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 is taken from the water itself and the water is cooled down. Assume that a pitcher contains $10kg$ of water and $0.2g$ 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 $5^{o}C$. Specific heat capacity of water ${=4200J{kg}^{-1}}^{}{C}^{-1}$ and latent heat of vaporization of water $=2\cdot 27\times {10}^{6}J{kg}^{-1}$.

Answer 1

Given:Specific heat of water,$S=4200$K${kg}^{-1}{C}^{-1}$

Latent heat of vapourisation of water, $L=2.27\times {10}^6$J${kg}^{-1}$

Mass,$M=0.2$g$=0.0002$kg

Let us first calculate the amount of energy required to decrease the temperature of $10$kg of water by $5^{\circ }$C

$U_1=10\times 4200$J/kg degee celcius$\times 5^{\circ }$C

$U_2=210000=21\times {10}^4$J

Let the time in which the temperature is decreases by $5^{\circ }$C be $t$

Energy required per second for evaporation of water(at the rate of $0.2$g/sec) is given by

$U_2=ML$

$=(2\times {10}^{-4})\times (2.27\times {10}^6)=454$J

Total energy required to decrease the temperature of the water$=454\times t=21\times {10}^4$J

Now, $t=\frac{21\times {10}^4}{454\times 60}=7.7$minutes

$\therefore$ The time required to decrease the temperature by $5^{\circ }$C is $7.7$ minutes.