Question

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 the energy needed for evaporation is taken from the water itself and the water is cool down assume that a pitcher contains 10 kg of water and 0.2 gram of water comes out for per second assuming no backward heat transfer from the atmosphere to the water calculate the time in which the temperature decreases by 5 degree Celsius specific heat capacity of water = 4200 j kg °c and and latent heat of vaporization of water = 2.27 * 10 kg

Answer 1

Given, mass of water contains in a pitcher , M = 10kg
water comes out per second, m = 0.2g = 0.0002kg
temperature of water , T = 5°C
we know, specific heat of water , s = 4200J/Kg/°C
amount of energy released to decrease temperature 5° to 0°C = Ms(T - 0)
= 10 × 4200 × 5
= 210000 J


Let t is the time taken to decrease temperature from 5°C to 0°C.
energy required per second for evaporation of water is given by, U = rate of water comes out per second × latent heat of vaporisation
= 0.0002 × 2.27 × 10^6 J
= 454J

so, total energy required to decrease the temperature of water = U × t = 454 × t

now, 454 × t = 210000

t = 210000/454 = 462.55 sec

hence, 462.55 sec required to decrease the temperature of water contains in a pitcher

Answer 2

Given, mass of water contains in a pitcher , M = 10kg

water comes out per second, m = 0.2g = 0.0002kg

temperature of water , T = 5°C

we know, specific heat of water , s = 4200J/Kg/°C

amount of energy released to decrease temperature 5° to 0°C = Ms(T - 0)

= 10 × 4200 × 5

= 210000 J

Let t is the time taken to decrease temperature from 5°C to 0°C.

energy required per second for evaporation of water is given by, U = rate of water comes out per second × latent heat of vaporisation

= 0.0002 × 2.27 × 10^6 J

= 454J

so, total energy required to decrease the temperature of water = U × t = 454 × t

now, 454 × t = 210000

t = 210000/454 = 462.55 sec

hence, 462.55 sec required to decrease the temperature of water contains in a pitcher

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