Question

The complete combustion of acetylene$\left[C_2{H}_2\left(g\right)\right]$ produces 1300 kJ of energy per mole of acetylene consumed. How many grams of acetylene must be burned to produce enough heat to raise the temperature of 1.00 gal of water by $10.0^{o}C$ if the process is $80.0\%$ efficient? Assume the density of water is $1.00g/c{m}^3$.

Short answer

Thus, the mass of acetylene that must be burned is 3.97 g.

Step-by-step solution

 Determination of mass of water

Conversion of volume of water from gal to cm³

$\begin{array}{rcl}1gal& =& 3.78541 L\\ 1L& =& 1000c{m}^3\\ Volume of water& =& 1.00gal\times 3.78541L=3.78541L\\ Volume of water& =& 3.78541L\times \frac{1000c{m}^3}{1L}=3785.41c{m}^3\\ & & \end{array}$

Mass of water is,

$\begin{array}{rcl}Mass& =& density\times volume\\ & =& 1.00g/c{m}^3\times 3785.41c{m}^3\\ & =& 3785.41g\\ & & \end{array}$

Determination of heat

Heat required to raise the temperature of water is,

$\begin{array}{rcl}Heat& =& mass\times heat capacity\times \left(increase in temperature\right)\\ & =& 3785.41g\times 4.18J/g^{o}C\times 10.0{}^{o}C\\ & =& 158230.14J\\ & =& 158.230kJ\\ & & \end{array}$

The efficiency of the process is 80%.

So, the heat required to raise the temperature is,

$Heat=\frac{158.230kJ}{80.0}\times 100=197.78kJ$

Determination of mass of acetylene burned

The combustion of 1 mol of acetylene produces 1300 kJ.

Number of moles of acetylene gives 197.78 kJ is,

$Moles=197.78kJ\times \frac{1mol}{1300kJ}=0.1523mol$

Mass of acetylene that should be burned is,

$\begin{array}{rcl}Mass& =& moles\times molar mass\\ & =& 0.1523mol\times 26.04g/mol\\ & =& 3.97g\\ & & \end{array}$

Thus, the mass of acetylene that must be burned is 3.97 g.