An earthen pitcher loses 1 g of water per minute due to evaporation. If the water equivalent of pitcher is 0.5 kg and the pitcher contains 9.5 kg of water, calculate the time required for the water in the pitcher to cool to 28^oC from its original temperature of 30^oC. Neglect radiation effects. Latent heat of vapourization of water in this range of temperature is 580 cal/g and specific heat of water is 1 k cal/gC^o

As water equivalent of pitcher is 0.5 kg, i.e., pitcher is equivalent to 0.5 kg of water, heat to be extracted from the system of water and pitcher for decreasing its temperature from 30 to 28^oC is

$\begin{array}{l}{\mathrm{Q}}_1=(\mathrm{m}+\mathrm{M})\mathrm{c\Delta T}\\ =(9.5+0.5)\mathrm{kg}\left(1\mathrm{kcal}/{\mathrm{kg}}^{\circ }\mathrm{C}\right)(30-28{)}^{\circ }\mathrm{C}=20\mathrm{kcal}\end{array}$

And heat extracted from the pitcher through evaporation in t minutes

${\mathrm{Q}}_2=\mathrm{mL}=\left[\frac{\mathrm{dm}}{\mathrm{dt}}\times \mathrm{t}\right]\mathrm{L}=\left[\frac{1\mathrm{g}}{min}\times \mathrm{t}\right]580\frac{\mathrm{cal}}{\mathrm{g}}=580\times \mathrm{tcal}$

According to given problem ${\mathrm{Q}}_2={\mathrm{Q}}_1,\text{ i.e., }580\times \mathrm{t}=20\times {10}^3$

        $\mathrm{t}=34.5\min$