Correct Answer - Option 2 : 6880 kcal/kg

**Concept:**

**Calorific values of fuel: **The calorific value of fuel is** the quantity of heat produced by its combustion **at constant pressure and under normal conditions. Calorific value mention in **kcal / kg.**

**The overall efficiency of the steam power station:** The overall efficiency of the steam power station is defined as t**he ratio of the power available at the generator terminal to the rate of energy released by the** **combustion of fuel. **It is given by,

\({\eta _{overall}} = \frac{{EO}}{{HC}}\)

Where EO is the Electrical output in the heat unit.

HC is Heat combustion.

**Calculation:**

**Given:** Overall efficiency = 25 % , 0.5 kg coal burnt.

Let x cal /kg be the calorific value of the fuel.

Heat equivalent of 1 kWh = 860 kcal

\({\eta _{overall}} = \frac{{EO}}{{HC}}\)

\(0.25 = \frac{{860}}{{0.5x}}\)

\(x = \frac{{860000}}{{25 \times 5}}\)

\(x = 6880\;kcal/kg\)