Question

A reversible heat engine operates on carnot cycle between source and sink temperature of 225C and 25C if engine receive 40kw from source Find the net work done ,heat rejected sink and efficiency of engine​

Answer 1

Answer:

225c is the answer I think

Step-by-step explanation:

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Answer 2

Given:

Temperature of the source (T1) = 225°C = 498K

Temperature of the sink (T2) = 25°C = 298K

The energy emitted from the source (Q1) = 40kW

To Find:

The work that was done on the engine (W)

The heat rejected to the sink (Q2)

The efficiency of the engine.

Solution:

The heat of the source and sink is related to their temperature as

⇒ Q1/Q2 = T1/T2

⇒ Q2 = (T2×Q1)÷ T1

⇒ Q2 = $\frac{298*40}{498} =23.94$ = 23.94kW

The work done on the engine (W) = Q1 - Q2 = 16.06kW

The efficiency of the heat engine = (Q1 - Q2) ÷ Q1 = $\frac{Q1-Q2}{Q1}$

⇒ Efficiency = work done on the engine ÷ heat emitted from the source

⇒ Efficiency = 16.06 ÷ 40 = 0.4015

Hence, the work done is 16.06kW, the heat rejected to the sink is 23.94kW and the efficiency of the system is 0.4015.