Question

QUESTION 2 Copy of A power plant was analysed based on Carnot cycle, the evaporator supplie 459 kJ of heat per cycle at 442°C and condenser cooling fluid was maintained at 33°C. Determine the amount of heat rejected in the condenser? Provide answer to no decimal place and insert the unit symbol in kilo.... 1 °C is 273K in this question if required.

Answer 1

To determine the amount of heat rejected in the condenser of a power plant operating on the Carnot cycle, we need to use the temperature difference between the evaporator and the condenser.

Given:
Temperature at the evaporator (T₁) = 442°C
Temperature at the condenser (T₂) = 33°C

To convert these temperatures to Kelvin, we add 273 to each value:
T₁ = 442 + 273 = 715 K
T₂ = 33 + 273 = 306 K

The amount of heat rejected in the condenser can be calculated using the formula:

Q = Qₕ - Qₗ

Where:
Q = Amount of heat rejected in the condenser
Qₕ = Heat supplied by the evaporator (459 kJ)
Qₗ = Heat absorbed by the condenser

Since the Carnot cycle is a reversible cycle, the heat rejected in the condenser is equal to the heat absorbed by the evaporator. Therefore:

Q = Qₕ - Qₗ = 459 kJ

So the amount of heat rejected in the condenser is 459 kilojoules (kJ)