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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
8 years ago
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233-2048_1.PNG
Tummala Vamshi
26 Points
3 years 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
2 years ago

 

The efficiency of the engine is defined by ε=W/Q1​ and is shown in the text to be
ε=T1​T1​−T2​​⇒Q1​W​=T1​T1​−T2​​
The coefficient of performance of the refrigerator is defined by K=Q4​/W and is shown in the text to be
K=T3​−T4​T4​​⇒WQ4​​=T3​−T4​T4​​
Now Q4​=Q3​−W, so
(Q3​−W)/W=T4​/(T3​−T4​)
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=(T1​−T2​)Q1​/T1​ and the substitution yields
T3​−T4​T4​​=WQ3​​−1=Q1​(T1​−T2​)Q3​T1​​−1
Solving for Q3​/Q1​, 
we obtain
Q1​Q3​​=(T3​−T4​T4​​+1)(T1​T1​−T2​​)=(T3​−T4​T3​​)(T1​T1​−T2​​)=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
2 years ago
The efficiency of the engine is defined by ε=W/Q 
1
​
  and is shown in the text to be
ε= 
T 
1
​
 
T 
1
​
 −T 
2
​
 
​
 ⇒ 
Q 
1
​
 
W
​
 = 
T 
1
​
 
T 
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= 
T 
3
​
 −T 
4
​
 
T 
4
​
 
​
 ⇒ 
W
Q 
4
​
 
​
 = 
T 
3
​
 −T 
4
​
 
T 
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
T 
3
​
 −T 
4
​
 
T 
4
​
 
​
 = 
W
Q 
3
​
 
​
 −1= 
Q 
1
​
 (T 
1
​
 −T 
2
​
 )
Q 
3
​
 T 
1
​
 
​
 −1
Solving for Q 
3
​
 /Q 
1
​
 , 
we obtain
Q 
1
​
 
Q 
3
​
 
​
 =( 
T 
3
​
 −T 
4
​
 
T 
4
​
 
​
 +1)( 
T 
1
​
 
T 
1
​
 −T 
2
​
 
​
 )=( 
T 
3
​
 −T 
4
​
 
T 
3
​
 
​
 )( 
T 
1
​
 
T 
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 
Q 
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= 
T 
1
​
 
T 
1
​
 −T 
2
​
 
​
 ⇒ 
Q 
1
​
 
W
​
 = 
T 
1
​
 
T 
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= 
T 
3
​
 −T 
4
​
 
T 
4
​
 
​
 ⇒ 
W
Q 
4
​
 
​
 = 
T 
3
​
 −T 
4
​
 
T 
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
T 
3
​
 −T 
4
​
 
T 
4
​
 
​
 = 
W
Q 
3
​
 
​
 −1= 
Q 
1
​
 (T 
1
​
 −T 
2
​
 )
Q 
3
​
 T 
1
​
 
​
 −1
Solving for Q 
3
​
 /Q 
1
​
 , 
we obtain
Q 
1
​
 
Q 
3
​
 
​
 =( 
T 
3
​
 −T 
4
​
 
T 
4
​
 
​
 +1)( 
T 
1
​
 
T 
1
​
 −T 
2
​
 
​
 )=( 
T 
3
​
 −T 
4
​
 
T 
3
​
 
​
 )( 
T 
1
​
 
T 
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 
Q 
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= 
T 
1
​
 
T 
1
​
 −T 
2
​
 
​
 ⇒ 
Q 
1
​
 
W
​
 = 
T 
1
​
 
T 
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= 
T 
3
​
 −T 
4
​
 
T 
4
​
 
​
 ⇒ 
W
Q 
4
​
 
​
 = 
T 
3
​
 −T 
4
​
 
T 
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
T 
3
​
 −T 
4
​
 
T 
4
​
 
​
 = 
W
Q 
3
​
 
​
 −1= 
Q 
1
​
 (T 
1
​
 −T 
2
​
 )
Q 
3
​
 T 
1
​
 
​
 −1
Solving for Q 
3
​
 /Q 
1
​
 , 
we obtain
Q 
1
​
 
Q 
3
​
 
​
 =( 
T 
3
​
 −T 
4
​
 
T 
4
​
 
​
 +1)( 
T 
1
​
 
T 
1
​
 −T 
2
​
 
​
 )=( 
T 
3
​
 −T 
4
​
 
T 
3
​
 
​
 )( 
T 
1
​
 
T 
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 
Q 
3
​
 /Q 
1
​
 =2.03

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