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 $1 \, cal/g/^\circ C$ .

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

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 $min$ )
$Q_{2}=m\times L$
$=\left(\frac{d m}{d t} \times t\right)L$
$=\left(1 \times t\right)\times 580$
$=580t \, cal$
So,
$Q_{1}=Q_{2}$
$580t=20\times 10^{3}$
$t=\frac{20000}{580}=34.5 \, min$