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Q34PE

Expert-verifiedFound in: Page 551

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.

**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. **

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

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.

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