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aqa a level physics

Tidal power could make a significant contribution to UK energy requirements. This question is about a tidal power station which traps sea water behind a tidal barrier at high tide and then releases the water through turbines 10.0 m below the high tide mark.

Calculate the mass of sea water covering an area of 120 km^2 and depth 10.0 m.

(i)
density of sea water = 1100 kg m^–3


(ii) Calculate the maximum loss of potential energy of the sea water in part (i) when it is released through the turbines.



I've done the first part right and got the mass to be 1.32x10^12 kg,


The next part I did Ep = mgh = (1.32x10^12) x 9.81 x 10

As the water drops 10m??


The mark scheme says the deltaH = 5m giving them a energies half that of mine!


Any help would be appreciated, the only thing i can think of is whether the mass is centred half way up the body of water? I really cant think of any other way it's 5m!
Sorry you've not had any responses about this. :frown: Are you sure you've posted in the right place? :smile: Here's a link to our subject forum which should help get you more responses if you post there. :redface:

You can also find the Exam Thread list for A-levels here and GCSE here. :dumbells:


Just quoting in Puddles the Monkey so she can move the thread if needed :h:

Spoiler

Original post by drmitz
Tidal power could make a significant contribution to UK energy requirements. This question is about a tidal power station which traps sea water behind a tidal barrier at high tide and then releases the water through turbines 10.0 m below the high tide mark.

Calculate the mass of sea water covering an area of 120 km^2 and depth 10.0 m.

(i)
density of sea water = 1100 kg m^–3


(ii) Calculate the maximum loss of potential energy of the sea water in part (i) when it is released through the turbines.



I've done the first part right and got the mass to be 1.32x10^12 kg,


The next part I did Ep = mgh = (1.32x10^12) x 9.81 x 10

As the water drops 10m??


The mark scheme says the deltaH = 5m giving them a energies half that of mine!


Any help would be appreciated, the only thing i can think of is whether the mass is centred half way up the body of water? I really cant think of any other way it's 5m!


It's more of an average change in height. Not all of the water falls 10m, some doesn't drop at all to hit the turbines while some drops the full 10m. So we use the mean height of 5m. (basically when calculating the GPE of a body, always use the centre of mass as the reference point).

To solve it fully you need to integrate:

E=Agρ∫010hp(h)dh \displaystyle E = Ag\rho\int^{10}_0 hp(h) \mathrm{d}h

where p(h)p(h) is a probability density distribution and in this case is equal to 0.1 everywhere.

This gives us the same answer as the mark scheme.
(edited 7 years ago)

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