How many grams of water can be cooled from 35^\circ {\rm C} to 16^\circ {\rm C} by the evaporation of 57{\rm g} of water? (The heat of vaporization of water in this temperature range is 2.4 {\rm kJ}/{\rm g}. The specific heat of water is 4.18 {\rm J}/{\rm g}\cdot {\rm K}.)

To determine the amount of water that can be cooled by the evaporation of 57 g of water, we need to calculate the heat lost during the cooling process.

First, let's calculate the heat released when water cools from 35°C to 16°C:

Q1 = m * c * ΔT1,

where m is the mass of water, c is the specific heat of water, and ΔT1 is the temperature change.

Given:
m = 57 g,
c = 4.18 J/g·K,
ΔT1 = (35°C - 16°C) = 19°C.

Substituting the values into the formula:

Q1 = 57 g * 4.18 J/g·K * 19°C = 43442.86 J.

Next, let's calculate the heat required to evaporate the 57 g of water:

Q2 = m * Lvap,

where m is the mass of water and Lvap is the heat of vaporization.

Given:
m = 57 g,
Lvap = 2.4 kJ/g = 2400 J/g.

Substituting the values into the formula:

Q2 = 57 g * 2400 J/g = 136800 J.

Since the heat lost during cooling (Q1) is equal to the heat required for evaporation (Q2), we can set up the equation:

Q1 = Q2.

43442.86 J = 136800 J.

Now we can solve for the mass of water that can be evaporated:

m = Q2 / Lvap.

m = 136800 J / 2400 J/g = 57 g.

Therefore, 57 grams of water can be cooled from 35°C to 16°C by the evaporation of 57 g of water.

To find the answer, we need to calculate the heat absorbed during evaporation and the heat released during cooling. Then we can determine the mass of water that can be cooled by the given amount of heat.

First, let's calculate the heat absorbed during evaporation of 57 g of water. The heat of vaporization of water is given as 2.4 kJ/g. Since the mass of water is 57 g, the heat absorbed during evaporation is:

Heat absorbed = Mass of water * Heat of vaporization
= 57 g * 2.4 kJ/g
= 136.8 kJ

Next, let's calculate the heat released during cooling. The temperature change is from 35°C to 16°C. The specific heat of water is 4.18 J/g·K. The formula to calculate the heat released is:

Heat released = Mass of water * Specific heat of water * Temperature change

To find the mass of water that can be cooled, we rearrange the formula:

Mass of water = Heat released / (Specific heat of water * Temperature change)

Substituting the values we have:

Mass of water = 136.8 kJ / (4.18 J/g·K * (35°C - 16°C))

Now we calculate the temperature change:

Temperature change = 35°C - 16°C
= 19°C

Substituting the values into the formula, we find:

Mass of water = 136.8 kJ / (4.18 J/g·K * 19°C)
≈ 136.8 kJ / (78.82 J/g)
≈ 1.734 g

Therefore, 1.734 grams of water can be cooled from 35°C to 16°C by the evaporation of 57 g of water.