Question

1. propane is a commonly used barbeque fuel. determine the efficiency of the barbeque as a heating device given the following information using the steps outlined below:

a) determine the energy input by calculating \\( \delta h_{comb}^{\circ} \\) using standard enthalpies of formation and a balanced chemical equation. you will have to use your data tables for this. \\( c_{3}h_{8(g)}+5o_{2(g)} \to 4h_{2}o_{(g)}+3co_{2(g)} \\) then determine the molar enthalpy of combustion of the propane. then determine the (theoretical) energy content of the propane used.
b) determine the energy output by calculating how much energy was absorbed by the pot and the water.
c) calculate the efficiency of the barbeque.

Explanation:

Step1: Determine energy input

• Balanced equation: \(C_{3}H_{8(g)}+5O_{2(g)}\to4H_{2}O_{(l)} + 3CO_{2(g)}\)
• Enthalpy of formation values (\(\Delta H_f^{\circ}\)):
• \(\Delta H_f^{\circ}(C_{3}H_{8})=-103.8\space kJ/mol\)
• \(\Delta H_f^{\circ}(O_{2}) = 0\space kJ/mol\)
• \(\Delta H_f^{\circ}(H_{2}O)=- 285.8\space kJ/mol\)
• \(\Delta H_f^{\circ}(CO_{2})=-393.5\space kJ/mol\)
• Using \(\Delta H_{comb}^{\circ}=\sum n\Delta H_f^{\circ}(products)-\sum n\Delta H_f^{\circ}(reactants)\)

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• Mass of propane used: \(m = 32.75 - 27.96=4.79\space g\)
• Molar mass of \(C_{3}H_{8}\): \(M=(3\times12 + 8\times1)=44\space g/mol\)
• Moles of propane: \(n=\frac{4.79}{44}=0.109\space mol\)
• Energy input (\(q_{in}\)): \(q_{in}=n\times\Delta H_{comb}^{\circ}=0.109\times2220 = 242\space kJ\)

Step2: Determine energy output

• Mass of water: \(m_{water}=610.15 - 210.15 = 400\space g\)
• \(\Delta T\): \(\Delta T=63.75 - 21.50 = 42.25^{\circ}C\)
• Using \(q = mc\Delta T\) for pot and water
• \(q_{pot}=m_{pot}c_{pot}\Delta T=210.15\times0.503\times42.25\approx4490\space J\)
• \(q_{water}=m_{water}c_{water}\Delta T=400\times4.19\times42.25\approx70700\space J\)
• Energy output (\(q_{out}\)): \(q_{out}=q_{pot}+q_{water}=4490 + 70700=75200\space J = 75.2\space kJ\)

Step3: Calculate efficiency

• Efficiency formula: \(\text{efficiency}=\frac{q_{out}}{q_{in}}\times100\%\)

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Answer:

The efficiency of the barbeque is \(31.1\%\)