Answer:
1.07×10⁹ J
3.76×10⁸ J
Explanation:
The efficiency of a Carnot engine is equal to the work output divided by the heat input. It is also equal to one minus the ratio of the cold temperature to the hot temperature.
[tex]\huge \text {$ \eta=\frac{\.W_{out}}{\.Q_{in}}=1-\frac{T_C}{T_H} $}[/tex]
Plug in values and solve for Q (make sure to use absolute temperatures).
[tex]\Large \text {$ \frac{193\ kW}{\.Q_{in}} =1-\frac{25+273\ K}{573+273\ K} $}\\\Large \text {$ \frac{193\ kW}{\.Q_{in}} =0.649 $}\\\Large \text {$ \.Q_{in} =298\ kW $}\\\Large \text {$ \.Q_{in} =298\frac{kJ}{s}\times\frac{3600\ s}{hr} $}\\\Large \text {$ \.Q_{in} = 1,071,300\frac{kJ}{hr} $}\\\Large \text {$ \.Q_{in} = 1.07\times10^9\frac{J}{hr} $}[/tex]
Energy is conserved, so the heat in must equal the sum of the heat out and the work out.
[tex]\Large \text {$ \.Q_{in}=\.Q_{out}+\.W_{out} $}\\\Large \text {$ 1.07\times10^9\frac{J}{hr} =\.Q_{out}+193,000\frac{J}{s}\times\frac{3600\ s}{hr} $}\\\Large \text {$ \.Q_{out}=3.76\times10^8\frac{J}{hr} $}[/tex]
