Question

In regions with dry climates, evaporative coolers are used to cool air. A typical electric air conditioner is rated at $\mathbf{1}\mathbf{.00}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}\mathbf{Btu}\mathbf{/}\mathbf{h}$( $\mathbf{1}\mathbf{}\mathbf{Btu}$, or British thermal unit, equals the amount of energy needed to raise the temperature of $\mathbf{1}\mathbf{lb}$of water by $\mathbf{18}\mathbf{F}$). How much water must be evaporated each hour to dissipate as much heat as a typical electric air conditioner?

Short answer

$4.7\mathrm{kg}{\mathrm{h}}^{-1}$ water must be evaporated each hour to dissipate as much heat as a typical electric air conditioner.

Step-by-step solution

Determine the number of moles.

Change thespecific heat capacity $\left(4.84{\mathrm{Jg}}^{-1}{}^{\circ }\mathrm{C}^{-1}\right)$of water into $\mathrm{Btu}$.

Theenthalpy of vaporization of water is$40700\mathrm{J}/\mathrm{mol}$.

$\frac{40700\mathrm{J}/\mathrm{mol}}{4.184{\mathrm{Jg}}^{-1}{}^{\circ }\mathrm{C}^{-1}}\times \frac{1\mathrm{Ib}}{453.45\mathrm{g}}\times \frac{{1.8}^{\circ }\mathrm{F}}{1^{\circ }\mathrm{C}}\times \frac{1\mathrm{Btu}}{{\mathrm{Ib}}^{\circ }\mathrm{F}}=38.61\mathrm{Btu}{\mathrm{mol}}^{-1}$

Now, the moles required:

$\begin{array}{l}\mathrm{n}=\frac{1.0\times {10}^4\mathrm{Btu}{\mathrm{h}}^{-1}}{38.61\mathrm{Btu}{\mathrm{mol}}^{-1}}\\ =259\mathrm{mol}{\mathrm{h}}^{-1}\end{array}$

Determine the amount of water evaporated each hour.

The mass requires to dissipate as much as an electric air conditioner:

$\mathrm{m}=\mathrm{n}\times \mathrm{molecular}\mathrm{weight}$ molecular weight of water

$\begin{array}{c}\mathrm{m}=259\mathrm{mol}{\mathrm{h}}^{-1}\times 18.2\mathrm{g}{\mathrm{mol}}^{-1}\\ =4.7\mathrm{kg}{\mathrm{h}}^{-1}\end{array}$