Question

A Carnot engine with an efficiency of 30% operates with a high-temperature reservoir at 188oC and exhausts 2000 J of heat each cycle. What are (a) the heat input per cycle and (b) the Celcius temperature of the low-temperature reservoir

Answer 1

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:

$\eta_{th} = 1-\frac{T_{L}}{T_{H}} = 1 - \frac{Q_{L}}{Q_{H}}$ (1)

Where:

$T_{L}$ - Low temperature reservoir, in Kelvin.

$T_{H}$ - High temperature reservoir, in Kelvin.

$Q_{L}$  - Heat output, in joules.

$Q_{H}$ - Heat input, in joules.

$\eta_{th }$ - Engine efficiency, no unit.

If we know that $\eta_{th} = 0.3$ and $Q_{L} = 2000\,J$, the heat input of the Carnot engine is:

$\eta_{th} = 1 - \frac{Q_{L}}{Q_{H}}$

$\frac{Q_{L}}{Q_{H}} = 1 - \eta_{th}$

$Q_{H} = \frac{Q_{L}}{1-\eta_{th}}$

$Q_{H} = \frac{2000\,J}{1-0.3}$

$Q_{H} = 2857.143\,J$

The heat input per cycle is 2857.143 joules.

b) If we know that $T_{H} = 461.15\,K$ and $\eta_{th} = 0.3$, then the temperature of the low-temperature reservoir:

$\eta_{th} = 1 - \frac{T_{L}}{T_{H}}$

$\frac{T_{L}}{T_{H}} = 1 - \eta_{th}$

$T_{L} = T_{H}\cdot (1-\eta_{th})$

$T_{L} = (461.15\,K)\cdot (1-0.3)$

$T_{L} = 322.805\,K$

$T_{L} = 49.655\,^{\circ}C$

The temperature of the low-temperature reservoir is 49.655 °C.