- Identify the total energy input as 800 J and wasted energy as 504 J.
- Calculate the useful energy output: \(800 \text{ J} - 504 \text{ J} = 296 \text{ J}\).
- Use the efficiency formula: \(\text{Efficiency} = \left( \frac{\text{Useful Energy Output}}{\text{Total Energy Input}} \right) \times 100\% = \left( \frac{296}{800} \right) \times 100\% = 37\%\).
- The final efficiency is \(\boxed{37\%}\).
Explanation
- Understanding the Problem and Data Analysis
We are given that a hair dryer has a total energy input of \(800 \text{ J}\) and that \(504 \text{ J}\) of this energy is transferred to wasted energy stores. We need to determine the efficiency of the hair dryer, which represents the percentage of the total energy input that is transformed into useful energy. We know that energy is conserved, so the total energy input must equal the sum of the useful energy output and the wasted energy output. Therefore, the useful energy output can be found by subtracting the wasted energy from the total input energy. Once we have the useful energy, we use the standard efficiency formula: \(\text{Efficiency} = \left( \frac{\text{Useful Energy Output}}{\text{Total Energy Input}} \right) \times 100\%\).
- Calculating Useful Energy Output
The useful energy output is the difference between the total energy input and the wasted energy output. Using the given values:
\[\text{Useful Energy Output} = 800 \text{ J} - 504 \text{ J} = 296 \text{ J}.\]
This represents the amount of energy actually used by the hair dryer for its intended purpose.
- Calculating Efficiency
Now that we have the useful energy output, we can calculate the efficiency by dividing this value by the total energy input and then multiplying by 100 to express it as a percentage:
\[\text{Efficiency} = \left( \frac{296 \text{ J}}{800 \text{ J}} \right) \times 100\%.\]
Dividing 296 by 800 gives:
\[\frac{296}{800} = 0.37.\]
Converting this decimal to a percentage:
\[0.37 \times 100\% = 37\%.\]
Thus, the efficiency of the hair dryer is 37%.
Examples
Efficiency is a fundamental concept in engineering and physics, helping us understand how effectively devices perform their tasks. For instance, consider a light bulb: if it uses 100 joules of electricity but only produces 10 joules of visible light, its efficiency is 10%, with the remaining 90% being lost as heat. Understanding these percentages allows engineers to design better, more energy-efficient appliances that consume less electricity while doing the same amount of work, ultimately saving money and reducing environmental impact.