Question 14 Marks
Indian style of cooling drinking water is to keep it in a pitcher having porous walls. Water comes to the outer surface very slowly and evaporates. Most of energy needed for evaporation is taken from the water itself and the water is cooled down. Assume that a pitcher contains 10kg of water and 0.2g of water comes out per second. Assuming no backward heat transfer from the atmosphere to the water, calculate the time in which the temperature decrease by 5°C. Specific heat capacity of water = 4200J kg-1°C-1 and latent heat of vaporization of water = 2.27 × 106 J kg-1.
Answer
View full question & answer→Given, Specific heat of water, S = 4200J kg-1 °C-1 Latent heat of vapourisation of water, L = 2.27 × 106 J kg-1 Mass, M = 0.2g = 0.0002kg Let us first calculate the amount of energy required to decrease the temperature of 10kg of water by 5°C. U1 = 10 × 4200J/kg° C × 5°C U1 = 210,000 = 21 × 104 J Let the time in which the temperature is decreased by 5°C be t. Energy required per second for evaporation of water (at the rate of 0.2g/sec) is given by U2 = ML U2 = (2 × 10-4 ) × (2.27 × 106) = 454J Total energy required to decrease the temperature of the water = 454 × t = 21 × 104 J Now,
$\text{t}=\frac{21\times10^4}{454}$ seconds
The time taken in minutes is given by,$\text{t}=\frac{21\times10^4}{454\times60}=7.7\text{ minute/s}$
$\therefore$ The time required to decrease the temperature by 5°C is 7.7 minutes.