recall and understand that the efficiency of a system is the ratio of useful energy output from the system to the total energy input

Energy Conservation

Energy cannot be created or destroyed; it can only be transferred or transformed from one form to another. This principle is known as the law of conservation of energy. In any physical process the total energy before the process equals the total energy after the process.

Mathematically, for a closed system:

\$\$

\sum E{\text{initial}} = \sum E{\text{final}}

\$\$

Useful vs. Lost Energy

During a transformation, part of the initial energy is converted into the desired (useful) form, while the remainder is dissipated as heat, sound, light, or other non‑useful forms. The proportion of useful energy to the total energy supplied is called the efficiency of the system.

Efficiency of a System

Definition

The efficiency, denoted by \$\eta\$, is the ratio of the useful energy output to the total energy input:

\$\$

\eta = \frac{E{\text{useful}}}{E{\text{input}}}

\$\$

It is usually expressed as a percentage:

\$\$

\eta{\%} = \left( \frac{E{\text{useful}}}{E_{\text{input}}} \right) \times 100\%

\$\$

Key Points to Remember

• Efficiency is always less than or equal to 100 %.
• Real systems always have losses (e.g., friction, air resistance, electrical resistance).
• Higher efficiency means less wasted energy and often lower operating costs.

Typical Efficiencies

Calculating Efficiency – Worked Example

1. Identify the total energy supplied to the system, \$E_{\text{input}}\$.
2. Determine the useful energy that the system delivers, \$E_{\text{useful}}\$.
3. Apply the efficiency formula:

   \$\eta = \frac{E{\text{useful}}}{E{\text{input}}}\$

4. Convert to a percentage if required.

Example: A 1500 W electric heater draws a current of 6.25 A from a 240 V supply for 10 minutes. The heater raises the temperature of water, delivering 80 % of the electrical energy as heat to the water. Calculate the efficiency of the heater.

Solution:

• Electrical power input: \$P = VI = 240\ \text{V} \times 6.25\ \text{A} = 1500\ \text{W}\$ (consistent with rating).
• Energy input over 10 min:

  \$E_{\text{input}} = P t = 1500\ \text{W} \times 600\ \text{s} = 9.0 \times 10^{5}\ \text{J}\$

• Useful energy delivered to water:

  \$E{\text{useful}} = 0.80 \times E{\text{input}} = 7.2 \times 10^{5}\ \text{J}\$

• Efficiency:

  \$\eta = \frac{7.2 \times 10^{5}}{9.0 \times 10^{5}} = 0.80 \; \text{or} \; 80\%\$

Common Misconceptions

• “Efficiency can be 100 %.” Only an ideal, loss‑free system could reach 100 %, which does not exist in practice.
• “If the output energy is larger than the input, the system is super‑efficient.” This would violate energy conservation; any apparent gain is due to additional energy sources not accounted for.
• “Efficiency is the same as power.” Power is the rate of energy transfer (\$P = \frac{E}{t}\$), whereas efficiency is a dimensionless ratio.

Practice Questions

1. A car engine receives 2.5 MJ of chemical energy from fuel and delivers 0.6 MJ as kinetic energy to the car. What is the engine’s efficiency? Express your answer as a percentage.
2. A solar panel has an area of 1.5 m² and receives solar irradiance of 800 W m⁻². If the panel produces 120 W of electrical power, calculate its efficiency.
3. During a physics experiment, a pendulum is released from a height of 0.45 m. Assuming no air resistance, the potential energy at the start is \$mgh\$. If the measured kinetic energy at the lowest point is only 85 % of the calculated \$mgh\$, what is the efficiency of the energy conversion in this experiment?

Suggested Diagram