Answer:1200
Explanation:
Given data
Upper Temprature[tex]\left ( T_H\right )=240^{\circ}\approx 513[/tex]
Lower Temprature [tex]\left ( T_L\right )=20^{\circ}\approx 293[/tex]
Engine power ouput[tex]\left ( W\right )=910 W[/tex]
Efficiency of carnot cycle is given by
[tex]\eta =1-\frac{T_L}{T_H}[/tex]
[tex]\eta =\frac{W_s}{Q_s}[/tex]
[tex]1-\frac{293}{513}=\frac{910}{Q_s}[/tex]
[tex]Q_s=2121.954 W[/tex]
[tex]Q_r=1211.954 W[/tex]
rounding off to two significant figures
[tex]Q_r=1200 W[/tex]
