Question

An earthen pitcher used in summer cools water in it essentially by evaporation of water from its porous surface. If a pitcher carries $4kg$ of water and the rate of evaporation is $20g$ 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

$=m=2h\times 20g/h$

$=40g=40\times 10^{-3}kg$

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 $\Delta T=\frac{m}{M}\times \frac{L}{S}$

$=\frac{40\times 10^{-3}}{4}\times 540$

$=5.4^{\circ }C$