An earthen pitcher loses 1 g of water per minute due to evaporation. If the water equivalent of the pitcher is 0.5 kg and the pitcher contains 9.5 kg of water, then calculate the time required for the water in pitcher to cool to 28 ° C from its original temperature of 30 ° C . Neglect the effect of radiation. Latent heat of vaporization of water in this range of temperature is 580 cal/g and specific heat of water is 1 c a l / g / ° C .

Question

An earthen pitcher loses 1 g of water per minute due to evaporation. If the water equivalent of the pitcher is 0.5 kg and the pitcher contains 9.5 kg of water, then calculate the time required for the water in pitcher to cool to 28 ° C from its original temperature of 30 ° C . Neglect the effect of radiation. Latent heat of vaporization of water in this range of temperature is 580 cal/g and specific heat of water is 1 c a l / g / ° C ., 30.5 min, 41.2 min, 38.6 min, 34.5 min

MARKS App · Accepted Answer

Heat lost by (Water + Pitcher) Q 1 = m + M × s × ∆ T = 9.5 + 0.5 × 10 3 × 30 - 28 = 20 × 10 3 c a l Heat gained for the water to evaporate: (Let t be time in min ) Q 2 = m × L = d m d t × t L = 1 × t × 580 = 580 t c a l So, Q 1 = Q 2 580 t = 20 × 10 3 t = 20000 580 = 34.5 m i n

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