InterviewSolution
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InterviewSolution
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The thermochemical equation for the combustion of ethylene gas, `C_(2)H_(4)`, is `C_(2)H_(4)(g) rarr 2CO_(2)(g) +2H_(2)O(l), DeltaH^(Theta) =- 337 kcal` Assuming `70%` efficiency, calculate the weight of water at `20^(@)C` that can be converted into system at `100^(@)C` by buring `1m^(3)` of `C_(2)H_(4)` gas measured at `STP`. The heat of vaporisation of water at `20^(@)C` and `100^(@)C` are `1.00 kcal kg^(-1)` and `540 kcal kg^(-1)` respectively. |
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Answer» Number of moles in `1m^(3)` of ethylene `= 44.6` mol ltbr `DeltaH` for `1m^(3)` of ethylene `(44.6 mol` of tthylene) `= n(C_(2)H_(4)) xx DeltaH (1 "mole")` `=- 1.50 xx 104 kcal` Therefore, the useful heat `= 1.05 xx 10^(4) cal` For the overall procee, consider two stages: `H_(2)O(l) 20^(@)C rarr H_(2)O(g) 100^(@)C`, `DeltaH = C_(P) DeltaT = (1.00 kcal kg^(-1),K) (80K) = 80 kcal kg^(-1)` `H_(2)O(l) 100^(@)C rarr H_(2)O(g) 100^(@)C, DeltaH = 540 kcal kg^(-1)` `:. Delta_("total")H = 620 kcal kg^(-1)` Therefore, weight of water converted into steam `=("Amount of heat avaiable")/("Heat required/kg") = (1.05 xx 10^(4))/(620) = 16.9 kg` |
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