1 a. A machine can never be 100% efficient. The out put energy will always be less than the input energy 1 a. i) see above ANS ii) see above ANS !!

2 a. Energy o/p x 100% = % efficiency Energy i/p 1 a. Energy wasted = Energy I/p - Energy o/p 1 a. Energy wasted = 60 J - 24 J 1 a. Energy wasted = 36 J 2b. 24 J x 100% = % efficiency 60 J 1 b. 24 J x 100% = 40 % efficient 60 J

3. Energy o/p x 100% = % efficiency Energy i/p 3. Energy o/p = 0.25 3200 3. Energy o/p = 0.25 x 3200 3. Energy o/p = 800 J

Work is done on the heater element by the electricity in the wires at the same rate as thermal energy radiates from the heater to raise the thermal energy store of the surroundings. There is no overall change in the thermal energy store of the heater . It has reached ‘Steady state equilibrium’. First law of thermodynamics: ∆U = ∆ Q + ∆ W ∆U = internal thermal energy store of object (heater ) ∆Q = thermal energy received by heater ∆ W = work done by electrical current on heater element Once the heater has been left on for a period of time it reaches equilibrium so ∆U = 0 Then 0 = ∆ Q + ∆ W ∆Q = negative ( element is giving out infra red waves) ∆ W = work done by current on heater element (positive)