1. The problem statement, all variables and given/known data Assuming 100% efficient energy conversion, how much water stored behind a hydroelectric plant with a head of 30m would be required to charge the battery? 2. Relevant equations ? MAYBE: w=1000QgH j\s, where Q is the discharge rate @ m^3\s, 1000Q kg\s, H = net water head in m (30m), g is 9.8 m\s^2 3. The attempt at a solution I don't even know where to begin with this question. My last question was a simple unit conversation question about how much energy was stored in a 12v 50Ah rated car battery, which I got 2,160,000 joules for my answer. I don't even know how to begin to tackle this question because the formula I gave was one I found on the internet that looked like it might apply. What I'm thinking with that question however, is that perhaps I can use my figure from the total energy of the battery equaling: 2,160,000W = 1000*Q*9.8m\s^2*30m j (I borrowed the seconds unit from the j\s to turn my 2160000 joules into watts on the left hand side.) 2,160kW = 294Q kg\s m^2\s^2 j 7346.938 = Q kg m^2/s^3 j The numbers look way off to me, and also the units have some serious funkiness going on with them which leads me to think this method isn't correct. Any help?