.125

therefore 3 half lives have elapsed

$t_{\frac{1}{2}}=\frac{4.3\times {10}^6}{3}=1.43\times {10}^6$ «≈ 1.4 × 10⁶» «y»

ALTERNATIVE 2

fraction of Be = $\frac{1}{8}$, 12.5%, or 0.125

$\frac{1}{8}={\text{e}}^{-\lambda }(4.3\times {10}^6)$ leading to λ = 4.836 × 10^–7 «y»^–1

$\frac{\ln 2}{\lambda }$ = 1.43 × 10⁶ «y»

Must see at least one extra sig fig in final answer.

[3 marks]

1.9 × 10¹¹

[1 mark]

emission of (infrared) electromagnetic/infrared energy/waves/radiation.

[1 mark]

the (peak) wavelength of emitted em waves depends on temperature of emitter/reference to Wein’s Law

so frequency/color depends on temperature

[2 marks]

$\lambda =\frac{2.90\times {10}^{-3}}{253}$

= 1.1 × 10^–5 «m»

Allow ECF from MP1 (incorrect temperature).

[2 marks]

correct units for Intensity (allow W, Nms^–1 OR Js^–1 in numerator)

rearrangement into proper SI units = kgs^–3

Allow ECF for MP2 if final answer is in fundamental units.

[2 marks]

Show that the intensity of the solar radiation at the location of Titan is 16 W m⁻²

Titan has an atmosphere of nitrogen. The albedo of the atmosphere is 0.22. The surface of Titan may be assumed to be a black body. Explain why the average intensity of solar radiation absorbed by the whole surface of Titan is 3.1 W m⁻²

Show that the equilibrium surface temperature of Titan is about 90 K.

The orbital radius of Titan around Saturn is $R$ and the period of revolution is $T$.

Show that $T^2=\frac{4{\mathrm{\pi }}^2{R}^{ 3}}{GM}$ where $M$ is the mass of Saturn.

The orbital radius of Titan around Saturn is 1.2 × 10⁹m and the orbital period is 15.9 days. Estimate the mass of Saturn.

incident intensity $\frac{1360}{9.3^2}$ OR $15.7\approx 16$ «W m⁻²» ✓

Allow the use of 1400 for the solar constant.

exposed surface is ¼ of the total surface ✓

absorbed intensity = (1−0.22) × incident intensity ✓

0.78 × 0.25 × 15.7  OR  3.07 «W m⁻²» ✓

Allow 3.06 from rounding and 3.12 if they use 16 W m⁻².

σT ⁴ = 3.07

OR

T = 86 «K» ✓

correct equating of gravitational force / acceleration to centripetal force / acceleration ✓

correct rearrangement to reach the expression given ✓

Allow use of $\sqrt{\frac{GM}{R}}=\frac{2\mathrm{\pi }R}{T}$ for MP1.

$T=15.9\times 24\times 3600$ «s» ✓

$M=\frac{4{\mathrm{\pi }}^2{\left(1.2\times {10}^9\right)}^3}{6.67\times {10}^{-11}\times {\left(15.9\times 24\times 3600\right)}^2}=5.4\times {10}^{26}$«kg» ✓

Award [2] marks for a bald correct answer.

Allow ECF from MP1.

Estimate the power input to the heating element. State an appropriate unit for your answer.

Outline whether your answer to (a) is likely to overestimate or underestimate the power input.

Discuss, with reference to the molecules in the liquid, the difference between milk at 11 °C and milk at 84 °C.

State how energy is transferred from the inside of the metal pipe to the outside of the metal pipe.

The missing section of insulation is 0.56 m long and the external radius of the pipe is 0.067 m. The emissivity of the pipe surface is 0.40. Determine the energy lost every second from the pipe surface. Ignore any absorption of radiation by the pipe surface.

Describe one other method by which significant amounts of energy can be transferred from the pipe to the surroundings.

energy required for milk entering in 1 s = mass x specific heat x 73 ✓

16 kW OR 16000 W ✓

MP1 is for substitution into mcΔT regardless of power of ten.

Allow any correct unit of power (such as J s^-1 OR kJ s^-1) if paired with an answer to the correct power of 10 for MP2.

