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A Carnot cycle receives heat at 327°C causing an increase in entropy equal to5 kJ/kg K. The engine delivers 2000 kJ/kg of work. Determine the efficiency of the cycle and temperature of the sink.​

User Matt Small
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Final answer:

To calculate the efficiency of a Carnot cycle with given parameters, the temperature of the source is converted to Kelvin and the formula for work done (W = S * (Tₖ - Tₗ)) is used to find the sink temperature. The efficiency is then calculated using the formula Η = 1 - (Tₗ / Tₖ), resulting in an efficiency of 66.68% and a sink temperature of 200.15 K.

Step-by-step explanation:

The question involves calculating the efficiency of a Carnot cycle which receives heat at 327°C and causes an entropy increase of 5 kJ/kg K while delivering 2000 kJ/kg of work. To determine the efficiency of the cycle, we use the following formula:

Η = 1 - Τₗ/Τₖ

Where Τₗ is the temperature of the sink (in Kelvin), and Τₖ is the temperature of the source (also in Kelvin). First, convert the source temperature from Celsius to Kelvin:

Τₖ = 327°C + 273.15 = 600.15 K

The work done by the engine is the product of the entropy increase and the temperature difference between the source and the sink:

W = S * (Τₖ - Τₗ)

Given W = 2000 kJ/kg, S = 5 kJ/kg K, and Τₖ = 600.15 K, we can now solve for Τₗ:

2000 = 5 * (600.15 - Τₗ)

400 = 600.15 - Τₗ

Τₗ = 600.15 - 400 K = 200.15 K

Now, we can calculate the efficiency using the formula:

Η = 1 - (200.15 / 600.15)

Η = 0.6668 or 66.68%

The efficiency of the cycle is 66.68%, and the temperature of the sink is 200.15 K.

User Daniel Leach
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