Answer:
(a) [tex]3*10^{6} W[/tex]
(b) [tex]4.9447*10^{6} W[/tex]
Explanation:
(a)
Minimum power to maintain desired temperature is given by
[tex]P_{min}=\frac {3W}{sq.ft} \times A=\frac {3W}{sq.ft} 1*10^{6} sq.ft=3*10^{6} W[/tex]
(b)
Efficiency of heat engine
Converting temperature to Kelvin
T1=31 degrees=31+273=304 K
T2=500 degree=500+273=773
[tex]Effieciency=1-\frac {T1}{T2}=1-\frac {304 K}{773 K}=0.6067[/tex]
Considering required output of [tex]3*10^{6} W[/tex] then the power needed to be supplied by the solar collector to heat engine is given by
[tex]P_{solar}=\frac {P_{min}}{Efficiency}=\frac {3*10^{6}}{0.6067}=4.9447*10^{6} W[/tex]