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Brums [2.3K]
4 years ago
15

A Carnot engine's operating temperatures are 240 ∘C and 20 ∘C. The engine's power output is 910 W . Part A Calculate the rate of

heat output. Express your answer using two significant figures.
Physics
1 answer:
scoray [572]4 years ago
5 0

Answer:1200

Explanation:

Given data

Upper Temprature\left ( T_H\right )=240^{\circ}\approx 513

Lower Temprature \left ( T_L\right )=20^{\circ}\approx 293

Engine power ouput\left ( W\right )=910 W

Efficiency of carnot cycle is given by

\eta =1-\frac{T_L}{T_H}

\eta =\frac{W_s}{Q_s}

1-\frac{293}{513}=\frac{910}{Q_s}

Q_s=2121.954 W

Q_r=1211.954 W

rounding off to two significant figures

Q_r=1200 W

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