Question

For many years drinking water has been cooled in hot climates by evaporating it from the surface of canvas bags or porous clay pots.How many grams of water can be cooled from 40 ∘C to 20 ∘C by the evaporation of 43 g of water? (The heat of vaporization of water in this temperature range is 2.4 kJ/g. The specific heat of water is 4.18 J/g⋅K.)

Answer 1

Answer : The mass of water cooled can be 1234.4 grams.

Explanation :

First we have to calculate the heat released during the reaction.

$\Delta H=\frac{q}{m}$

where,

$\Delta H$ = enthalpy change or heat of vaporization = 2.4 kJ/g = 2400 J/g

q = heat released = ?

m = mass of water vapor = 43 g

Now put all the given values in the above formula, we get:

$2400J/g=\frac{q}{43g}$

$q=103200J$

Thus, the heat released during the reaction is 103200 J

Now we have to calculate the initial mass of water.

$q=m\times c\times (T_{final}-T_{initial})$

where,

m = initial mass of water = ?

q = heat absorbed = heat released = 103200 J

c = specific heat = $4.18J/g.K$

$T_{final}$ = final temperature = $40^oC=273+40=313K$

$T_{initial}$ = initial temperature = $20^oC=273+20=293K$

Now put all the given values in the above formula, we get:

$103200J=m\times 4.18J/g.K\times (313-293)K$

$m=1234.4g$

Therefore, the mass of water cooled can be 1234.4 grams.

Answer 2

The mass of water that can be cooled  from 40 ∘C to 20 ∘C by the evaporation of 43 g of water is: 1234.4 grams.

Meaning of water cooling

Water can be defined as any liquid that is odorless, tasteless, colorless.
Water cooling is a process of extracting or removing all the heat in a water molecule.

Water cooling can also be said to be a process of reducing the temperature of water.

In conclusion, The mass of water that can be cooled  from 40 ∘C to 20 ∘C by the evaporation of 43 g of water is: 1234.4 grams.

Learn more about water cooling: https://brainly.com/question/11759755

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