A traditional method of cooling drinking water is to keep it in a pitcher having porous walls. Water reaches the outer surface very slowly and evaporates. Most of the energy required for evaporation is extracted from the water itself, thereby cooling it down. Assume a pitcher holds 10 kg of water and 0.2 g of water seeps out per second. Assuming no backward heat transfer from the atmosphere to the water, determine th

Step-by-step solution

Let t be the time required for the temperature to drop by 5^ C . The rate of heat loss due to evaporation is given by: dQ dt = dm dt L Substituting the given values: dQ dt = (0.2 10⁻³ kg s ⁻¹) (2.27 10^6 J kg ⁻¹) = 454 J s ⁻¹ The total heat to be extracted from the water to cool it by 5^ C is: Q = M s T Q = 10 kg 4200 J kg ⁻¹ ^ C⁻¹ 5^ C = 210000 J Equating the total heat extracted to the heat lost over time t : Q = ( dQ dt ) t 210000 = 454 t t = 210000 454 s 462.55 s Converting the time into minutes: t = 462.55 60