Question

An earthen pitcher loses $1\text{g}$ of water per minute due to evaporation. If the water equivalent of the pitcher is $0.5\text{kg}$ and the pitcher contains $9.5\text{ kg}$ of water, then calculate the time required for the water in pitcher to cool to $28^{\circ }C$ from its original temperature of $30^{\circ }C$ . Neglect the effect of radiation. Latent heat of vaporization of water in this range of temperature is $580\text{cal/g}$ and specific heat of water is $1cal/g{/}^{\circ }C$ .

• A. $30.5min$
• B. $41.2min$
• C. $38.6min$
• D. $34.5min$

Answer 1

Answer: (D)

Heat lost by (Water + Pitcher)
$Q_1=\left(m+M\right)\times s\times \Delta T$
$=\left(9.5+0.5\right)\times {\left(10\right)}^3\times \left(30-28\right)$
$=20\times 10^{3}cal$
Heat gained for the water to evaporate:
(Let $t$ be time in $min$ )
$Q_2=m\times L$
$=\left(\frac{dm}{dt}\times t\right)L$
$=\left(1\times t\right)\times 580$
$=580tcal$
So,
$Q_1=Q_2$
$580t=20\times 10^3$
$t=\frac{20000}{580}=34.5min$