A Carnot engine operates between 1350 °F and 125 °F. If it rejects 55 Btu as heat, determine the work output. a. W =68.54 Btu
b. W=145.19 Btu
c. W = 115.12 Btu
d. W=235.7 Blu

C.

Explanation:

Thermodynamically speaking, a Carnot engine represents an entirely reversible thermal process and its energy efficiency represents the maximum theoretical efficiency that thermal machines can reach. The efficiency of the ideal thermal process (), no unit, is:

(1)

Where:

- Temperature of the cold reservoir, measured in Rankine.

- Temperature of the hot reservoir, measured in Rankine.

If we know that and , then the energy efficiency of the ideal thermal process is:

By First Law of Thermodynamics, we calculate the work output:

(By definition of efficiency)

(2)

Where:

- Heat received by the engine, measured in Btu.

- Heat rejected by the engine, measured in Btu.

- Work output, measured in Btu.

If we know that and , then the work output of the Carnot engine is:

The work output of the Carnot engine is 115.807 Btu. ( C)

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west