   Chapter 12, Problem 38P

Chapter
Section
Textbook Problem

A lawnmower engine ejects 1.00 × 104 J each second while running with an efficiency of 0.200. Find the engine’s horse-power rating, using the conversion factor 1 hp = 746 W.

To determine
The horse power rating of engine.

Explanation

Section1:

To determine: To determine the energy absorbed from the hot reservoir.

Answer: The energy absorbed from the hot reservoir is 1.25×104J

Explanation:

Given Info:

Efficiency of the engine is 0.200 .

The energy expelled to the cold reservoir is 1×104J .

From First Law of Thermodynamics,

e=1|Qc||Qh|

• Qc is the energy expelled to the cold reservoir
• Qh is the energy absorbed from the hot reservoir

Formula to calculate the energy absorbed from the hot reservoir is,

|Qh|=|Qc|1e

Substitute 1×104J for Qc and 0.200 for e to find the energy absorbed from the hot reservoir.

|Qh|=(1.00×104)J10.200=1.25×104J

Thus, |Qh| is 1.25×104J .

Section 2:

To determine: the work done by the engine in each second.

Answer: The work done by the engine in each second is 2.50×103J .

Explanation:

Given Info:

The energy expelled to the cold reservoir is 1×104J .

The energy absorbed from the hot reservoir is 1

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