Question

Furnace oil has a heat of combustion of 44 MJ/kg. assuming that 70% of the heat is useful, how many kilograms of oil are required to raise the temperature of 2000 kg of water from 20 degree Celsius to 99 degree celsius

Answer 1

\[0.7 *44*10^6=\frac{C_w\Delta T m_w}{m_o}\]
\[m_o=\frac{ 4189,9*79*2000 }{0,7*44*10^6 }\]

Answer 2

combustion is the energy per mass of oil. We know that 70% of that is useful so we multiply it by 0,7. The heat is C_w * mass of water * increase of temparature, and we have to divide that by mass of an oil to get the same unit as on left side.

Answer 3

70%=0,7

Answer 4

how did you convert MJ to J?

Answer 5

MJ=10^6 J

Answer 6

oh.. can you please explain how did you come up with the formula..

Answer 7

combustion=heat produced \ mass of oil

Answer 8

thanks. i get it now
:)

Answer 9

yw