IGCSE Physics: Energy Efficiency Calculations and Sankey Diagrams

Efficiency Calculations in Physics

Definition of Efficiency

  • Efficiency is a measure of how much useful energy output is obtained from the total energy input.

  • It is calculated using the formula: Efficiency=(Useful Energy OutputTotal Energy Input)×100%\text{Efficiency} = \left( \frac{\text{Useful Energy Output}}{\text{Total Energy Input}} \right) \times 100\%

Mathematical Skill Practice: Efficiency Equations
Example 1: Calculating Efficiency of a Light Bulb

  • Problem: What is the efficiency of a light bulb that transfers 600J600 \text{J} by light radiation to the surroundings when it is supplied with 950J950 \text{J} electrically?

  • Solution:

    • Useful Energy Output (light radiation) =600J= 600 \text{J}

    • Total Energy Input (electrical) =950J= 950 \text{J}

    • Efficiency=(600J950J)×100%=0.63157…×100%≈63.2%\text{Efficiency} = \left( \frac{600 \text{J}}{950 \text{J}} \right) \times 100\% = 0.63157… \times 100\% \approx 63.2\%

Example 2: Calculating Efficiency of a TV

  • Problem: A TV is supplied with 1000J1000 \text{J} electrically. It transfers out 300J300 \text{J} as sound radiation and 375J375 \text{J} as light. What is the efficiency of the TV?

  • Solution:

    • Identify useful energy output. For a TV, both sound and light are typically considered useful outputs.

    • Useful Energy Output =300J (sound)+375J (light)=675J= 300 \text{J (sound)} + 375 \text{J (light)} = 675 \text{J}

    • Total Energy Input (electrical) =1000J= 1000 \text{J}

    • Efficiency=(675J1000J)×100%=0.675×100%=67.5%\text{Efficiency} = \left( \frac{675 \text{J}}{1000 \text{J}} \right) \times 100\% = 0.675 \times 100\% = 67.5\%

Example 3: Calculating Useful Energy Output (Given Efficiency & Input)

  • Problem: A microwave is 75%75\% efficient. If it is supplied with 2000J2000 \text{J} electrically, how much energy will it usefully transfer to thermal energy?

  • Solution:

    • Efficiency =75%=0.75= 75\% = 0.75

    • Total Energy Input =2000J= 2000 \text{J}

    • Using the rearranged formula: Useful Energy Output=Efficiency×Total Energy Input\text{Useful Energy Output} = \text{Efficiency} \times \text{Total Energy Input}

    • Useful Energy Output=0.75×2000J=1500J\text{Useful Energy Output} = 0.75 \times 2000 \text{J} = 1500 \text{J}

Example 4: Calculating Useful Energy Output (Given Efficiency & Input - Small Efficiency)

  • Problem: A plant gets 200J200 \text{J} by light radiation from the sun. If the plant is 0.5%0.5\% efficient, how much of that energy goes into photosynthesis?

  • Solution:

    • Efficiency =0.5%=0.005= 0.5\% = 0.005

    • Total Energy Input =200J= 200 \text{J}

    • Useful Energy Output=0.005×200J=1J\text{Useful Energy Output} = 0.005 \times 200 \text{J} = 1 \text{J}

Example 5: Calculating Total Energy Input (Given Efficiency & Useful Output)

  • Problem: A light bulb is rated as 40%40\% efficient. How much energy does the bulb need to be supplied with to give out 80J80 \text{J} by light radiation?

  • Solution:

    • Efficiency =40%=0.4= 40\% = 0.4

    • Useful Energy Output =80J= 80 \text{J}

    • Rearranging the formula: Total Energy Input=Useful Energy OutputEfficiency\text{Total Energy Input} = \frac{\text{Useful Energy Output}}{\text{Efficiency}}

    • Total Energy Input=80J0.4=200J\text{Total Energy Input} = \frac{80 \text{J}}{0.4} = 200 \text{J}

Example 6: Calculating Total Energy Input (Given Efficiency & Useful Output - Toaster)

  • Problem: A toaster is rated as 85%85\% efficient. If you need 700J700 \text{J} of thermal energy to toast your bread, how much energy do you need to give the toaster?

  • Solution:

    • Efficiency =85%=0.85= 85\% = 0.85

    • Useful Energy Output =700J= 700 \text{J}

    • Total Energy Input=700J0.85≈823.5J\text{Total Energy Input} = \frac{700 \text{J}}{0.85} \approx 823.5 \text{J}

Example 7: Calculating Efficiency (Given Total Input and Wasted Energy)

  • Problem: A light bulb is supplied with 800J800 \text{J} of energy. If it wastes 500J500 \text{J} as thermal energy, how efficient is the light bulb?

  • Solution:

    • Total Energy Input =800J= 800 \text{J}

    • Wasted Energy =500J= 500 \text{J}

    • Useful Energy Output =Total Energy Input−Wasted Energy=800J−500J=300J= \text{Total Energy Input} - \text{Wasted Energy} = 800 \text{J} - 500 \text{J} = 300 \text{J}

    • Efficiency=(300J800J)×100%=0.375×100%=37.5%\text{Efficiency} = \left( \frac{300 \text{J}}{800 \text{J}} \right) \times 100\% = 0.375 \times 100\% = 37.5\%

Example 8: Calculating Efficiency (Kettle with multiple wasted energies)

  • Problem: A kettle is supplied with 3kJ3 \text{kJ} electrically. If 100J100 \text{J} of energy is wasted via sound radiation, how efficient is the kettle?

