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 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 decrease by 5°C. Specific heat capacity of water = 4200 J kg⁻¹ °C⁻¹ and latent heat of vaporization of water = 2.27 × 10⁶ J kg⁻¹.

Answer 1

Given:
Specific heat of water ,S = 4200 J kg⁻¹ °C⁻¹
Latent heat of vapourisation of water ,L = 2.27 × 10⁶ J kg⁻¹
Mass, M = 0.2 g = 0.0002 kg

Let us first calculate the amount of energy required to decrease the temperature of 10 kg of water by 5°C.

U₁ = 10 × 4200 J/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 0.2 g/sec) is given by
U₂ = ML
U₂ = (2 × 10⁻⁴ )× (2.27 × 10⁶) = 454 J

Total energy required to decrease the temperature of the water = 454 × t
= 21 × 10⁴ J
Now,
t $=\frac{21\times {10}^4}{454}$ seconds

The time taken in minutes is given by
t$=\frac{21\times {10}^4}{454\times 60}=7.7\mathrm{minute}s$

∴ The time required to decrease the temperature by 5°C is 7.7 minutes.