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InterviewSolution
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a. A cylinder of gas is assumed to contain `11.2 kg` of butane. If a normal family needs `20000kJ` of energy per day for cooking, how long will the cylinder last? Given that the heat of combustion of butane is `2658 kJ mol^(-1)`. b. If the air supply of the burner is insufficient (i.e. you have a yellow instead of a blue flame), a portion of the gas escape without combustion. Assuming that `33%` of the gas is wasted due to this inefficiency, how long would the cylinder last? |
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Answer» a. `1mol` of butane, i.e., `C_(4)H_(10) (= 58g)` gives heat `= 2658 kJ` `:. 58g of C_(4)H_(10)` gives heat `=2658 kJ` `11.2 xx 1000 g` gives `rArr (2658 xx 11.2 xx 1000)/(58) rArr 513268.96 kJ` `:. 20000 kJ `of heat is required for `1` day. `:. 513268.96 kJ` of heat is required for `= (513268.96)/(20000) = 25.66 days` `~~ 26 days` b. `33%` of heat is wasted, therefore, `67%` of heat is utilised. `:.` Heat utilised `= (513268.96xx67)/(100) = 343890 kJ` Number of days `= (343890)/(20000) = 17.19 days` `~~ 17 days`. |
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