Underestimate / more energy or power required ✓

because energy transferred as heat / thermal energy is lost «to surroundings or electrical components» ✓

Do not allow general term “energy” or “power” for MP2.

the temperature has increased so the internal energy / « average » KE «of the molecules» has increased OR temperature is proportional to average KE «of the molecules». ✓

«therefore» the «average» speed of the molecules or particles is higher OR more frequent collisions « between molecules » OR spacing between molecules has increased OR average force of collisions is higher OR intermolecular forces are less OR intermolecular bonds break and reform at a higher rate OR molecules are vibrating faster. ✓

conduction/conducting/conductor «through metal» ✓

use of $P=e\sigma A{T}^4$ where T = 357 K ✓

use of $A=2\pi r l$ « = 0.236 m²» ✓

P = 87 «W» ✓

Allow 85 – 89 W for MP3.

Allow ECF for MP3.

convection «is likely to be a significant loss» ✓

«due to reduction in density of air near pipe surface» hot air rises «and is replaced by cooler air from elsewhere»

OR

«due to» conduction «of heat or thermal energy» from pipe to air ✓

Most candidates recognized that this was a specific heat question and set up a proper calculation, but many struggled to match their answer to an appropriate unit. A common mistake was to leave the answer in some form of an energy unit and others did not match the power of ten of the unit to their answer (e.g. 16 W).

Many candidates recognized that this was an underestimate of the total energy but failed to provide an adequate reason. Many gave generic responses (such as "some power will be lost"/not 100% efficient) without discussing the specific form of energy lost (e.g. heat energy).

This was generally well answered. Most HL candidates linked the increase in temperature to the increase in the kinetic energy of the molecules and were able to come up with a consequence of this change (such as the molecules moving faster). SL candidates tended to focus more on consequences, often neglecting to mention the change in KE.

Many candidates recognized that heat transfer by conduction was the correct response. This was a "state" question, so candidates were not required to go beyond this.

Candidates at both levels were able to recognize that this was a blackbody radiation question. One common mistake candidates made was not calculating the area of a cylinder properly. It is important to remind candidates that they are expected to know how to calculate areas and volumes for basic geometric shapes. Other common errors included the use of T in Celsius and neglecting to raise T ^4. Examiners awarded a large number of ECF marks for candidates who clearly showed work but made these fundamental errors.

A few candidates recognized that convection was the third source of heat loss, although few managed to describe the mechanism of convection properly for MP2. Some candidates did not read the question carefully and instead wrote about methods to increase the rate of heat loss (such as removing more insulation or decreasing the temperature of the environment).

Show that the intensity radiated by the oceans is about 400 W m^-2.

Explain why some of this radiation is returned to the oceans from the atmosphere.

5.67 × 10⁻⁸ × 289⁴

OR

= 396 «W m⁻²» ✔

«≈ 400 W m⁻²»

«most of the radiation emitted by the oceans is in the» infrared ✔

«this radiation is» absorbed by greenhouse gases/named greenhouse gas in the atmosphere ✔

«the gases» reradiate/re-emit ✔

partly back towards oceans/in all directions/awareness that radiation in other directions is also present ✔

This was well answered with candidates scoring the mark for either a correct substitution or an answer given to at least one more sf than the show that value. Some candidates used 298 rather than 289.

For many this was a well-rehearsed answer which succinctly scored full marks. For others too many vague terms were used. There was much talk about energy being trapped or reflected and the ozone layer was often included. The word ‘albedo’ was often written down with no indication of what it means and ‘the albedo effect also featured.

A black body is on the Moon’s surface at point A. Show that the maximum temperature that this body can reach is 400 K. Assume that the Earth and the Moon are the same distance from the Sun.

Another black body is on the Moon’s surface at point B.

Outline, without calculation, why the aximum temperature of the black body at point B is less than at point A.

The albedo of the Earth’s atmosphere is 0.28. Outline why the maximum temperature of a black body on the Earth when the Sun is overhead is less than that at point A on the Moon.

Outline why a force acts on the Moon.

Outline why this force does no work on the Moon.

T = ${\left(\frac{1360}{\sigma }\right)}^{0.25}$   ✔

390 «K» ✔

Must see 1360 (from data booklet) used for MP1.

