A Carnot engine (E) is working between two temperatures 473 K and 273 K. In a new system two engines - engine works between 473 K to 373 K and engine works between 373 K to 273 K. If , and are the efficiencies of the engines E, and , respectively, th

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

Q.
A Carnot engine (E) is working between two temperatures 473 K and 273 K. In a new system two engines - engine $E_{1}$ works between 473 K to 373 K and engine $E_{2}$ works between 373 K to 273 K. If $η_{12}$ , $η_{1}$ and $η_{2}$ are the efficiencies of the engines E, $E_{1}$ and $E_{2}$ , respectively, then [2025]

1 $η_{12} < η_{1} + η_{2}$

2 $η_{12} = η_{1} η_{2}$

3 $η_{12} = η_{1} + η_{2}$

4 $η_{12} \geq η_{1} + η_{2}$

Ans.
(1)

Efficiencies of a Carnot engine $η = 1 - \frac{T_{sink}}{T_{source}}$

$\Rightarrow η_{1} = 1 - \frac{373 K}{473 K} = \frac{100}{473}$

$η_{2} = 1 - \frac{273 K}{373 K} = \frac{100}{373}$

$η_{12} = 1 - \frac{273 K}{473 K} = \frac{200}{473}$

$η_{12} - η_{1} = \frac{200}{473} - \frac{100}{473} = \frac{100}{473} < \frac{100}{373}$

$\Rightarrow η_{12} - η_{1} < η_{2}$

or $η_{12} < η_{1} + η_{2}$

Solution

1 $η_{12} < η_{1} + η_{2}$

2 $η_{12} = η_{1} η_{2}$

3 $η_{12} = η_{1} + η_{2}$

4 $η_{12} \geq η_{1} + η_{2}$

Want Us to Email you About Special Offers & Updates?

Solution

Q. A Carnot engine (E) is working between two temperatures 473 K and 273 K. In a new system two engines - engine E1 works between 473 K to 373 K and engine E2 works between 373 K to 273 K. If η12, η1 and η2 are the efficiencies of the engines E, E1 and E2, respectively, then [2025]

1 η12<η1+η2

2 η12=η1η2

3 η12=η1+η2

4 η12≥η1+η2

Ans. (1) Efficiencies of a Carnot engine η=1–TsinkTsource ⇒ η1=1–373 K473 K=100473 η2=1–273 K373 K=100373 η12=1–273 K473 K=200473 η12–η1=200473–100473=100473<100373 ⇒ η12–η1<η2 or η12<η1+η2

Want Us to Email you About Special Offers & Updates?

1 $η_{12} < η_{1} + η_{2}$

2 $η_{12} = η_{1} η_{2}$

3 $η_{12} = η_{1} + η_{2}$

4 $η_{12} \geq η_{1} + η_{2}$