91Ó°ÊÓ

\(q = 1000 \times 4.18 \times 65 = 271700 \, \mathrm{J}\) #tag_title#Step 2: Determine the amount of methane needed to produce the required heat#tag_content#The combustion of methane can be represented by the balanced chemical equation: \(\mathrm{CH}_{4}(g) + 2\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g) + 2\mathrm{H}_{2}\mathrm{O}(l)\) The heat of combustion of methane is -890.8 kJ/mol, meaning that 890.8 kJ of heat is released when 1 mol of methane is combusted. First, we need to convert the heat required to kilojoules: \(271700 \, \mathrm{J} = 271.7 \, \mathrm{kJ}\) Now we can determine the amount of methane needed by setting up a proportion: \(\frac{890.8 \, \mathrm{kJ}}{1 \, \mathrm{mol} \, \mathrm{CH}_{4}} = \frac{271.7 \, \mathrm{kJ}}{x \, \mathrm{mol} \, \mathrm{CH}_{4}}\) Solving for x, we get: \(x = \frac{271.7 \, \mathrm{kJ} \times 1 \, \mathrm{mol} \, \mathrm{CH}_{4}}{890.8 \, \mathrm{kJ}} = 0.305 \, \mathrm{mol} \, \mathrm{CH}_{4}\) Finally, we convert the moles of methane to grams by using the molar mass of methane (16.04 g/mol): \(0.305 \, \mathrm{mol} \, \mathrm{CH}_{4} \times \frac{16.04 \, \mathrm{g}}{1 \, \mathrm{mol} \, \mathrm{CH}_{4}} = 4.89 \, \mathrm{g} \, \mathrm{CH}_{4}\) So, 4.89 grams of methane must be combusted to heat 1.00 kg of water from 25.0 °C to 90.0 °C, assuming 100 % efficiency in heat transfer.