How many grams of water can be cooled from 31 C to 20 C by the evaporation of 37 grams 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.
heat water cooled = heat water lost.
mass x specific heat water x delta T = delta H x 37 g H2O.
mass x 4.18 x (31-20) = 2.4 x 10^3 x 37
solve for mass.
Check my thinking.
1931
To calculate the amount of water that can be cooled by the evaporation of 37 grams of water, we need to consider two factors: the heat required to cool the water from 31°C to 20°C and the heat required for the evaporation.
First, let's calculate the heat required to cool the water from 31°C to 20°C using the specific heat formula Q = m * c * ΔT, where Q is the heat, m is the mass, c is the specific heat, and ΔT is the temperature change.
Q1 = m1 * c1 * ΔT1
= 37 g * 4.18 J/g*K * (20°C - 31°C)
Now let's calculate the heat required for the evaporation. We'll use the heat of vaporization formula Q = m * H, where Q is the heat, m is the mass, and H is the heat of vaporization.
Q2 = m2 * H
= 37 g * 2.4 kJ/g
To find the total heat required for both cooling and evaporation, we add the two calculated values.
Total heat required = Q1 + Q2
Finally, to find the mass of water that can be cooled by the evaporation of 37 grams of water, we divide the total heat required by the heat of vaporization.
Mass of water cooled = (Q1 + Q2) / H
Plug in the values into the equations and calculate to find the result.