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A Carnot engine performs 2.5 * 104 J of work in each cycle and has an efficiency of 66%. (a) How much heat does the engine extract from its heat source in each cycle? (b) If the engine exhausts heat at room temperature 120.0°C2, what is the temperature of its heat source? A Carnot engine performs 2.5 * 104 J of work in each cycle and has an efficiency of 66%. (a) How much heat does the engine extract from its heat source in each cycle? (b) If the engine exhausts heat at room temperature (20.0°C), what is the temperature of its heat source?

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a.

The heat the engine extracts from its heat source in each cycle is 3.79 × 10⁴ J

The efficiency of the Carnot engine, ε = W/Q where W = work done per cycle = 2.5 × 10⁴ J and Q = heat extracted in each cycle.

Making Q subject of the formula, we have

Q = W/ε

Since ε = 66% = 0.66, we substitute the values of the variables into the equation for Q.

So, Q = W/ε

Q = 2.5 × 10⁴ J/0.66

Q = 3.79 × 10⁴ J

So, the heat the engine extracts from its heat source in each cycle is 3.79 × 10⁴ J

b.

The temperature of its heat source is 588.76 °C

Also, the efficiency of the Carnot engine, ε = 1 - T/T' where T = temperature of exhaust heat = 20° C = 273 + 20 = 293 K and T' = temperature of heat source.

Making T' subject of the formula, we have

T' = T/(1 - ε)

Since ε = 66 % = 0.66

Substituting the values of the variables into the equation, we have

T' = T/(1 - ε)

T' = 293 K/(1 - 0.66)

T = 293 K/0.34

T = 861.76 K

We convert this to Celsius.

T(K) = T(°C) + 273

T(°C) = T(K) - 273

T(°C) = 861.76 - 273

T(°C) = 588.76 °C

So, the temperature of its heat source is 588.76 °C

Learn more about Carnot engine here:

https://brainly.com/question/13170743

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