use the concept of efficiency to solve problems

Energy Conservation and Efficiency

The law of conservation of energy states that energy cannot be created or destroyed, only transferred or transformed. In real‑world systems some of the input energy is inevitably lost as heat, sound, or other non‑useful forms. The concept of efficiency quantifies how effectively a system converts input energy into a desired output.

Key Concepts

• Conserved quantity: total energy of an isolated system remains constant.
• Useful energy: the portion of the input energy that appears as the intended form of output (e.g., mechanical work, electrical power).
• Energy loss: energy transformed into forms that do not contribute to the intended output, usually dissipated as heat.
• Efficiency ($\eta$): the ratio of useful output energy to input energy, expressed as a percentage.

Definition of Efficiency

Symbol	Definition	Typical Units
$\eta$	Efficiency = $\dfrac{\text{Useful output energy}}{\text{Input energy}}$	Dimensionless (often expressed as %)
$E_{\text{in}}$	Energy supplied to the system	Joules (J)
$E_{\text{out}}$	Useful energy obtained from the system	Joules (J)
$E_{\text{loss}}$	Energy dissipated as heat, sound, etc.	Joules (J)

Using Efficiency in Problem Solving

1. Identify the form of energy supplied to the system ($E_{\text{in}}$).
2. Determine the required useful output energy ($E_{\text{out}}$) or the desired output quantity (e.g., work, kinetic energy).
3. Read or calculate the efficiency $\eta$ of the device or process.
4. Apply the efficiency relation: $$E_{\text{out}} = \eta \, E_{\text{in}}$$ or, if $E_{\text{out}}$ is known, $$E_{\text{in}} = \frac{E_{\text{out}}}{\eta}$$
5. Account for any additional energy transformations (e.g., gravitational potential, kinetic energy) using the appropriate kinematic or dynamic equations.
6. Check that the final answer respects the conservation of energy and that the calculated losses are physically reasonable.

Worked Example

Problem: A 1500 W electric kettle is used to heat 0.5 kg of water from 20 °C to its boiling point (100 °C). The kettle’s efficiency is 80 %. Calculate:

1. The energy supplied by the kettle.
2. The time required to bring the water to boil.

Solution:

1. Calculate the useful energy needed to raise the water temperature: $$E_{\text{useful}} = m c \Delta T$$ where $m = 0.5\ \text{kg}$, $c = 4186\ \text{J kg}^{-1}\text{K}^{-1}$, $\Delta T = 100-20 = 80\ \text{K}$. $$E_{\text{useful}} = 0.5 \times 4186 \times 80 = 1.674 \times 10^{5}\ \text{J}$$
2. Relate useful energy to the input energy using efficiency $\eta = 0.80$: $$E_{\text{in}} = \frac{E_{\text{useful}}}{\eta} = \frac{1.674 \times 10^{5}}{0.80} = 2.0925 \times 10^{5}\ \text{J}$$
3. Power of the kettle is $P = 1500\ \text{W} = 1500\ \text{J s}^{-1}$. Time required: $$t = \frac{E_{\text{in}}}{P} = \frac{2.0925 \times 10^{5}}{1500} \approx 139.5\ \text{s}$$ So the water will boil in about $2.3\ \text{min}$.

Common Pitfalls

• Forgetting to convert efficiency from a percentage to a decimal before using it in calculations.
• Assuming 100 % efficiency for devices that are known to have losses (e.g., engines, transformers).
• Neglecting the energy required for phase changes when heating or cooling substances.
• Mixing up input and output powers when dealing with continuous processes; remember $P = \frac{E}{t}$.

Summary

Efficiency provides a practical bridge between the idealised conservation of energy and real‑world applications where losses occur. By incorporating the efficiency factor into energy calculations, students can accurately predict the performance of devices, estimate required input energy, and evaluate the feasibility of engineering solutions.