The efficiency of a Carnot's engine, working between steam point and ice point, will be :

CONCEPT:

The carnot efficiency is proportional to the ratio of temperature of the sink to the temperature of source.

\(η = 1 - \frac{T_2}{T_1}\)

Here we have \(η\) is the efficiency , T₂ is the temperature of sink and T₁ is the temperature of source.

CALCULATION:

Now as we know that the efficiency of the Carnot engine we have;

\(η = 1 - \frac{T_2}{T_1}\) -----(1)

and percentage form of efficiency is written as;

 η % = \(1 - \frac{T_2}{T_1}× 100\)

The temperature of the source of steam point, T₁ = 100^o C

= 273 + 100 = 373 K

and the temperature of the sink of the ice, T₂ =  0^o = 0 + 273 K = 273 K

Now, from the equation (1)  putting the given values we have;

\(η = 1 - \frac{273}{373}\)

⇒ ​\(η = \frac{100}{373}\)​

⇒ η  = 0.2681

The percentage of efficiency is written as;

η % = 0.2681 ×  100

⇒ η % = 26.81 %

Hence option 1) is the correct answer.