Question

An India pitcher has 10 kg of water. Water cools by means of evaporation through the pores. Find the time in which the temperature of water falls by 5${}^o$C. The rate of vaporisation is 5g min${}^{-1}$ :

• A. 20 min 10 s
• B. 18 min 26 s
• C. 14 min 12 s
• D. none of these

Answer 1

Answer: (B)

18 min 26 s

Let $\frac{dm}{dt}$ be the rate of evaporation and
$M=M_0-t\frac{dm}{dt}$ is the amount of liquid left in the pitcher at any instant.
$MCd\theta =\left(\frac{dm}{dt}t\right)L$
$(M_0-\frac{dm}{dt}t)Cd\theta =Lt\frac{dm}{dt}$
or $(10000-\frac{5}{60}t)1\times 5=\frac{5}{60}t\left(540\right)$
or $50000=45t+\frac{5}{12}t$
or $t=\frac{50,000\times 12}{545}=18min26s$