Question

< Question 20 of 20 > Suppose a boil water notice is sent out advising all residents in the area to boil their water before drinking it or using it for cooking. You need to boil 16.5 L of water using your natrual gas (primarily methane) stove. What volume of natural gas is needed to boil the water if only 17.7% 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 24.1 °C. Enthalpy of formation values can be found in this table. Assume that gaseous water is formed in the combustion of methane. volume of methane: Expand

Answer 1

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Volume of water required to boil,$V= 16.5 \;L=16500 \;mL$

Therefore, mass of water $m=volume\times density=V\times D=16500 \;mL\times1\;g/mL=16500\;g$

Hence, heat required to boil water is given by,

$H=mC(T_{f}-T_{i})=16500\;g\times 4.184 J/g.^{o}C\times(100^{o}C-24.1^{o}C)$

$\therefore H=5239832.4\;J$

We know that, $1\;J=10^{-3}\;kJ$

$\therefore H=5239.8324\;kJ$

Heat used by water from combustion,

$\therefore H'=\frac{H}{\%}=\frac{5239.8324\;kJ}{17.7\%}$

$\therefore H'=29603.57288\;kJ$

We have, enthalpy of combustion of methane, $\Delta H=802.5 \;kJ/mol$

Hence, no.of moles of methane combusted is given by,

$n=\frac{H'}{\Delta H}=\frac{29603.57288\;kJ}{802.5\;kJ}=36.889\;moles$

Hence, mass of methane combusted,

$m'=n\times molar \; mass=36.889\;moles\times 16.04\;g/mol$

$\therefore m'=591.702\;g$

Therefore volume of methane used is given by,

$Volume=\frac{mass}{density}=\frac{591.702\;g}{0.660\;g/L}$

$\mathbf{\therefore Volume=896.52\;L}$

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