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 is taken from the water itself and the water is cooled down. If pitcher has $10 kg$ of water, find the time in which the temperature of water falls by $5^{\circ} C$ . The approximate rate of vaporisation is $5 {g min}^{- 1}$ $(C_{w} = 1 {cal/g-}^{\circ} C, L_{v} = 540 cal/g)$

20 min 10 s

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18 min 20 s

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14 min 12 s

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none of these

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Solution

The correct option is B $18 min 20 s$
Let $\frac{d m}{d t}$ be the rate of evaporation and $M = M_{0} - t \frac{d m}{d t}$ is the amount of water left in the pitcher at any instant.
Heat lost by left over water=Heat gained for evaporation
$M C_{w} d θ = (\frac{d m}{d t} t) L_{v}$
$(M_{0} - \frac{d m}{d t} . t) C_{w} d θ = L_{v} t \frac{d m}{d t}$
$(10000 - \frac{5}{60} . t) 1 \times 5 = \frac{5}{60} t (540)$
or $50000 = 45 t + \frac{5}{12} t$
or $t = \frac{50, 000 \times 12}{545} = 18 min 26 s$

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Similar questions

Q. 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 is taken from the water itself and the water is cooled down. If pitcher has

10 kg

of water, find the time in which the temperature of water falls by

5^{\circ} C

. The approximate rate of vaporisation is

5 {g min}^{- 1}

(C_{w} = 1 {cal/g-}^{\circ} C, L_{v} = 540 cal/g)

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 cooled down. Assume that a pitcher contains 10 kg of water and 0.2 g of water comes out per second. Assuming no backward heat transfer from the atmosphere to the water, calculate the time in which the temperature decreases by $5^{\circ} C$ . Specific heat capacity of water $= 4200 J k g^{- 1}$ $^{\circ} C^{- 1}$ and latent heat of vaporization of water $= 2.27 \times 10^{6} J k g^{- 1}$

10 k g

of water and

0.2 g

of water comes out per second. Assuming no backward heat transfer from the atmosphere to the water, calculate the time in which the temperature decreases by

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and latent heat of vaporization of water

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