Chapter 5, Problem 110SCQ

Peanuts and peanut oil are organic materials and bum in air. How many burning peanuts does it take to provide the energy to boil a cup of water (250 mL of water)? To solve this problem, we assume each peanut, with an average mass of 0.73 g, is 49% peanut oil and 21% starch; the remainder is noncombustible We further assume peanut oil is palmitic acid, C₁₆H₃₂O₂, with an enthalpy of formation of −848.4 kJ/mol. Starch is a long chain of C₆H₁₀O₅ units, each unit having an enthalpy of formation of −960 kJ.

Interpretation Introduction

Interpretation:

To identify how many burning peanuts are required to boil $\text{250mL}$ of water

Concept Introduction:

Heat energy required to raise the temperature of 1g of substance by 1K.Energy gained or lost can be calculated using the below equation.

  $\text{q=C×m×ΔT}$

Where, q= energy gained or lost for a given mass of substance (m), C =specific heat capacity,$\Delta T$= change in temperature.

The enthalpy of combustion can be calculated as

                                               ${\text{Δ}}_{\text{c}}{\text{H}}^{\text{0}}$${\text{ΣnΔ}}_{\text{f}}{\text{H}}^{\text{0}}{\text{(reactants)-ΣnΔ}}_{\text{f}}{\text{H}}^{\text{0}}\text{(products)}$

Answer to Problem 110SCQ

The number of peanuts required is $\text{41}\text{.6}$ peanuts

Explanation of Solution

Given $\text{49\%}$ of the peanut is palmitic acid.

  $\text{0}\text{.73g}\text{×}\frac{\text{49}}{\text{100}}$ = $\text{0}\text{.3577g}$ of palmitic acid.

Given $\text{21\%}$ of the starch is palmitic acid.

  $\text{0}\text{.73g}\text{×}\frac{\text{21}}{\text{100}}$=$\text{0}\text{.153g}$ of starch

Molecular weight of starch=$\text{162g/mol}$

So we have $\frac{\text{0}\text{.15}}{\text{162}}$ = $\text{0}\text{.000925}$ moles of starch.

Enthalpy of combustion of palmitic acid.is calculated as,

  ${\text{C}}_{\text{16}}{\text{H}}_{\text{32}}{\text{O}}_{\text{2}}\text{+}{\text{23O}}_{\text{2}}\to {\text{16O}}_{\text{2}}\text{+16}{\text{H}}_{\text{2}}\text{O}$

Substitute in ${\text{Δ}}_{\text{c}}{\text{H}}^{\text{0}}$${\text{ΣnΔ}}_{\text{f}}{\text{H}}^{\text{0}}{\text{(reactants)-ΣnΔ}}_{\text{f}}{\text{H}}^{\text{0}}\text{(products)}$ as

  ${\text{Δ}}_{\text{c}}{\text{H}}^{\text{0}}$=$\text{(848}\text{.4}\text{+}\text{0)-(16× -393}\text{.5}\text{+}\text{16}\text{×}-\text{241}\text{.8)}$=$\text{-9315}\text{.6kJ/mol}$

Enthalpy of combustion of starch.is calculated as

  ${\text{C}}_{\text{6}}{\text{H}}_{\text{10}}{\text{O}}_{\text{5}}{\text{+6O}}_{\text{2}}\to {\text{6O}}_{\text{2}}\text{+5}{\text{H}}_{\text{2}}\text{O}$

  ${\text{Δ}}_{\text{c}}{\text{H}}^{\text{0}}$=$\text{(-960+}\text{0)-(6× -393}\text{.5}\text{+}\text{5×}\text{-241}\text{.8)}$=$\text{-2610kJ/mol}$

Since we have $\text{0}\text{.000925}$ moles of starch = $\text{0}\text{.000925}$ moles $\text{×}$ $\text{-2610kJ/mol}$

                                                                =$\text{-2}\text{.41kJ}$ per peanut

We have $\text{0}\text{.3577g}$ of palmitic acid=$\text{0}\text{.3577g}$$÷$$\text{256}\text{.4 g/mol}$ =$\text{0}\text{.0014}$ moles

Since we have $\text{0}\text{.0014}$ moles of palmitic acid =$\text{0}\text{.0014}$ moles $\text{×}$ $\text{-9315}\text{.6kJ/mol}$

                                                               =$\text{-13}\text{.04kJ}$ per peanut

Mass of $\text{250mL}$=$\text{250g of water}$

Substitute in $\text{q=C×m×ΔT}$

                     $\text{=}\text{250g×}\text{4}\text{.184J/gK×}\text{75K}$

                     $\text{q}$=$\text{78540J}$ =$\text{78kJ}$

Energy required to raise the temperature to ${\text{100}}^{\text{0}}\text{C}$ , $\text{=Heat of vaporization ×}\text{Mass}$

                                                                                    =$\text{2}\text{.25kJ/g×250g}$

                                                                                    =$\text{562}\text{.5kJ}$

The total energy =$\text{78kJ}$$+$$\text{562}\text{.5kJ}$=$\text{642}\text{.73kJ}$

The total amount of energy released per peanut=$\text{-2}\text{.41-13}\text{.04}$ =$\text{-15}\text{.45kJ}$ per peanut

Therefore, $\text{642}\text{.73/15}\text{.45}$=$\text{41}\text{.6}$ peanuts

Conclusion

The number of peanuts are required to boil $\text{250mL}$  of water was calculated.