A Carnot engine receives 4000 J of heat from a reservoir at 500 K, does some work, and rejects 2000 J heat to a reservoir at 200 K.
(a) How much work does it do?
(b) What is its efficiency?
Answer (2)
In this question, we are dealing with a Carnot engine, which is a theoretical thermodynamic cycle proposed by Nicolas Léonard Sadi Carnot. It is considered the most efficient cycle possible and sets an upper limit on the efficiency of all real heat engines.
To solve this problem, we need to find two things: (a) the work done by the engine and (b) its efficiency.
(a) Work Done by the Engine
The work done by a Carnot engine, or any heat engine, can be found using the First Law of Thermodynamics, which is expressed as follows:
W = Q 1 − Q 2
Where:
W is the work done by the engine.
Q 1 is the heat absorbed from the hot reservoir.
Q 2 is the heat rejected to the cold reservoir.
In this case:
Q 1 = 4000 J
Q 2 = 2000 J
Calculating the work done:
W = 4000 J − 2000 J = 2000 J
So, the engine does 2000 J of work.
(b) Efficiency of the Engine
The efficiency ( e p s i l o n ) of a Carnot engine is calculated using the temperatures of the hot and cold reservoirs. The formula for efficiency is:
e p s i l o n = 1 − T H T C
Where:
T C is the absolute temperature of the cold reservoir in Kelvin.
T H is the absolute temperature of the hot reservoir in Kelvin.
Given:
T H = 500 K
T C = 200 K
Substitute the values into the efficiency formula:
e p s i l o n = 1 − 500 200
e p s i l o n = 1 − 0.4 = 0.6
Therefore, the efficiency of the Carnot engine is 0.6, or 60%.
Summary:
The Carnot engine does 2000 J of work.
Its efficiency is 60%.
The Carnot engine does 2000 J of work and has an efficiency of 60%.
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