Answer
The cook will need $7.791 grams$ of the gas to heat the water.
Work Step by Step
Using the equation $q=mcΔT$, we can solve for the heat. The specific heat capacity of water is $4.184\frac{J}{g\times^{\circ}C}$ and the mass of the water in this case is 1.35 kg or 1350 grams. Also the change in temperature of the water can be calculated by taking the final temperature and subtracting the initial temperature; $100^{\circ}C - 32^{\circ}C = 68^{\circ}C$
$q=(1350 grams×4.184\frac{J}{g\times^{\circ}C}×68^{\circ}C)=384091.2 J$ Now we must convert the units of joules into kilojoules by simply dividing by 1000; $384091.2 J J\div 1000 = 384.0912 kJ$
Now to find the mass of the gas we can divide the heat required for the water to increase in temperature by the heat of the gas/unit mass;
$384.0912 kJ\times\frac{1 g}{49.3 kJ} = 7.791 grams$
