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Sergeu [11.5K]
3 years ago
6

A Carnot engine with an efficiency of 30% operates with a high-temperature reservoir at 188oC and exhausts 2000 J of heat each c

ycle. What are (a) the heat input per cycle and (b) the Celcius temperature of the low-temperature reservoir
Physics
1 answer:
Reika [66]3 years ago
6 0

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.

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