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Expert-verified Found in: Page 551 ### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000 # Would you be willing to financially back an inventor who is marketing a device that she claims has 25 kJ of heat transfer at 600 K, has heat transfer to the environment at 300 K, and does 12 kJ of work? Explain your answer.

The device has a maximum efficiency of 50%. The calculated efficiency is 48% which is very close to maximum efficiency. Therefore, I would back the inventor financially.

See the step by step solution

## Step 1: Introduction

The ratio of amount of work done by the engine to the amount of heat intake from the source gives the efficiency of the heat engine.

## Step 2:  Given parameters and formula for efficiency of heat engine

Temperature of cold reservoir $${T_c} = 300\;{\rm{K}}$$

Temperature of hot reservoir $${T_h} = 600\;{\rm{K}}$$

Work done $$W = 12\;{\rm{kJ}}$$

Heat transferred $$Q = 25\;{\rm{kJ}}$$

The efficiency of the Carnot engine $$\eta = 1 - \frac{{{T_c}}}{{{T_h}}}$$

Here,

$$\eta$$ - efficiency of the engine.

$${T_c}$$- the temperature of the cold reservoir.

$${T_h}$$- temperature of the hot reservoir

## Step 3:  Calculate maximum efficiency and real efficiency of the device

Maximum efficiency can be calculated by calculating Carnot efficiency as

\begin{aligned}{}h &= 1 - \frac{{{T_c}}}{{{T_h}}}\\ &= 1 - \frac{{300\,{\rm{k}}}}{{600\,{\rm{k}}}}\\& = 0.5\\ &= 50\% \end{aligned}

Real efficiency can be calculated as

\begin{aligned}{}h &= \frac{W}{Q}\\ &= \frac{{12\,{\rm{J}}}}{{25\,{\rm{J}}}}\\ &= 0.48\\ &= 48\% \end{aligned}

Therefore, the maximum efficiency is $$50\%$$ for the device. The efficiency which device is giving is $$48\%$$ which is very close to maximum efficiency. Therefore, I would be willing to financially back the inventor. ### Want to see more solutions like these? 