Question

An earthen pitcher loses $1g$ of water per minute due to evaporation. If the water equivalent of pitcher is $0.5kg$ and the pitcher contains $9.5kg$ 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. $38.6min$
• B. $30.5min$
• C. $34.5min$
• D. $41.2min$

Answer 1

Answer: (C)

As water equivalent of pitcher is $0.5kg$,
i..e., pitcher is equivalent to $0.5kg$ of water, heat to be extracted from the system of water and pitcher for decreasing its temperature from $30$ to $28^{\circ }C$ is
$Q_1=m+Mc\Delta T$
$=9.5+0.5kg1kcal/kg{C}^{\circ }30-28^{\circ }C$
$=20kcal$
And heat extracted from the pitcher through evaporation in t minutes
$Q_2=mL=\frac{dm}{dt}\times tL=\frac{1g}{\min }\times t580\frac{cal}{g}$
$=580\times tcal$
According to given problem $Q_2=Q_1$,
i.e., $580\times t=20\times 10^3$
$t=34.5\min$