Chapter 15, Problem 127A

Interpretation Introduction

Interpretation:

The mass of canola oil required to produce the given amount of energy is to be calculated.

Concept introduction:

The enthalpy of combustion gives the energy produced by burning unit mass of the substance.

Answer to Problem 127A

1.363 grams of canola oil will be required to provide the energy to vaporize the given amount of water.

Explanation of Solution

$\begin{array}{l}\text{Enthalpy of vaporization of water }=40.7kJ/mol\hfill \\ \text{Number of moles in }25\text{g of water}=\frac{25}{18(molarmass)}=1.38mol\hfill \\ \text{Energy required to vaporize }1.38\text{mol of water}=1.38\times 40.7=56.52kJ\hfill \\ \text{Energy produced by 1g of canola oil}=41.45kJ\hfill \\ \text{Mass of canola oil required }=\frac{56.52}{41.45}=1.363g\hfill \end{array}$

Conclusion

The amount of canola oil required was calculated by using the given energy data and the vaporization enthalpy of water.