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

Step-by-step solution

Water evaporated in two hours = m = 2   h × 20   g / h = 40   g = 40 × 10 - 3   k g Heat absorbed by water during evaporation is Q = Mass evaporated × Latent heat Q = m   L ...( i ) Assuming this heat is taken entirely from water in earthen pot, if Δ   T is decrease of temperature of pot then, Q = M s   Δ   T ...(ii) where, M = mass of water in pot and s = specific heat of water. Eguating Egs. ( i ) and ( ii ), we get m L = M s   Δ   T