An I.C. engine consumes high speed diesel oil at the rate of 0.5 kg/h. What is the power of the engine?
a) 6.51 KW
b) 6.48 KW
c) 6.10 KW
d) 6.15 KW

Answer: d
Explanation: 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