3. 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 !!

4. 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 = J J
1 a Energy wasted = J
2b J x 100% = % efficiency
60 J
1 b J x 100% = % efficient
60 J

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

6. 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 = ∆ Q + ∆ W ∆Q = negative ( element is giving out infra red waves)
∆ W = work done by current on heater element (positive)