Question

A bucket full of water is place in a room at ${15}^{\circ }C$ with initial relative humidity $40$%. The volume of the room is $50m^3$. (a) How much water will evaporate? (b) If the room temperature is increased by $5^{\circ }C$ how much more water will evaporate? The saturation vapour pressure of water at ${15}^{\circ }C$ and ${20}^{\circ }C$ are $1.6kPa$ and $2.4kPa$ respectively.

Answer 1

a) Rel. humidity is given as:

$=\frac{VP}{SVPat{15}^{\circ }C}$

$\Rightarrow 0.4=\frac{VP}{1.6\times {10}^3}$

$\Rightarrow VP=0.4\times 1.6\times {10}^3$

The evaporation occurs as along as the atmosphere does not become saturated

Net pressure change $=1.6\times {10}^3-0.4\times 1.6\times {10}^3=(1.6-0.4\times 1.6){10}^3=0.96\times {10}^3$

Net mass of water evaporated

$m\Rightarrow 0.96\times {10}^3\times 50=\frac{m}{18}\times 8.3\times 288$

$\Rightarrow m=\frac{0.96\times 50\times 18\times {10}^3}{8.3\times 288}=361.45\approx 361g$

b) At ${20}^{\circ }CSVP=2.4KPa$,

At ${15}^{\circ }C$ Saturated Vapor Pressure $=1.6KPa$

Net pressure charge $=(2.4-1.6)\times {10}^{3}Pa=0.8\times {10}^{3}Pa$

Mass of water evaporated

$m^{\prime }=0.8\times {10}^{3}50=\frac{m^{\prime }}{18}\times 8.3\times 293$

$\Rightarrow m^{\prime }=\frac{0.8\times 50\times 18\times {10}^3}{83\times 293}=296.06\approx 296grams$