State the difference between renewable and non-renewable energy sources.

(i)    The basin empties over a six hour period. Show that about \(6000{\text{ }}{{\text{m}}^3}\) of water flows through the turbines every second.

(ii)     Show that the average power that the water can supply over the six hour period is about 0.2 GW.

(iii)     La Rance tidal power station has an energy output of \(5.4 \times {10^8}{\text{ kW}}\,{\text{h}}\) per year. Calculate the overall efficiency of the power station. Assume that the water can supply 0.2 GW at all times.

Energy resources such as La Rance tidal power station could replace the use of

fossil fuels. This may result in an increase in the average albedo of Earth.

(iv)     State two reasons why the albedo of Earth must be given as an average value.

(i)     State the principle of conservation of momentum.

(ii)     Show that the speed of the neutron immediately after the collision is about \(8.0 \times {10^6}{\text{ m}}\,{{\text{s}}^{ - 1}}\).

(iii)     Show that the fractional change in energy of the neutron as a result of the collision

is about 0.3.

(iv)     Estimate the minimum number of collisions required for the neutron to reduce its initial energy by a factor of \({10^6}\).

(v)     Outline why the reduction in energy is necessary for this type of reactor to function.

only non-renewable is depleted/cannot re-generate whereas renewable can / consumption rate of non-renewables is greater than formation rate and consumption rate of renewables is less than formation rate;

Do not allow “cannot be used again”.

(i)     volume released = \((22 \times {10^6} \times 6 = ){\text{ }}1.32 \times {10^8}{\text{ (}}{{\text{m}}^3}{\text{)}}\);

volume per second \(\frac{{1.32 \times {{10}^8}}}{{6 \times 3600}}{\text{ }}( = 6111{\text{ }}{{\text{m}}^3})\);

(ii)     use of average depth for calculation (3 m);

gpe lost \(6100 \times 1000 \times 9.81 \times 3\);

0.18 (GW);

Accept g = 10 m\(\,\)s^–2.

Award [1 max] if 6 m is used and an “average” is used at end of solution without mention of average depth.

(iii)     converts/states output with units; } (allow values quoted from question without unit)

converts/states input with units; } (allow values quoted from question without unit)

calculates efficiency from \(\frac{{{\text{output}}}}{{{\text{input}}}}\) as 0.31;

Award [3] for bald correct answer.

eg:

power output \(\frac{{5.4 \times {{10}^8}}}{{365 \times 24 \times 3600}}{\text{ }}\left( { = 17{\text{ kW}}\,{\text{h}}\,{{\text{s}}^{ - 1}}} \right)\);

\( = 17 \times 3600000 = 6.16 \times {10^7}{\text{ (W)}}\);

efficiency = \(\left( {\frac{{6.16 \times {{10}^7}}}{{2.0 \times {{10}^8}}} = } \right){\text{ }}31\% \)\(\,\,\,\)or\(\,\,\,\)0.31;

or

0.2 GW is \(1.752 \times {10^9}{\text{ (kW}}\,{\text{h}}\,{\text{yea}}{{\text{r}}^{ - 1}}{\text{)}}\);

\(\frac{{5.4 \times {{10}^8}}}{{1.752 \times {{10}^9}}}\);

efficiency \( = 0.31\);

(iv)     cloud cover / weather conditions;

latitude;

time of year / season;

nature/colour of surface;

(i)     (total) momentum unchanged before and after collision / momentum of a system is constant; } (allow symbols if explained)

no external forces / isolated system / closed system;

Do not accept “conserved”.

(ii)     final momentum of neutron \( = {\text{neutron mass}} \times 9.8 \times {10^6} - 1{\text{u}} \times 12 \times 1.5 \times {10^6}\); } (allow any appropriate and consistent mass unit)

final speed of neutron \( = 8.0\)\(\,\,\,\)or\(\,\,\,\)\(8.2 \times {10^6}{\text{ (m}}\,{{\text{s}}^{ - 1}}{\text{)}}\);

\(\left( { \approx {\text{8.0}} \times {\text{1}}{{\text{0}}^6}{\text{ (m}}\,{{\text{s}}^{ - 1}}{\text{)}}} \right)\)

Allow use of 1 u for both masses giving an answer of 8.2 \( \times \) 10⁶ (m\(\,\)s^–1).

(iii)     initial energy of neutron \( = 8.04 \times {10^{ - 14}}{\text{ (J)}}\) and final energy of neutron \( = 5.36 \times {10^{ - 14}}{\text{ (J)}}\); } (both needed)

fractional change in energy \( = \left( {\frac{{(8.04 - 5.36)}}{{8.04}} = } \right){\text{ }}0.33\);

or

fractional change \( = \left( {\frac{{\frac{1}{2}mv_{\text{i}}^2 - \frac{1}{2}mu_{\text{f}}^2}}{{\frac{1}{2}mv_{\text{i}}^2}}} \right)\); }(allow any algebra that shows a subtraction of initial term from final term divided by initial value)

\(\left( { = \frac{{{{(9.8 \times {{10}^6})}^2} - {{(8.0 \times {{10}^6})}^2}}}{{{{(9.8 \times {{10}^6})}^2}}}} \right)\) (allow omission of 10⁶)

\( = 0.33\); (allow 0.30 if 8.2 used)

Do not allow ECF if there is no subtraction of energies in first marking point.

(iv)     \({(0.33)^n} = {10^{ - 6}}\);

\(n = 13\); (allow n = 12 if 0.3 is used)

(v)     neutrons produced in fission have large energies;

greatest probability of (further) fission/absorption (when incident neutrons have thermal energy or low energy);

Do not accept “reaction” for “fission reaction”.

Many candidates continue to give weak responses to questions in which they are asked to compare renewable and non-renewable resources. Although the “it cannot be used again” answer has largely disappeared, many candidates still fail to appreciate that the issue is about the rate at which the resource can be replaced.

(i)     This was often well done, although occasional recourse was made to inappropriate physics (see bii). Candidates should note that in questions where the final answer is quoted (typically “Show that …..” questions) candidates are strongly advised to quote answers to one more significant figure than in the question.

(ii)     The rare candidate who understood the physics here was able to give a clear account of the solution. Many failed to spot the factor of a half in the water level change and introduced a factor of two later and arbitrarily. Others completely misunderstood the (simple) nature of the problem and used a random equation from the data booklet (usually \(1/3\rho A{v^3}\)). This of course gained no marks. A simple initial diagram would have helped many to avoid errors.

(iii)      As in question 1 there were far too many candidates who clearly do not understand and have not practised the problem of converting between energy units. Effective use of units would have made this an easy calculation. Explanations were few and candidates were clearly struggling with this aspect of energy.

(iv)     Many candidates were able to give one coherent reason but two distinct answers were rare.

(i)     As is often the case with this question, candidates state that “momentum is conserved” and fail to explain what this means. There was much confusion with energy conservation rules.

(ii)     Calculations of the final speed of the neutron were confused with little or no explanation of the equations. It was often not clear what mass values (if any) were being used in the solution.

(iii)     There were few clear solutions to this problem. Some candidates did not appreciate the meaning of fractional energy change and others were still travelling along the momentum route from an earlier part, scoring few, if any, marks.

(iv)     Candidates had evidently not considered the mechanical issues of moderation in their learning. There was little recognition that the change in fractional energy is 0.33\(n\) where \(n\) is the number of collisions. The most frequent answer was that the change is 0.33\(n\).

(v)     There was more clarity about the reasons for moderation but even so, answers were poorly expressed. Only a minority recognised that the probability of absorption is greatest at low neutron incident energy.