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tekilochka [14]

2 years ago

5

A heat engine performs (245 + A) J of work in each cycle while also delivering (142 + B) J of heat to the cold reservoir. Find t

he efficiency of the heat engine. Give your answer in percent (%) and with 3 significant figures.
A=14, B=72

1 answer:

Ganezh [65]2 years ago

5 0

Answer:

The value is $\eta = 54.4 \%$

Explanation:

From the question we are told that

The work input is $W = ( 245 + A ) \ J$

The heat delivered is $Q = (142 + B) \ J$

The value of A is A = 14

The value of B is B = 72

Generally the efficiency of the heat engine is mathematically represented as

$\eta = \frac{W}{Q_t}$

Here $Q_t$ is the total out energy produce by the heat engine and this is mathematically represented as

$Q_t= Q + W$

=> $Q_t= 245 + A + 142 + B$

=> $Q_t= 390 + A+B$

So

$\eta = \frac{245 + A }{390 + A+ B}$

=> $\eta = 0.544$

=> $\eta = 0.544 *100$

=> $\eta = 54.4 \%$

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2 years ago

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3 years ago

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