Question #97164
You need to boil 12.5 L of water using your natrual gas (primarily methane) stove. What volume of natural gas is needed to boil the water if only 15.6% of the heat generated goes towards heating the water. Assume the density of methane is 0.660g/L , the density of water is 1.00 g/mL , and that the water has an initial temperature of 23.4 °C
1
Expert's answer
2019-10-23T07:01:21-0400

Energy required for boiling of 12.5L12.5 L of water:

1. \ Energy required to increase it's temperature from 23.4°C100°C23.4\degree C\to100\degree C

2. \ Energy required for boiling(Latent heat of evaporation)

=mC(T2T1)+mLv==mC(T_2-T_1)+mL_v= 12.5×4.184×(10023.4)+12.5×2256=12.5 \times 4.184\times (100-23.4)+12.5\times 2256= 32206.18KJ32206.18KJ

Let VV be the volume of methane required,

Mass of methane will be=d×V=0.66Vg=d\times V=0.66V g (where d is the density)

Calorific value of methane=55KJ/g=55KJ/g

15.615.6 % of Total energy of methane == 32206.18KJ32206.18 KJ

    15.6100×55×0.66×V=32206.18\implies \frac{15.6}{100}\times 55\times 0.66\times V=32206.18

V=5687.32LV=5687.32 L


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