   Chapter 12, Problem 42P

Chapter
Section
Textbook Problem

Two heat engines are operated in series so that part of the energy expelled from engine A is absorbed by engine B with |QhB| = 0.750|QcA|. Engines A and B have efficiencies eA = eB = 0.250 and engine A performs work WA = 275 J. Find the overall efficiency of the two-engine combination, given by e = W A + W B | Q b A | .

To determine
The overall efficiency of two engine combination.

Explanation

Section1:

To determine: The heat absorbed from hot reservoir by engine A

Solution: |QhA| is 1.10×103J .

Explanation:

Given Info:

The efficiency of engine A is 0.250 .

The work done by engine A is 275J .

Formula to calculate |QhA| ,

|QhA|=WAeA

• WA is the work done by engine A
• eA is the efficiency of engine A

Substitute 275J for WA and 0.250 for eA to find |QhA| ,

|QhA|=275J0.250=1.10×103J

Thus, |QhA| is 1.10×103J .

Section 2:

To determine: the total energy ejected to the cold reservoir by engine A.

Solution: |QcA| is 825J .

Explanation:

Given Info:

The efficiency of engine A is 0.250 .

The work done by engine A is 275J .

From First law of Thermodynamics,

|QcA|=|QhA|WA

• |QhA| is the energy absorbed from the hot reservoir by engine A,
• WA is the work done by engine A,

Thus,

QcA=WAeAWA=WA(1eAeA)

• eA is the efficiency of engine A

Substitute 0.250 for eA and 275J for WA to find the energy ejected to the cold reservoir by engine A,

|QcA|=275J(10

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