Right answer is (d) 6.15 KW

For explanation I would say: Heat value = quantity * calorific value

= 0.5 * 10550 (calorific value of high speed diesel = 10550 kcal/kg)

1 calorie = 4.2 joules

1 joule/sec = 1 watt

5275 kcal/h = 5275 * 4.2 k joules/h

= \(\frac{5275*4.2*1000}{60*60}\) joules/sec = \(\frac{5275*4.2*1000}{60*60} \)watts

= \(\frac{5275*4.2*1000}{60*60*1000} \) KW

Power of engine = 6.15 KW