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Chapter 15: Problem 17

The efficiency of an engine is \(0.21 .\) For every \(1.00 \mathrm{kJ}\) of heat absorbed by the engine, how much (a) net work is done by it and (b) heat is released by it?

Short Answer

Expert verified
Answer: The net work done by the engine is 0.21 kJ, and the heat released is 0.79 kJ.

Step by step solution

01

Recall the formula for efficiency

In thermodynamics, the efficiency of an engine can be calculated using the formula: $$Efficiency = \frac{Work\, output}{Heat\, absorbed}$$
02

Calculate the net work output

In this case, we are given the efficiency of the engine (0.21) and the amount of heat absorbed (1.00 kJ). We can rearrange the efficiency equation to find the work output: $$Work\, output = Efficiency × Heat\, absorbed$$ By plugging in the given values, we get: $$ Work\, output = 0.21 × 1.00 \, kJ = 0.21\, kJ $$
03

Calculate the heat released

The relationship between heat absorbed, work output, and heat released can be expressed as: $$Heat\, absorbed = Work\, output + Heat\, released$$ Now, we can rearrange the equation to solve for the heat released, which will look like: $$Heat\, released = Heat\, absorbed - Work\, output$$ Plugging in the values we have already found: $$ Heat\, released = 1.00\, kJ - 0.21\, kJ = 0.79\, kJ $$
04

Results

(a) The net work done by the engine is 0.21 kJ. (b) The heat released by the engine is 0.79 kJ.

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Most popular questions from this chapter

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 \(6.71 \times 10^{6} \mathrm{J} / \mathrm{kg}\).
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