Answered: The hot reservoir for a Carnot engine…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Question

The hot reservoir for a Carnot engine has a temperature of 938 K, while the cold reservoir has a temperature of 362 K. The heat input for this engine is 7110 J. The 362-K reservoir also serves as the hot reservoir for a second Carnot engine. This second engine uses the rejected heat of the first engine as input and extracts additional work from it. The rejected heat from the second engine goes into a reservoir that has a temperature of 213 K. Find the total work delivered by the two engines.

Expert Solution

Step 1

Given,

Temperature of hot reservoir for Engine 1, $T_{H 1} = 938 K$ , Temperature of cold reservoir for Engine 1, $T_{C 1} = 362 K$ , Heat input in Engine 1, $Q_{1} = 7110 J$ , Temperature of hot reservoir for Engine 2, $T_{H 2} = 362 K$ , Temperature of cold reservoir for Engine 2, $T_{C 2} = 213 K$ , Heat input in Engine 1, $Q_{2} = Q_{1} - W_{1}$ .

Step 2

The efficiency of the 1st engine is given as,

$\begin{array}{rcl} η_{1} & = & 1 - \frac{T_{C 1}}{T_{H 1}} \\ = & 1 - \frac{362 K}{938 K} \\ = & \frac{938 - 362}{938} \\ = & \frac{576}{938} \\ = & 61.40 % \end{array}$

The work done by the 1st engine,

$\begin{array}{rcl} W_{1} & = & η_{1} \times Q_{1} \\ = & 61.40 % \times 7110 J \\ = & 4365.54 J \end{array}$

The efficiency of the 2nd engine is given as,

$\begin{array}{rcl} η_{2} & = & 1 - \frac{T_{C 2}}{T_{H 2}} \\ = & 1 - \frac{213 K}{362 K} \\ = & \frac{362 - 213}{362} \\ = & \frac{149}{362} \\ = & 41.16 % \end{array}$

The work done by the 2nd engine,

$\begin{array}{rcl} W_{2} & = & η_{2} \times Q_{2} \\ = & 41.16 % \times (Q_{1} - W_{1}) \\ = & 41.16 % \times (7110 J - 4365.54 J) \\ = & 1129.62 J \end{array}$

Thus, the total work done is,

$\begin{array}{rcl} W & = & W_{1} + W_{2} \\ = & 4365.54 J + 1129.62 J \\ = & 5495.16 J \end{array}$

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