Energy required for boiling of 12.5L of water:
1. \ Energy required to increase it's temperature from 23.4°C→100°C
2. \ Energy required for boiling(Latent heat of evaporation)
=mC(T2−T1)+mLv= 12.5×4.184×(100−23.4)+12.5×2256= 32206.18KJ
Let V be the volume of methane required,
Mass of methane will be=d×V=0.66Vg (where d is the density)
Calorific value of methane=55KJ/g
15.6 % of Total energy of methane = 32206.18KJ
⟹10015.6×55×0.66×V=32206.18
V=5687.32L
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