Answer:
2.104 L fuel
Explanation:
Given that:
Volume of water = 35 L = 35 × 10³ mL
initial temperature of water = 25.0 ° C
The amount of heat needed to boil water at this temperature can be calculated by using the formula:
[tex]q_{boiling} = mc \Delta T[/tex]
where
specific heat of water c= 4.18 J/g° C
[tex]q_{boiling} = 35 \times 10^{3} \times \dfrac{1.00 \ g}{1 \ mL} \times 4.18 \ J/g^0 C \times (100 - 25)^0 C[/tex]
[tex]q_{boiling} = 10.9725 \times 10^6 \ J[/tex]
Also; Assume that the fuel has an average formula of C7 H16 and 15% of the heat generated from combustion goes to heat the water;
thus the heat of combustion can be determined via the expression
[tex]q_{combustion} =- \dfrac{q_{boiling}}{0.15}[/tex]
[tex]q_{combustion} =- \dfrac{10.9725 \times 10^6 J}{0.15}[/tex]
[tex]q_{combustion} = -7.315 \times 10^{7} \ J[/tex]
[tex]q_{combustion} = -7.315 \times 10^{4} \ kJ[/tex]
For heptane; the equation for its combustion reaction can be written as:
[tex]C_7H_{16} + 11O_{2(g)} -----> 7CO_{2(g)}+ 8H_2O_{(g)}[/tex]
The standard enthalpies of the products and the reactants are:
[tex]\Delta H _f \ CO_{2(g)} = -393.5 kJ/mol[/tex]
[tex]\Delta H _f \ H_2O_{(g)} = -242 kJ/mol[/tex]
[tex]\Delta H _f \ C_7H_{16 }_{(g)} = -224.4 kJ/mol[/tex]
[tex]\Delta H _f \ O_{2{(g)}} = 0 kJ/mol[/tex]
Therefore; the standard enthalpy for this combustion reaction is:
[tex]\Delta H ^0= \sum n_p\Delta H^0_{f(products)}- \sum n_r\Delta H^0_{f(reactants)}[/tex]
[tex]\Delta H^0 =( 7 \ mol ( -393.5 \ kJ/mol) + 8 \ mol (-242 \ kJ/mol) -1 \ mol( -224.4 \ kJ/mol) - 11 \ mol (0 \ kJ/mol))[/tex]
[tex]\Delta H^0 = (-2754.5 \ \ kJ - 1936 \ \ kJ+224.4 \ \ kJ+0 \ \ kJ)[/tex]
[tex]\Delta H^0 = -4466.1 \ kJ[/tex]
This simply implies that the amount of heat released from 1 mol of C7H16 = 4466.1 kJ
However the number of moles of fuel required to burn [tex]7.315 \times 10^{4} \ kJ[/tex] heat released is:
[tex]n_{fuel} = \dfrac{q}{\Delta \ H^0}[/tex]
[tex]n_{fuel} = \dfrac{-7.315 \times 10^{4} \ kJ}{-4466.1 \ kJ}[/tex]
[tex]n_{fuel} = 16.38 \ mol \ of \ C_7 H_{16[/tex]
Since number of moles = mass/molar mass
The mass of the fuel is:
[tex]m_{fuel } = 16.38 mol \times 100.198 \ g/mol}[/tex]
[tex]m_{fuel } = 1.641 \times 10^{3} \ g[/tex]
Given that the density of the fuel is = 0.78 g/mL
and we know that :
density = mass/volume
therefore making volume the subject of the formula in order to determine the volume of the fuel ; we have
volume of the fuel = mass of the fuel / density of the fuel
volume of the fuel = [tex]\dfrac{1.641 \times 10^3 \ g }{0.78 g/mL} \times \dfrac{L}{10^3 \ mL}[/tex]
volume of the fuel = 2.104 L fuel
