Question

a. A cylinder of gas is assumed to contain $11.2kg$ 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 $2658kJmo{l}^{-1}$.
b. If the air supply of the burner is insufficient (i.e. you have a yellow instead of blue flame), a portion of the gas escapes without combustion. Assuming that $30\%$ of the gas is wasted due to this inefficiency, how long would the cylinder last?

• A. a. $26$ days , b. $18$ days
• B. a. $17$ days , b. $26$ days
• C. a. $16$ days , b. $17$ days
• D. None of these

Answer 1

Answer: (A)

a. $26$ days , b. $18$ days

Molecular mass of butane,

$C_4{H}_{10}=58$g $mo{l}^{-1}$

$58$g of butane on combustion give $2658$ kJ of heat

$\therefore 11.2$kg or $11.2\times {10}^3$g of butane would yield

$\frac{2658kJ\times 11.2\times {10}^{3}g}{58g}=513268.97$kJ

The daily requirement of energy $=20,000$ kJ/day

$\therefore$ Number of days the cylinder would last

$=\frac{513268.97kJ}{20000kJ/day}=26$days.

Since $30$ per cent of the gas is wasted due to incomplete combustion, the amount of heat produced per cylinder.

$=513268.97\times \frac{70}{100}kJ=359288.28$kJ

$\therefore$ Number of days the cylinder would last

$=\frac{359288.28kJ}{20000kJ/day}=18$days.