Question 16.2: A Carnot engine In this example we will examine the heat flo...
A Carnot engine
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?
SET UP AND SOLVE From Equation 16.7, the heat QC discarded by the engine is
\mathrm{\frac{Q_C}{Q_H}=-\frac{T_C}{T_H} } . (16.7)
\mathrm{Q_C=-Q_H\frac{T_C}{T_H}=-(2000 J)\frac{350 K}{500 K} }
= -1400 J.
Then, from the first law, the work W done by the engine is
\mathrm{W=Q_H+Q_C=2000 J+(-1400 J)}
= 600 J
From Equation 16.8, the thermal efficiency is
\mathrm{e_{Caront}=1-\frac{T_C}{T_H}=\frac{T_H-T_C}{T_H} } . (16.8)
\mathrm{e_{Caront}=1-\frac{T_C}{T_H}=1-\frac{350 K}{500 K} = 0.30 = 30 } %.
Alternatively, from the basic definition of thermal efficiency,
\mathrm{e=\frac{W}{Q_H}=\frac{600 J}{2000 J} = 0.30 = 30 } %.
REFLECT The efficiency of any engine is always less than 1. For the Carnot engine, the greater the ratio of \mathrm{T_H to T_C,} the greater is the efficiency.
Practice Problem: What temperature \mathrm{T_H} would the high-temperature reservoir need to have in order to increase the efficiency to e = 0.50? Answer: 700 K.