Conservation of Energy & efficiency

Fluency

Energy cannot be created or destroyed, only transformed from one form to another. In a closed system the total energy stays constant.
Useful energy: output that fulfils the device’s purpose. Wasted energy: output in forms we cannot or do not use (usually heat, sometimes sound). Efficiency: the fraction of input energy that becomes useful output.
(a) $\eta = 150/200 \times 100 = 75\%$ . (b) Useful $= 400 - 80 = 320$ kJ. $\eta = 320/400 \times 100 = 80\%$ .
$\eta = 210/250 \times 100 = 84\%$ .
Sankey: input band (say 100 J) splits into useful (70 J heated air flow) and wasted (30 J heat + sound). Widths must add to input width.
Chemical -> electrical: battery. Electrical -> light: LED. Kinetic -> electrical: generator (wind turbine or hydro).

Reasoning

Incandescent: $\eta = 3/60 = 5\%$ . LED: $\eta = 6/10 = 60\%$ . LED gives twice the light for one-sixth the electricity, so the yearly saving at 4 h/day is about $(60 - 10) \times 4 \times 365 = 73\,000$ Wh $= 73$ kWh per bulb. Across a house that is a large reduction in bill and emissions.
The engine’s wasted energy leaves the car mainly as hot exhaust gas and heat carried away by the radiator; these are the largest loss paths, so any calculation of efficiency must include them to be realistic.
Electricity $= 0.40 \times 1000 = 400$ MJ. The rest ( $600$ MJ) is mostly waste heat lost at the boiler and cooling towers, plus smaller losses at the turbine and generator.
Each bounce loses some kinetic energy to heat, sound, and air resistance. That energy is not returned, so kinetic energy on the way up is smaller each time, reducing the bounce height. Total energy is still conserved (the lost energy warms the ball, floor, and air).
Kinetic energy -> heat (via friction between skin surfaces). Some sound is also produced.

Problem solving

$\eta = 196\,000 / 1.8 \times 10^{7} \times 100 \approx 1.09\%$ . That is low because 5 kWh is a huge amount of energy for a 20 m lift — probably this is a single lift of many, or most energy is elsewhere. (Or the numbers represent all the runs over an hour, not one.)
$\eta = 180/1000 \times 100 = 18\%$ .
The other $200$ W is lost as heat in the motor and casing, and some sound. $\eta = 1800/2000 = 90\%$ .
$\eta = (500 - 8)/500 \times 100 = 492/500 \times 100 = 98.4\%$ .

Reasoning

Overall $= 0.40 \times 0.92 \times 0.70 = 0.2576$ , or about $25.8\%$ . Only about a quarter of the coal’s chemical energy becomes useful output; most is lost as heat at the power station, with smaller losses in transmission and in the appliance.
A heat pump does not create energy; it moves heat from a colder outside to a warmer inside using electrical work. Total energy into the house = electrical input + heat taken from outside = heat delivered. Conservation holds; COP just describes the ratio of heat delivered to work input.
Counter-points: (i) Electricity is often more expensive per joule than gas heat, so using electric bulbs for heating is wasteful financially. (ii) The light that leaves through windows is not returned; part of the energy leaves the house rather than warming it. (iii) In summer the heat is unwanted and may drive more air-conditioning. (iv) Switching to LEDs + efficient heating saves more energy and emissions overall.
Overall $= 0.40 \times 0.92 \times 0.50 = 0.184$ or $18.4\%$ of the original coal energy becomes visible light. Sankey: coal 100 -> power station (40 to electricity, 60 waste heat) -> transmission (36.8 delivered, 3.2 line losses) -> LED bulb (18.4 light, 18.4 heat).

Year 9 answers