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

26 days approx, 17 days approx.

Explanation

(a) 1 Mole `C_(4)H_(10) = 58 g ` `:. ` Heat produced from `11200 g = ( 2658) /( 58) xx 11200 = 513268.9kJ`
No. of days which it will last ` = 513268.9 // 20,000= 25.7 `days `= 26 ` days
(b) After wastage, heat available `= ( 67)/( 100) xx 513268.9 = 343890J`
No. of days for which it will last `= 343890 // 20000 = 17` days