A Carnot heat engine has an efficiency of 0.400. If it operates between a deep lake with a constant temperature of 298.0k and a hot reservoir, what is the temperature of the hot reservoir?

Answers

Answer 1
Answer:

Answer:

496.7 K

Explanation:

The efficiency of a Carnot engine is given by the equation:

\eta = 1 - (T_H)/(T_L)

where:

T_H is the temperature of the hot reservoir

T_C is the temperature of the cold reservoir

For the engine in the problem, we know that

\eta = 0.400 is the efficiency

T_C = 298.0 K is the temperature of the cold reservoir

Solving for T_H, we find:

(T_C)/(T_H)=1-\eta\nT_H = (T_C)/(1-\eta) =(298.0)/(1-0.400)=496.7 K


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Answers

Answer:

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