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. If pitcher has $10\text{kg}$ of water, find the time in which the temperature of water falls by $5^{\circ }\text{C}$. The approximate rate of vaporisation is $5{\text{g min}}^{-1}$ $(C_w=1{\text{cal/g-}}^{\circ }\text{C},L_v=540\text{cal/g})$

• A. $20\text{min}10\text{s}$
• B. $18\text{min}20\text{s}$
• C. $14\text{min}12\text{s}$
• D. $\text{none of these}$

Answer 1

The correct option is B $18\text{min}20\text{s}$

Let $\frac{dm}{dt}$ be the rate of evaporation and $M=M_0-t\frac{dm}{dt}$ is the amount of water left in the pitcher at any instant.
Heat lost by left over water=Heat gained for evaporation
$M{C}_{w}d\theta =\left(\frac{dm}{dt}t\right)L_v$
$(M_0-\frac{dm}{dt}.t)C_{w}d\theta =L_{v}t\frac{dm}{dt}$
$(10000-\frac{5}{60}.t)1\times 5=\frac{5}{60}t\left(540\right)$
or $50000=45t+\frac{5}{12}t$
or $t=\frac{50,000\times 12}{545}=18\text{min}20\text{s}$