Question

Chapter 16 Part 2: Question 16.3

An ideal Carnot 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. Let’s see what happens if the Carnot engine described is run backward as a refrigerator.What is its performance coefficient?

SOLUTION

**SET UP AND SOLVE** We know that a refrigerator
will act like an engine but with the signs for *Q*H,
*Q*C, and *W*reversed because the cycle is run
backwards. From the equation,
*Q*C*Q*H=?*T*C*T*H, the heat
*Q*C discarded by a Carnot engine that runs between these
two temperatures is

*Q*C=?*Q*H*T*C*T*H=?(2000J)350K500K=?1400J

Then from the first law, the work *W* done by the engine
is

*W*=*Q*H+*Q*C=2000J+(?1400J)=600J

Therefore, for our refrigerator, we will use the following
values *Q*C=?1400J, *Q*H=2000J, *W*=600J, with
all the signs reversed because the cycle is run backward:

*Q*C=+1400J, *Q*H=?2000J, *W*=?600J

From the equation, *K*=*Q*C|*W*|, the
performance coefficient *K* is

*K*=*Q**C*|*W*|=1400J600J=2.33

Because the cycle is a Carnot cycle, we may also use the
equation *K*carnot=*T*C*T*H?*T*C,

*K*=*T*C*T*H?*T*C=350K500K?350K=2.33

**REFLECT** For a Carnot cycle, *e* and
*K* depend only on the temperatures, as shown by the
equations
*e*carnot=1?*T*C*T*H=*T*H?*T*C*T*H
and *K*carnot=*T*C*T*H?*T*C, and we
don't need to calculate *Q* and *W*. For cycles
containing irreversible processes, however, these equations are not
valid, and more detailed calculations are necessary.

**Part A -** **Practice Problem:**

Suppose you want to increase the performance coefficient
*K*, would it be better to decrease *T*H by 40 K or
to increase *T*C by 40 K ?

Suppose you want to increase the performance coefficient , would it be better to decrease by 40 or to increase by 40 ?

Decrease TH by 40 K . |

Increase TC by 40 K . |

Answer #1

given ideal carnot engine

Qin = 2000 J

Th = 500 K

Tc = 350 K

a. for reverse cycle

Qrej = 2000 J

Th = 500 K

Tc = 350 K

Qext = ?

now

Qext = Qc

Qrej = Qh

(Qc - Qh)/Qh = (Tc - Th)/Th

Qc/2000 - 1 = 350/500 - 1

Qc = 4*350 = 1400 J

Work done W = Qh - Qc = 600 J

K = Qc/W = 1400/600 = 7/3 = 2.33

for carnott cycle, Kcarnott = K = 2.33

A. to increase K

K = Qc/(Qh - Qc) = 1/(Qh/Qc - 1) = 1/(Th/Tc - 1)

k1 = 1/((Th - 40)/Tc - 1) = Tc/(Th - Tc - 40)

k2 = 1/(Th/(Tc + 40) - 1) = (Tc + 40)/(Th - Tc - 40)

hence

k2 > k1

hence

increasing Tc by 40 K will improve perfprmance coeficient more

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