Question

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?

Answer 1

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`.