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Chapter 15: Problem 14

What is the efficiency of an electric generator that produces $1.17 \mathrm{kW}\(.h per \)\mathrm{kg}$ of coal burned? The heat of combustion of coal is \(6.71 \times 10^{6} \mathrm{J} / \mathrm{kg}\).

Short Answer

Expert verified
Answer: The efficiency of the electric generator is 62.74%.

Step by step solution

01

Convert electrical energy to joules

First, we need to convert the electrical energy from kilowatt-hours to joules. 1.17 kW.h equals 1.17 * 1000 W * 3600 s = 4212000 J.
02

Calculate the efficiency

Now we can find the efficiency using the formula: Efficiency = (Output energy / Input energy) × 100 The output energy is the electrical energy generated, which is 4212000 J, while the input energy is the energy obtained from burning coal, which is \(6.71 \times 10^{6} \mathrm{J}\) per kg. Efficiency = (4212000 J / \(6.71 \times 10^{6} \mathrm{J}\)) × 100 = 0.6274 * 100 = 62.74% The efficiency of the electric generator is 62.74%.

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Most popular questions from this chapter

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