Question

An earthen pitcher used in summer cools water in it essentially by evaporation of water from its porous surface. If a pitcher carries $4 \,kg$ of water and the rate of evaporation is $20\, g$ per hour, temperature of water in it decreases by $\Delta\,T$ in two hours. The value of $\Delta\,T$ is close to (ratio of latent of evaporation to specific heat of water is $540^{\circ}C)$

• A. $2.7^{\circ} C$
• B. $4.2^{\circ}C$
• C. $5.4 ^{\circ} C$
• D. $10.8 ^{\circ}C$

Answer 1

Answer: (C)

Water evaporated in two hours
$=m=2\, h \times 20\, g/h$
$=40\,g=40 \times 10^{-3}\,kg$
Heat absorbed by water during evaporation is
$Q$ = Mass evaporated $\times$ Latent heat
$Q=mL$ ...(i)
Assuming this heat is taken entirely from water in earthen pot, if $\Delta\,T$ is decrease of temperature of pot then,
$Q=Ms \Delta\,T$ ...(ii)
where, M = mass of water in pot
and s = specific heat of water.
Equating Eqs. (i) and (ii), we get
$mL=Ms \Delta\,T$
or $\Delta\,T= \frac{m}{M}\times \frac{L}{S}$
$=\frac{40\times 10^{-3}}{4}\times 540$
$=5.4^{\circ}C$