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Energy - Law of Conservation of Energy

Grade 6Cambridge (IGCSE)

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

🔑Concepts

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The Law of Conservation of Energy states that energy cannot be created or destroyed; it can only be transferred from one object to another or transformed from one form to another.

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In any closed system, the total amount of energy remains constant: Total Energyinitial=Total Energyfinal\text{Total Energy}_{initial} = \text{Total Energy}_{final}.

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Energy transformation refers to energy changing from one type to another (e.g., Chemical energy in a battery changing into Electrical energy).

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Energy transfer refers to energy moving from one place to another (e.g., Thermal energy moving from a hot cup to a cold hand).

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While energy is conserved, it is often 'dissipated' into the surroundings as 'wasted' energy, usually in the form of heat (QQ) or sound, making it less useful for doing work.

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The total energy output is always equal to the total energy input, expressed as: Input Energy=Useful Energy+Wasted Energy\text{Input Energy} = \text{Useful Energy} + \text{Wasted Energy}.

📐Formulae

Total Energyinput=Total Energyoutput\text{Total Energy}_{input} = \text{Total Energy}_{output}

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

Etotal=KE+GPE+QE_{total} = KE + GPE + Q

💡Examples

Problem 1:

A light bulb is supplied with 100 J100\text{ J} of electrical energy. It produces 20 J20\text{ J} of useful light energy. How much energy is wasted as thermal energy?

Solution:

80 J80\text{ J}

Explanation:

According to the Law of Conservation of Energy, Total Input=Useful Output+Wasted Output\text{Total Input} = \text{Useful Output} + \text{Wasted Output}. By substituting the values: 100 J=20 J+Wasted Energy100\text{ J} = 20\text{ J} + \text{Wasted Energy}. Therefore, Wasted Energy=100 J−20 J=80 J\text{Wasted Energy} = 100\text{ J} - 20\text{ J} = 80\text{ J}.

Problem 2:

A roller coaster car has 5000 J5000\text{ J} of Gravitational Potential Energy (GPEGPE) at the top of a hill. As it rolls down to the lowest point, it converts most of this into Kinetic Energy (KEKE). If 500 J500\text{ J} is lost to friction as heat, how much KEKE does the car have at the bottom?

Solution:

4500 J4500\text{ J}

Explanation:

The total energy at the start is 5000 J5000\text{ J}. At the bottom, this energy is split between KEKE and wasted thermal energy (QQ). Using the formula Initial GPE=KE+Q\text{Initial } GPE = KE + Q, we get 5000 J=KE+500 J5000\text{ J} = KE + 500\text{ J}. Thus, KE=5000 J−500 J=4500 JKE = 5000\text{ J} - 500\text{ J} = 4500\text{ J}.

Problem 3:

A battery stores 10 J10\text{ J} of chemical energy. When used in a toy car, 3 J3\text{ J} is converted to kinetic energy and 2 J2\text{ J} is converted to sound. The rest is heat. Calculate the heat energy produced.

Solution:

5 J5\text{ J}

Explanation:

The total energy must remain 10 J10\text{ J}. So, 10 J=3 J(KE)+2 J(Sound)+Heat10\text{ J} = 3\text{ J} (KE) + 2\text{ J} (Sound) + \text{Heat}. This means 10 J=5 J+Heat10\text{ J} = 5\text{ J} + \text{Heat}. Subtracting 5 J5\text{ J} from both sides gives Heat=5 J\text{Heat} = 5\text{ J}.