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A Carnot engine works between temperature T 1 and T 2 . it drives a Carnot refrigerator that works between two different temperature T 3 and T 4 (see Fig. 24-21). Find the ratio |Q 3 |/|Q 1 | in terms of the four temperature.

A Carnot engine works between temperature T1 and T2 . it drives a Carnot refrigerator    that works between two different temperature T3 and T4 (see Fig. 24-21). Find the ratio |Q3|/|Q1| in terms of the four temperature.     
 

Grade:upto college level

4 Answers

Deepak Patra
askIITians Faculty 471 Points
6 years ago
233-2196_1.PNG
233-2048_1.PNG
Tummala Vamshi
26 Points
one year ago
A carnot engine that works between temperatures T1=400K and T2=150K and drives a carnot refrigerator that works between temperatures T3= 325K and T4=225K find the ratio between Q3/Q2
Anomous
14 Points
one month ago

 

The efficiency of the engine is defined by ε=W/Q1 and is shown in the text to be
ε=T1T1T2Q1W=T1T1T2
The coefficient of performance of the refrigerator is defined by K=Q4/W and is shown in the text to be
K=T3T4T4WQ4=T3T4T4
Now Q4=Q3W, so
(Q3W)/W=T4/(T3T4)
The work done by the engine is used to drive the refrigerator, so W is the same for the two. Solve the engine equation for W and substitute the resulting expression into the refrigerator equation. 
The engine equation yields W=(T1T2)Q1/T1 and the substitution yields
T3T4T4=WQ31=Q1(T1T2)Q3T11
Solving for Q3/Q1, 
we obtain
Q1Q3=(T3T4T4+1)(T1T1T2)=(T3T4T3)(T1T1T2)=1(T4/T3)1(T2/T1)
With T1=400K,T2=150K,T3=325K, and T4=225K, the ratio becomes 
Q3/Q1=2.03
Anomous
14 Points
one month ago
The efficiency of the engine is defined by ε=W/Q 
1
  and is shown in the text to be
ε= 
1
 
1
 −T 
2
 
 ⇒ 
1
 
W
 = 
1
 
1
 −T 
2
 
 
The coefficient of performance of the refrigerator is defined by K=Q 
4
 /W and is shown in the text to be
K= 
3
 −T 
4
 
4
 
 ⇒ 
W
4
 
 = 
3
 −T 
4
 
4
 
 
Now Q 
4
 =Q 
3
 −W, so
(Q 
3
 −W)/W=T 
4
 /(T 
3
 −T 
4
 )
The work done by the engine is used to drive the refrigerator, so W is the same for the two. Solve the engine equation for W and substitute the resulting expression into the refrigerator equation. 
The engine equation yields W=(T 
1
 −T 
2
 )Q 
1
 /T 
1
  and the substitution yields
3
 −T 
4
 
4
 
 = 
W
3
 
 −1= 
1
 (T 
1
 −T 
2
 )
3
 T 
1
 
 −1
Solving for Q 
3
 /Q 
1
 , 
we obtain
1
 
3
 
 =( 
3
 −T 
4
 
4
 
 +1)( 
1
 
1
 −T 
2
 
 )=( 
3
 −T 
4
 
3
 
 )( 
1
 
1
 −T 
2
 
 )= 
1−(T 
4
 /T 
3
 )
1−(T 
2
 /T 
1
 )
 
With T 
1
 =400K,T 
2
 =150K,T 
3
 =325K, and T 
4
 =225K, the ratio becomes 
3
 /Q 
1
 =2.03The efficiency of the engine is defined by ε=W/Q 
1
  and is shown in the text to be
ε= 
1
 
1
 −T 
2
 
 ⇒ 
1
 
W
 = 
1
 
1
 −T 
2
 
 
The coefficient of performance of the refrigerator is defined by K=Q 
4
 /W and is shown in the text to be
K= 
3
 −T 
4
 
4
 
 ⇒ 
W
4
 
 = 
3
 −T 
4
 
4
 
 
Now Q 
4
 =Q 
3
 −W, so
(Q 
3
 −W)/W=T 
4
 /(T 
3
 −T 
4
 )
The work done by the engine is used to drive the refrigerator, so W is the same for the two. Solve the engine equation for W and substitute the resulting expression into the refrigerator equation. 
The engine equation yields W=(T 
1
 −T 
2
 )Q 
1
 /T 
1
  and the substitution yields
3
 −T 
4
 
4
 
 = 
W
3
 
 −1= 
1
 (T 
1
 −T 
2
 )
3
 T 
1
 
 −1
Solving for Q 
3
 /Q 
1
 , 
we obtain
1
 
3
 
 =( 
3
 −T 
4
 
4
 
 +1)( 
1
 
1
 −T 
2
 
 )=( 
3
 −T 
4
 
3
 
 )( 
1
 
1
 −T 
2
 
 )= 
1−(T 
4
 /T 
3
 )
1−(T 
2
 /T 
1
 )
 
With T 
1
 =400K,T 
2
 =150K,T 
3
 =325K, and T 
4
 =225K, the ratio becomes 
3
 /Q 
1
 =2.03The efficiency of the engine is defined by ε=W/Q 
1
  and is shown in the text to be
ε= 
1
 
1
 −T 
2
 
 ⇒ 
1
 
W
 = 
1
 
1
 −T 
2
 
 
The coefficient of performance of the refrigerator is defined by K=Q 
4
 /W and is shown in the text to be
K= 
3
 −T 
4
 
4
 
 ⇒ 
W
4
 
 = 
3
 −T 
4
 
4
 
 
Now Q 
4
 =Q 
3
 −W, so
(Q 
3
 −W)/W=T 
4
 /(T 
3
 −T 
4
 )
The work done by the engine is used to drive the refrigerator, so W is the same for the two. Solve the engine equation for W and substitute the resulting expression into the refrigerator equation. 
The engine equation yields W=(T 
1
 −T 
2
 )Q 
1
 /T 
1
  and the substitution yields
3
 −T 
4
 
4
 
 = 
W
3
 
 −1= 
1
 (T 
1
 −T 
2
 )
3
 T 
1
 
 −1
Solving for Q 
3
 /Q 
1
 , 
we obtain
1
 
3
 
 =( 
3
 −T 
4
 
4
 
 +1)( 
1
 
1
 −T 
2
 
 )=( 
3
 −T 
4
 
3
 
 )( 
1
 
1
 −T 
2
 
 )= 
1−(T 
4
 /T 
3
 )
1−(T 
2
 /T 
1
 )
 
With T 
1
 =400K,T 
2
 =150K,T 
3
 =325K, and T 
4
 =225K, the ratio becomes 
3
 /Q 
1
 =2.03

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