Here, mass of water heated,

m=3000 g min−1 Volume of water heated

=3.0 lit min

−1Rise in temperature

ΔT=(77−27)°C=50°CSpecific heat of water

c=4.2Jg−1K−1Amount of heat used

ΔQ=mcΔTNow,

=cΔT⇒=×4.2×50;=63000Jmin−1Heat of consumption

(ΔQ)=ΔmL=.L⇒630000 J min−1= ( )×4×104Jg−1Rate of consumption of fuel

==15.75 ≈16g min−1