# 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.

Deepak Patra
askIITians Faculty 471 Points
8 years ago

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=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
2 years 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