Model A converts every 1000 J of electrical energy into 600 J kinetic, 300 J heat, and 100 J sound; Model B converts every 1500 J into 800 J kinetic, 400 J heat, and 300 J sound. What is the energy efficiency of each model, and which is more efficient?

What “energy efficiency” means for an electric beater

An electric beater’s job is to produce motion, so the useful energy output is the kinetic energy. Heat and sound are wasted outputs.

Model A: compute the useful fraction

Input electrical energy: $1000\ \text{J}$

Useful (kinetic) output: $600\ \text{J}$

$$\eta_A = \frac{E_{\text{useful}}}{E_{\text{in}}}\times 100\% = \frac{600}{1000}\times 100\% = 60\%$$

Model B: compute the useful fraction

Input electrical energy: $1500\ \text{J}$

Useful (kinetic) output: $800\ \text{J}$

$$\eta_B = \frac{800}{1500}\times 100\% \approx 53.3\%$$

Compare and interpret

• Model A: $60\%$ efficient
• Model B: $53.3\%$ efficient

So, Model A is more efficient because a greater percentage of its input energy becomes kinetic energy.

Quick check using energy conservation

For each model, the outputs add to the input:

• A: $600 + 300 + 100 = 1000\ \text{J}$
• B: $800 + 400 + 300 = 1500\ \text{J}$