Question

The African bombardier beetle (Stenotomus insignis) can emit a jet ofdefensive spray from the movable tip of its abdomen (Fig. P17.91). The beetle’s body has reservoirs containing two chemicals; when the beetle is disturbed, these chemicals combine in a reaction chamber, producing a compound that is warmed from 20ºC to 100ºC by the heat of reaction. The high pressure produced allows the compound to be sprayed out at speeds up to 19 m/s (68 km/h), scaring away predators of all kinds. (The beetle shown in Fig. 1 below is 2 cm long.) Calculate the heat of reaction of the two chemicals (in J/kg). Assume that the specific heat of the chemicals and of the spray is the same as that of water, $\mathbf{4}\mathbf{.}\mathbf{19}\mathbf{\times }\mathbf{10}\mathbf{^}\mathbf{3}\mathbf{\hspace{0.33em}}\mathbf{J}\mathbf{/}\mathbf{kg}\mathbf{\times }\mathbf{K}$, and that the initial temperature of the chemicals is 20ºC.

Fig. 1

Short answer

(a) The heat of reaction of two chemicals is$335,380\mathrm{J}/\mathrm{kg}$

Step-by-step solution

Concept of First Law of Thermodynamics.

Statement of the First Law of Thermodynamics infers the law of conservation energy in a thermodynamic process.

Mathematically in thiscontext,

${\mathbf{Q}}_{\mathbf{reaction}}\mathbf{}\mathbf{}\mathbf{=}\mathbf{W}\mathbf{}\mathbf{+}\mathbf{}{\mathbf{\Delta U}}_{\mathbf{spray}}$ .....(1)

 Determination of the heat produced when the two chemicals react.

The reaction between the chemicals produces heat and can be expressed as,

${\mathrm{Q}}_{\mathrm{reaction}}={\mathrm{mL}}_{\mathrm{reaction}}$ ....(II)

Here, ${\mathrm{L}}_{\mathrm{reaction}}$denotes the heat produced of the reaction of two chemicals and m is the mass.

For the spray with mass m, the work done is,

$$\begin{gathered} \mathrm{W}=\frac{1}{2}{\mathrm{mv}}^2 \\ =\frac{1}{2}\mathrm{m}\left(19\mathrm{m}/\mathrm{s}\right) \\ =\left(180.5\mathrm{J}/\mathrm{Kg}\right)\mathrm{m} \end{gathered}$$

Now,

$$\begin{gathered} {\mathrm{\Delta U}}_{\mathrm{spray}}={\mathrm{Q}}_{\mathrm{spray}} \\ =\mathrm{mC}∆\mathrm{T} \\ =\mathrm{m}\left(4190\mathrm{J}/\mathrm{Kg}.\mathrm{K}\right)\left(100°\mathrm{C}-20°\mathrm{C}\right) \\ =\left(335,200\mathrm{J}/\mathrm{Kg}\right)\mathrm{m}. \end{gathered}$$

Substitute all values in equation (i),

$$\begin{gathered} \mathrm{Q}=\left(180\mathrm{J}/\mathrm{Kg}+335,200\mathrm{J}/\mathrm{Kg}\right)\mathrm{m} \\ =\left(335,380\mathrm{J}/\mathrm{Kg}\right)\mathrm{m}. \end{gathered}$$

Equate the above value with equation (ii),

$$\begin{gathered} {\mathrm{mL}}_{\mathrm{reaction}}=\mathrm{m}(335,380\mathrm{J}/\mathrm{kg}) \\ {\mathrm{L}}_{\mathrm{reaction}}=(335,380\mathrm{J}/\mathrm{kg}) \end{gathered}$$

Thus, the heat produced by the chemicals reacting in the insect is $335,380\mathrm{J}/\mathrm{kg}$.