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Energy and Heat - Energy Transformations and Conservation

Grade 7IB

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

🔑Concepts

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Energy is defined as the capacity to do work and is measured in Joules (JJ).

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The Law of Conservation of Energy states that energy cannot be created or destroyed, only transformed from one form to another or transferred between objects. The total energy in a closed system remains constant (Etotal=constantE_{total} = \text{constant}).

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Kinetic Energy (KEKE) is the energy of an object due to its motion. It depends on the mass (mm) and the square of the velocity (v2v^2).

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Gravitational Potential Energy (GPEGPE) is the energy stored in an object due to its height above a reference point. It is calculated using mass (mm), gravitational acceleration (g≈9.8 m/s2g \approx 9.8 \text{ m/s}^2), and height (hh).

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Energy Transformation refers to the change of energy from one form to another (e.g., a battery converting chemical energy into electrical energy).

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Heat is the transfer of thermal energy from a system at a higher temperature to one at a lower temperature until thermal equilibrium is reached.

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Efficiency is a measure of how much 'useful' energy is produced compared to the total energy input. No machine is 100%100\% efficient as some energy is always dissipated as waste heat (QwasteQ_{waste}).

📐Formulae

KE=12mv2KE = \frac{1}{2}mv^2

GPE=mghGPE = mgh

Etotal=KE+PEE_{total} = KE + PE

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

💡Examples

Problem 1:

A 2 kg2\text{ kg} ball is held at a height of 5 m5\text{ m} above the ground. Calculate its Gravitational Potential Energy (GPEGPE). (Assume g=9.8 m/s2g = 9.8\text{ m/s}^2)

Solution:

GPE=mgh=2 kg×9.8 m/s2×5 m=98 JGPE = mgh = 2\text{ kg} \times 9.8\text{ m/s}^2 \times 5\text{ m} = 98\text{ J}

Explanation:

Using the formula for potential energy, we multiply the mass, the acceleration due to gravity, and the height to find the energy stored in the object.

Problem 2:

An electric motor takes in 500 J500\text{ J} of electrical energy. It produces 350 J350\text{ J} of useful mechanical work. What is the efficiency of the motor?

Solution:

Efficiency=350 J500 J×100%=70%\text{Efficiency} = \frac{350\text{ J}}{500\text{ J}} \times 100\% = 70\%

Explanation:

Efficiency is calculated by dividing the useful output energy by the total input energy and multiplying by 100100 to get a percentage. The remaining 150 J150\text{ J} is likely lost as heat.

Problem 3:

A toy car with a mass of 0.5 kg0.5\text{ kg} is moving at a velocity of 4 m/s4\text{ m/s}. Calculate its Kinetic Energy (KEKE).

Solution:

KE=12×0.5 kg×(4 m/s)2=0.25×16=4 JKE = \frac{1}{2} \times 0.5\text{ kg} \times (4\text{ m/s})^2 = 0.25 \times 16 = 4\text{ J}

Explanation:

Applying the kinetic energy formula, we square the velocity first (1616), then multiply by the mass and the factor of 0.50.5.