Must see at least 2 s.f.

energy/Power/Intensity lower at B ✔

connection made between energy/power/intensity and temperature of blackbody ✔

(28 %) of sun’s energy is scattered/reflected by earth’s atmosphere OR only 72 % of incident energy gets absorbed by blackbody ✔

Must be clear that the energy is being scattered by the atmosphere.

Award [0] for simple definition of “albedo”.

gravitational attraction/force/field «of the planet/Moon» ✔

Do not accept “gravity”.

the force/field and the velocity/displacement are at 90° to each other OR there is no change in GPE of the moon ✔

Award [0] for any mention of no net force on the satellite.

Do not accept acceleration is perpendicular to velocity.

Many candidates struggled with this question. A significant portion attempted to apply Wein’s Law and simply stated that a particular wavelength was the peak and then used that to determine the temperature. Some did use the solar constant from the data booklet and were able to calculate the correct temperature. As part of their preparation for the exam candidates should thoroughly review the data booklet and be aware of what constants are given there. As with all “show that” questions candidates should be reminded to include an unrounded answer.

This is question is another example of candidates not thinking beyond the obvious in the question. Many simply said that point B is farther away, or that it is at an angle. Some used vague terms like “the sunlight is more spread out” rather than using proper physics terms. Few candidates connected the lower intensity at B with the lower temperature of the blackbody.

This question was assessing the understanding of the concept of albedo. Many candidates were able to connect that an albedo of 0.28 meant that 28 % of the incident energy from the sun was being reflected or scattered by the atmosphere before reaching the black body.

This was generally well answered, although some candidates simply used the vague term “gravity” rather than specifying that it is a gravitational force or a gravitational field. Candidates need to be reminded about using proper physics terms and not more general, “every day” terms on the exam.

Some candidates connected the idea that the gravitational force is perpendicular to the velocity (and hence the displacement) for the mark. It was also allowed to discuss that there is no change in gravitational potential energy, so therefore no work was being done. It was not acceptable to simply state that the net displacement over one full orbit is zero. Unfortunately, some candidates suggested that there is no net force on the moon so there is no work done, or that the moon is so much smaller so no work could be done on it.

Determine the maximum power that can be extracted from the wind by this turbine.

Suggest why the answer in (a) is a maximum.

$P=\frac{1}{2}\times A\times 1.2\times 8^3$ OR $P=\frac{1}{2}\times A\times 1.32\times 4^3$  ✔

$P$ «in incoming beam» = 1.4×105 «W»
OR
$P$ «in outgoing beam» = 1.9×104 «W» ✔

subtracts both P to obtain 1.2×105 «W» ✔

NOTE: Condone use of a wrong area or use of circumference in MP1.
Allow ECF from MP2.
Award [1] max for any attempt to use the formula for wind power which cubes the difference of velocities
Award [3] for a bald correct answer

because some power is lost due to inefficiencies in the system/transfers to the surroundings ✔

NOTE: Accept power or energy indistinctly

Estimate the specific energy of water in this storage system, giving an appropriate unit for your answer.

Show that the average rate at which the gravitational potential energy of the water decreases is 2.5 GW.

The storage system produces 1.8 GW of electrical power. Determine the overall efficiency of the storage system.

After the upper lake is emptied it must be refilled with water from the lower lake and this requires energy. Suggest how the operators of this storage system can still make a profit.

Average height = 127 «m»

Specific energy «= $\frac{mg\overline{h}}{m}=g\overline{h}$ = 9.81 × 127» = 1.2 × 10³ J kg^–1

Unit is essential

Allow g = 10 gives 1.3 × 10³ J kg^–1

Allow ECF from 110 m

(1.1 × 10³ J kg^–1) or 144 m

(1.4 × 10³ J kg^–1)

[2 marks]

mass per second leaving dam is $\frac{1.2\times {10}^5}{60}$ × 10³ = «2.0 × 10⁶ kg s^–1»

rate of decrease of GPE is = 2.0 × 10⁶ × 9.81 × 127

= 2.49 × 10⁹ «W» /2.49 «GW»

Do not award ECF for the use of 110 m or 144 m

Allow 2.4 GW if rounded value used from (a)(i) or 2.6 GW if g = 10 is used

[3 marks]

efficiency is «$\frac{1.8}{2.5}$ =» 0.72 / 72%

[1 mark]

water is pumped back up at times when the demand for/price of electricity is low

[1 mark]