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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 ........... $^{\circ} C$ (ratio of latent of evaporation to specific heat of water is $540^{\circ} C$
$2.7$
$4.2$
$5.4$
$10.8$
Solution
(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
$Q=m L…(i)$
Assuming this heat is taken entirely from water in earthen pot, if $\Delta T$ is decrease of temperature of pot then,
$Q=M s \Delta T \dots(ii)$
where, $M=$ mass of water in pot and $s=$ specific heat of water.
Equating Eqs. $(i)$ and $(ii)$, we get
$m L=M s \Delta T$
$\text { or } \Delta T=\frac{m}{M} \times \frac{L}{s}=\frac{40 \times 10^{-3}}{4} \times 540=5.4^{\circ} C$
