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_{(l)}+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: Calculate mass of propane used

Mass of propane \(m = 32.75 - 27.96=4.79\space g\)

Step2: Calculate mass of water

Mass of water \(m_{water}=610.15 - 210.15 = 400\space g\)

Step3: Calculate energy absorbed by pot and water

Energy absorbed \(Q = Q_{pot}+Q_{water}\)
\(Q_{pot}=m_{pot}c_{pot}\Delta T\), \(\Delta T=63.75 - 21.50 = 42.25^{\circ}C\), \(m_{pot} = 210.15\space g\), \(c_{pot}=0.503\space J/g\cdot^{\circ}C\)
\(Q_{pot}=210.15\times0.503\times42.25\)
\(Q_{water}=m_{water}c_{water}\Delta T\), \(m_{water} = 400\space g\), \(c_{water}=4.19\space J/g\cdot^{\circ}C\)
\(Q_{water}=400\times4.19\times42.25\)
\(Q=(210.15\times0.503 + 400\times4.19)\times42.25\)
\(=(105.70545+1676)\times42.25\)
\(=1781.70545\times42.25\)
\(Q = 75277.15\space J\)

Step4: Calculate molar enthalpy of combustion of propane

From standard enthalpy of formation data (assuming \(\Delta H_f^{\circ}(C_3H_8)=-103.8\space kJ/mol\), \(\Delta H_f^{\circ}(CO_2)=-393.5\space kJ/mol\), \(\Delta H_f^{\circ}(H_2O)=-285.8\space kJ/mol\), \(\Delta H_f^{\circ}(O_2) = 0\space kJ/mol\))
\(\Delta H_{comb}^{\circ}=\sum n\Delta H_f^{\circ}(products)-\sum m\Delta H_f^{\circ}(reactants)\)
\(\Delta H_{comb}^{\circ}=(4\times(- 285.8)+3\times(-393.5))-(-103.8)\)
\(\Delta H_{comb}^{\circ}=(-1143.2-1180.5)+103.8\)
\(\Delta H_{comb}^{\circ}=-2220\space kJ/mol\)
Molar mass of \(C_3H_8=(3\times12 + 8\times1)=44\space g/mol\)
Moles of propane \(n=\frac{4.79}{44}\space mol\)
Theoretical energy content \(Q_{input}=n\times\vert\Delta H_{comb}^{\circ}\vert=\frac{4.79}{44}\times2220\times1000\space J\)
\(Q_{input}=\frac{4.79\times2220\times1000}{44}\space J\approx241931.82\space J\)

Step5: Calculate efficiency

Efficiency \(\eta=\frac{Q_{output}}{Q_{input}}\times100\%\)
\(\eta=\frac{75277.15}{241931.82}\times100\%\approx31.1\%\)

Answer:

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