Energy Efficiency and Real-Life Use

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Energy Efficiency and Real-Life Use — Free MYP3 Sciences Practice Questions

1QuestionCalculating Efficiency PercentConcept Practice

2 marks~3 minCriterion A

A hair dryer receives

1000 \text{ J}

of electrical energy. It produces

800 \text{ J}

of heat energy and

200 \text{ J}

of sound energy.

Using the terms input energy, useful output, and wasted energy, describe how electrical energy is transformed as it passes through the hair dryer. [2]

Solutions

2QuestionCalculating Efficiency PercentAssessment Practice

3 marks~5 minCriterion C

The graph below shows how the efficiency of an electric kettle changes as the volume of water inside it increases.

\text{Efficiency} (\%) = \frac{\text{useful heat energy transferred to water}}{\text{total electrical energy input}} \times 100\%

Describe the trend shown in the graph. [1]

Explain why efficiency is lower when the volume of water is small. Use the concept of heat loss to the surroundings in your answer. [1]

The graph shows that efficiency levels off at large volumes. Analyse why increasing the volume further produces little additional change in efficiency. [1]

Solutions

3QuestionCalculating Efficiency PercentAssessment Practice

6 marks~9 minCriterion B

A solar panel receives

500 \text{ J}

of light energy and produces

180 \text{ J}

of useful electrical energy. The remaining energy is lost as heat.

The table below shows the efficiency of three solar panels at different light intensities.

Light intensity ( $\text{W/m}^2$ )	200	500	800
Panel A efficiency (\%)	30	36	34
Panel B efficiency (\%)	25	32	38
Panel C efficiency (\%)	28	36	30

Calculate the efficiency of the solar panel described above. [2]

Calculate the average efficiency of all three panels at each light intensity and identify which light intensity gives the highest average efficiency. [2]

Analyse the data in the table to describe the relationship between light intensity and solar panel efficiency. [2]

Solutions

4QuestionCalculating Efficiency PercentAssessment Practice

8 marks~12 minCriterion D

A school is replacing 200 old bulbs (each 60 W, 10% efficient) with 200 LED bulbs (each 12 W, 80% efficient). The bulbs operate for 5 hours each day. Electricity costs 0.15 USD per kilowatt-hour (kWh).

Calculate the useful power output of one old bulb and one LED bulb. [2]

Calculate the daily electricity cost of running all 200 old bulbs and all 200 LED bulbs. [3]

A student claims: "The LED bulbs are the better choice because they are more efficient." Using the data above and one factor beyond efficiency, analyse whether this claim is fully supported. [3]

Solutions

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