Question

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.

Answer 1

1200

Step-by-step explanation:

Answer 2

Answer:1200

Step-by-step 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-(T_L)/(T_H)`

`\eta =(W_s)/(Q_s)`

`1-(293)/(513)=(910)/(Q_s)`

`Q_s=2121.954 W`

`Q_r=1211.954 W`

rounding off to two significant figures

`Q_r=1200 W`