Answer:
Efficiency = 0.273 = 27.3%
Step-by-step explanation:
The ideal energy that can be produced by certain mass of diesel fuel is given as follows:
![E_(ideal) = (HV)(m)](https://img.qammunity.org/2021/formulas/physics/high-school/muqymia8riy4zdk6b4zkrdkft7zjr23wu8.png)
where,
HV = Heating Value of Diesel = 42 - 46 MJ = 44 MJ (Taking average value)
m = mass of diesel fuel = 0.3 kg
E(ideal) = Ideal energy produced by this mass of fuel = ?
Therefore,
![E_(ideal) = (44\ MJ/kg)(0.3\ kg)\\E_(ideal) = 13.2\ MJ\\](https://img.qammunity.org/2021/formulas/physics/high-school/v8e5j0hvajayzrmyxadfn35v783jr1voa1.png)
So, the efficiency can be given as:
![Efficiency = (E_(actual))/(E_(ideal))\\\\Efficiency = (3.6\ MJ)/(13.2\ MJ)](https://img.qammunity.org/2021/formulas/physics/high-school/2cridpdl1qilip9ezkddsl5jxfy7pq4bhz.png)
Efficiency = 0.273 = 27.3%