Answer:
Step-by-step explanation:
Heat released by one litre of oil = 2.8 x 10⁷ J .
Let m gm be the mass of water required .
heat absorbed in increasing the temperature to 100 degree
= mass x specific heat x rise in temp
= m x 4.2 x ( 100 - 20 ) = 336 m
heat required in boiling of all the water
= mass x latent heat of evaporation
= m x 2268 = 2268 m
heat required to raise the temperature of steam from 100 to 300 degree
= mass x specific heat of steam x rise in temp
= m x 2.03 x 200
= 406 m
Total heat absorbed = 3010 m J
So 3010 m J = 2.8 x 10⁷ J
m = 9.30 x 10³ g
density of water = 10³ g per litre
volume of water in litre = 9.3 x 10³ / 10³
= 9.3 litre .