Answer:
a) The heat input per cycle is 2857.143 joules.
b) The temperature of the low-temperature reservoir is 49.655 °C.
Explanation:
a) The efficiency of the Carnot engine is defined by the following formula:
[tex]\eta_{th} = 1-\frac{T_{L}}{T_{H}} = 1 - \frac{Q_{L}}{Q_{H}}[/tex] (1)
Where:
[tex]T_{L}[/tex] - Low temperature reservoir, in Kelvin.
[tex]T_{H}[/tex] - High temperature reservoir, in Kelvin.
[tex]Q_{L}[/tex] - Heat output, in joules.
[tex]Q_{H}[/tex] - Heat input, in joules.
[tex]\eta_{th }[/tex] - Engine efficiency, no unit.
If we know that [tex]\eta_{th} = 0.3[/tex] and [tex]Q_{L} = 2000\,J[/tex], the heat input of the Carnot engine is:
[tex]\eta_{th} = 1 - \frac{Q_{L}}{Q_{H}}[/tex]
[tex]\frac{Q_{L}}{Q_{H}} = 1 - \eta_{th}[/tex]
[tex]Q_{H} = \frac{Q_{L}}{1-\eta_{th}}[/tex]
[tex]Q_{H} = \frac{2000\,J}{1-0.3}[/tex]
[tex]Q_{H} = 2857.143\,J[/tex]
The heat input per cycle is 2857.143 joules.
b) If we know that [tex]T_{H} = 461.15\,K[/tex] and [tex]\eta_{th} = 0.3[/tex], then the temperature of the low-temperature reservoir:
[tex]\eta_{th} = 1 - \frac{T_{L}}{T_{H}}[/tex]
[tex]\frac{T_{L}}{T_{H}} = 1 - \eta_{th}[/tex]
[tex]T_{L} = T_{H}\cdot (1-\eta_{th})[/tex]
[tex]T_{L} = (461.15\,K)\cdot (1-0.3)[/tex]
[tex]T_{L} = 322.805\,K[/tex]
[tex]T_{L} = 49.655\,^{\circ}C[/tex]
The temperature of the low-temperature reservoir is 49.655 °C.