  • Solution:

    • First, ensure consistent units: 3kJ=3000J3 \text{kJ} = 3000 \text{J}

    • Total Energy Input =3000J= 3000 \text{J}

    • Wasted Energy (sound) =100J= 100 \text{J}

    • Useful Energy Output (thermal for heating water) =Total Energy Input−Wasted Energy=3000J−100J=2900J= \text{Total Energy Input} - \text{Wasted Energy} = 3000 \text{J} - 100 \text{J} = 2900 \text{J}

    • Efficiency=(2900J3000J)×100%≈0.9666…×100%≈96.7%\text{Efficiency} = \left( \frac{2900 \text{J}}{3000 \text{J}} \right) \times 100\% \approx 0.9666… \times 100\% \approx 96.7\%

Example 9: Calculating Efficiency of a Car (Large Energy Values)

  • Problem: Petrol supplies a car with 25MJ25 \text{MJ} of chemical energy. The car turns 600kJ600 \text{kJ} of this into useful kinetic energy. How efficient is the car?

  • Solution:

    • Ensure consistent units: 25MJ=25,000kJ25 \text{MJ} = 25,000 \text{kJ}

    • Total Energy Input =25,000kJ= 25,000 \text{kJ}

    • Useful Energy Output =600kJ= 600 \text{kJ}

    • Efficiency=(600kJ25,000kJ)×100%=0.024×100%=2.4%\text{Efficiency} = \left( \frac{600 \text{kJ}}{25,000 \text{kJ}} \right) \times 100\% = 0.024 \times 100\% = 2.4\%

Sankey Diagrams
Concept of Sankey Diagrams

  • Sankey diagrams are visual representations of energy transfers, illustrating the flow of energy from input to useful output and wasted energy.

  • The width of the arrows is proportional to the amount of energy they represent.

Example 10: Labeling a Sankey Diagram for a Car

  • Description: A car transfers energy from its chemical store into the thermal store of the surroundings and the kinetic store of the car.

  • Diagram Components:

    • Input: Chemical Energy Input (from petrol)

    • Useful Output: Useful Kinetic Energy (motion of the car)

    • Wasted Output: Wasted Thermal Energy (heat lost to surroundings, engine heat, friction)

Example 11: Calculating Efficiency of a Hairdryer from a Sankey Diagram

  • Problem: Using the information from the Sankey diagram below, calculate the efficiency of the hairdryer (careful, think what we want the hairdryer to do!).

  • Diagram Data:

    • Electrical energy transfer (Input) =750J= 750 \text{J}

    • Thermal energy (Output) =550J= 550 \text{J}

    • Kinetic energy (Output) =125J= 125 \text{J}

    • Sound energy transfer (Output) =150J= 150 \text{J}

  • Analysis: For a hairdryer, its primary purpose is to produce heat (thermal energy) and move air (kinetic energy). Sound energy is wasted.

  • Solution:

    • Thermal energy (useful) =550J= 550 \text{J}

    • Kinetic energy (useful) =125J= 125 \text{J}

    • Total Useful Energy Output =550J+125J=675J= 550 \text{J} + 125 \text{J} = 675 \text{J}

    • Total Energy Input =750J= 750 \text{J}

    • Efficiency=(675J750J)×100%=0.9×100%=90%\text{Efficiency} = \left( \frac{675 \text{J}}{750 \text{J}} \right) \times 100\% = 0.9 \times 100\% = 90\%

Example 12: Calculating Efficiency of a Power Station from a Sankey Diagram

  • Problem: Below is a Sankey diagram for a power station. Add the energy values to the diagram for input and each remaining output. Calculate the efficiency of the power station.

  • Given Data:

    • Energy to customers (useful output) =450kJ= 450 \text{kJ}

    • Energy used in transmission (wasted) =150kJ= 150 \text{kJ}

    • Energy used in power station (wasted) =150kJ= 150 \text{kJ}

    • Thermal energy loss in power station (wasted) =750kJ= 750 \text{kJ}

    • Total Input is currently missing from diagram, but one value is given: 1500kJ1500 \text{kJ}. This implies the Total Energy Input is 1500kJ1500 \text{kJ}.

  • Diagram Completion:

    • Input: (Chemical/Fuel) Energy Input (Total) =1500kJ= 1500 \text{kJ}

    • Outputs:

      • Energy to customers (Useful) =450kJ= 450 \text{kJ}

      • Energy used in transmission (Wasted) =150kJ= 150 \text{kJ}

      • Energy used in power station (Wasted) =150kJ= 150 \text{kJ}

      • Thermal energy loss in power station (Wasted) =750kJ= 750 \text{kJ}

  • Verification of Energy Conservation (optional but good practice):

    • Useful Output + Wasted Outputs =450kJ+150kJ+150kJ+750kJ=1500kJ= 450 \text{kJ} + 150 \text{kJ} + 150 \text{kJ} + 750 \text{kJ} = 1500 \text{kJ}

    • This matches the Total Energy Input, confirming the diagram values are consistent.

  • Solution:

    • Useful Energy Output (energy to customers) =450kJ= 450 \text{kJ}

    • Total Energy Input =1500kJ= 1500 \text{kJ}

    • Efficiency=(450kJ1500kJ)×100%=0.3×100%=30%\text{Efficiency} = \left( \frac{450 \text{kJ}}{1500 \text{kJ}} \right) \times 100\% = 0.3 \times 100\% = 30\%