Question

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

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

Answer 1

Answer: (C)

As water equivalent of pitcher is $0.5 \,kg$,
i..e., pitcher is equivalent to $0.5 \,kg$ of water, heat to be extracted from the system of water and pitcher for decreasing its temperature from $30$ to $28^{\circ} C$ is
$Q _1 = m + Mc \Delta T $
$=9.5+0.5 \,kg \,1 \,k \, cal / kg C ^{\circ} 30-28^{\circ} C$
$=20 \, kcal$
And heat extracted from the pitcher through evaporation in t minutes
$Q_2= mL=\frac{ dm }{ dt } \times tL =\frac{1 g }{\min } \times t 580 \frac{ cal }{ g } $
$=580 \times t \,cal$
According to given problem $Q _2= Q _1$,
i.e., $580 \times t =20 \times 10^3$
$t=34.5 \min$