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Physics - Energy Transfers and Transformations

Grade 7IGCSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Energy is measured in Joules (JJ). It cannot be created or destroyed, only transferred from one store to another (The Law of Conservation of Energy).

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Energy stores include: Kinetic (EkE_k), Gravitational Potential (EpE_p), Chemical, Elastic, Thermal, Magnetic, Electrostatic, and Nuclear.

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Energy transfers occur through four main pathways: Mechanically (by forces), Electrically (by current), by Heating (conduction, convection, radiation), and by Radiation (light and sound waves).

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Efficiency is a measure of how much 'useful' energy is produced compared to the total energy input. It is often expressed as a percentage (%).

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Sankey Diagrams are used to represent energy transfers visually, where the width of the arrow represents the amount of energy in Joules (JJ).

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Thermal energy transfer in solids occurs via conduction, where particles vibrate and pass energy to neighbors. In fluids (liquids/gases), it occurs via convection due to density changes.

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Thermal radiation (Infrared) is the only method of energy transfer that can travel through a vacuum, such as the space between the Sun and Earth.

📐Formulae

Efficiency=Useful Energy OutputTotal Energy Input×100%Efficiency = \frac{\text{Useful Energy Output}}{\text{Total Energy Input}} \times 100\%

W=F×dW = F \times d

ΔEp=m×g×Δh\Delta E_p = m \times g \times \Delta h

Ek=12mv2E_k = \frac{1}{2} m v^2

Total Energy In=Useful Energy Out+Wasted Energy Out\text{Total Energy In} = \text{Useful Energy Out} + \text{Wasted Energy Out}

💡Examples

Problem 1:

An electric motor takes in 200 J200\,J of electrical energy. It transfers 150 J150\,J into useful kinetic energy to lift a weight. Calculate the efficiency of the motor.

Solution:

Efficiency=150 J200 J×100=75%Efficiency = \frac{150\,J}{200\,J} \times 100 = 75\%

Explanation:

To find the efficiency, divide the useful energy output (150 J150\,J) by the total energy input (200 J200\,J) and multiply by 100100 to get a percentage.

Problem 2:

A ball with a mass of 0.5 kg0.5\,kg is held at a height of 2 m2\,m. Calculate its Gravitational Potential Energy (EpE_p). (Assume g=10 N/kgg = 10\,N/kg)

Solution:

Ep=0.5 kg×10 N/kg×2 m=10 JE_p = 0.5\,kg \times 10\,N/kg \times 2\,m = 10\,J

Explanation:

The formula for gravitational potential energy is mass (mm) multiplied by the gravitational field strength (gg) and the height (hh).

Problem 3:

A light bulb is supplied with 100 J100\,J of energy. 20 J20\,J is transferred as useful light energy. How much energy is 'wasted' as thermal energy?

Solution:

Wasted Energy=100 J−20 J=80 JWasted\,Energy = 100\,J - 20\,J = 80\,J

Explanation:

According to the Law of Conservation of Energy, the total energy input must equal the total energy output. Therefore, 100 J=20 J (light)+thermal energy100\,J = 20\,J \text{ (light)} + \text{thermal energy}.