Question

Two heat engines A and B are operating at 700 K,
300 K and 800 K, 400 K respectively. If na and me
are their efficiencies then
(1) nA = nB
(2) NA = 2 MB
(4) NA = 1 1 1 1 3
(3) NA = 18​

Answer 1

The required option is (4). $n_{A}$ = 1.113 ×  Efficiency. $n_{A}$.

Explanation:

Two heat engines A and B are operating at 700 K,

300 K.

Efficiency, $n_{A}$ = ( 1 - $\dfrac{T_{2} }{T_{1}}$)

= ( 1 - $\dfrac{300}{700}$)

= 1 - 0.4285

=  0.571

Efficiency, $n_{B}$ = ( 1 - $\dfrac{T_{2} }{T_{1}}$)

= ( 1 - $\dfrac{800}{400}$)

= 1 - 0.5

=  0.5

∴ Efficiency, $n_{A}$ = 1.113 ×  Efficiency. $n_{A}$

Hence, the required option is (4). $n_{A}$ = 1.113 ×  Efficiency. $n_{A}$.