If a carnot engine is working between steam point and ice point, then its efficiency will be:
A. 24.9%
B. 25.7%
C. 26.8%
D. 28.8%

Hint: The given problem is based on thermodynamic cycles. A carnot cycle is an ideal cycle which efficiency is maximum for given specific temperature limits. Carnot cycle efficiency is proportional to the ration of sink temperature to source temperature.

Complete step by step answer:
As we know that the carnot cycle is an ideal cycle which consists of four reversible processes. The loss of energy will be minimum in the case of reversible processes. So a carnot cycle has maximum efficiency between given two specific temperature limits.
Also we know that the carnot efficiency is proportional to the ratio of temperature of sink to temperature of source. The carnot efficiency is given by
$\eta = 1 - \dfrac{{{T_2}}}{{{T_1}}}$
Also it can be represented in percentage form as
$\eta \% = \left( {1 - \dfrac{{{T_2}}}{{{T_1}}}} \right) \times 100$
Here, ${T_2}$= Temperature of sink
${T_1}$= Temperature of source

The source temperature will be always greater than sink temperature in carnot operations. So the given data in question is
 ${T_1}$= Steam Point= 100℃= 373K
${T_2}$= Ice point= 0℃= 273K
As we know that the steam point and ice point for water are ${100^ \circ }C$ and ${0^ \circ }C$ respectively.

To calculate the efficiency of cycle, we will put these values in efficiency formula as
$
  \eta \% = \left( {1 - \dfrac{{273}}{{373}}} \right) \times 100 \\
   \Rightarrow \eta \% = (1 - 0.731) \times 100 \\
  \therefore \eta \% = 26.8\% \\
 $
This is the maximum efficiency that a thermodynamic engine has for these temperature limits.
So for given values, the carnot efficiency is equal to the $26.8\% $.

Hence, the correct answer is option (C).

Note: The maximum efficiency of the carnot cycle denotes that the maximum extracted heat from the source is converted into mechanical work, Which improves the efficiency of the thermodynamic engine.