Question

Indian style of cooling drinking water is to keep it in a pitcher having porous walls. Water comes to the outer surface very slowly and evaporates. Most of the energy needed for evaporation is taken from the water itself and the water is cooled down. Assume that a pitcher contains 10 kg of water and 0.2 g of water comes out per second. Assuming no backward heat transfer from the atmosphere to the water, calculate the time in which the temperature decreases by $5^{\circ }C$. Specific heat capacity of water $=4200Jk{g}^{-1}$ ${}^{\circ }{C}^{-1}$ and latent heat of vaporization of water $=2.27\times {10}^{6}Jk{g}^{-1}$

Answer 1

Energy required to decreases the temp of 10 kg. of water to $5^{\circ }C.$

$U=10\times 4200J/k{g}^{\circ }C\times 5^{\circ }C$

$=210,000=21\times {10}^{4}J.$

Energy required for evaporation of water to 0.2g/sec.

$=2\times {10}^{-4}\times 2.27\times {10}^6$

$=454.$

454 J energy losing system per second

$=\frac{21\times {10}^4}{454}$

$21\times {10}^{4}J$ energy losing system during one minute

$=\frac{21\times {10}^4}{454\times 60}=7.7$ minute

\therefore The time required to decrease temp by $5^{\circ }C$ is 7.7 minute