Practice Problem 16.2 In this example we will examine the heat flow and efficien

Question

Practice Problem 16.2 In this example we will examine the heat flow and efficiency of an ideal Carnot engine. The engine takes 2000 J of heat from a reservoir at 500 K, does some work, and discards some heat to a reservoir at 350 K. How much work does it do, how much heat is discarded, and what is its efficiency? SOLUTION SET UP AND SOLVE From the equation, QCQH=TCTH, the heat QC discarded by the engine is QC==QHTCTH=(2000J)350K500K1400J Then, from the first law, the work W done by the engine is W==QH+QC=2000J+(1400J)600J From the equation below, the thermal efficiency is eCarnot=1TCTH=1350K500K=0.30=30% Alternatively, from the basic definition of thermal efficiency, e=WQH=600J2000J=0.30=30% REFLECT The efficiency of any engine is always less than 1. For the Carnot engine, the greater the ratio of TH to TC, the greater is the efficiency. Part A - Practice Problem: What temperature TH would the high-temperature reservoir need to have in order to increase the efficiency to e= 0.55? Express your answer in kelvins as an integer.

Explanation / Answer

e = 1-Tc/Th

Tc = 350 K

e = 0.55

0.55 = 1- Tc/Th

Tc/Th = 1-0.55

Th = 777.778 K

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