Answer and Explanation:
To determine the amount of water that can be cooled from 44 °C to 15 °C by the evaporation of 40 g of water, we need to calculate the heat exchanged in the process.
First, we calculate the heat required to cool down the water from 44 °C to 15 °C using the specific heat formula:
q1 = m * C * ΔT1
where:
q1 = heat required to cool the water (in Joules)
m = mass of water (in grams) = 40 g
C = specific heat of water = 4.18 J/g.K
ΔT1 = change in temperature = (15 °C - 44 °C) = -29 °C
Plugging in the values:
q1 = 40 g * 4.18 J/g.K * (-29 °C)
q1 = -4793.6 J
Next, we calculate the heat of vaporization for the evaporation of water:
q2 = m * Hvap
where:
q2 = heat required for evaporation (in Joules)
m = mass of water (in grams) = 40 g
Hvap = heat of vaporization of water = 2.4 kJ/g = 2400 J/g
Plugging in the values:
q2 = 40 g * 2400 J/g
q2 = 96000 J
To find the total heat exchanged, we add the heat required to cool the water (q1) and the heat required for evaporation (q2):
q_total = q1 + q2
q_total = -4793.6 J + 96000 J
q_total = 91206.4 J
Now, we need to convert the heat exchanged to the amount of water evaporated using the equation:
q_total = m_evaporated * Hvap
Rearranging the equation to solve for m_evaporated:
m_evaporated = q_total / Hvap
m_evaporated = 91206.4 J / 2400 J/g
m_evaporated = 38 g
Therefore, approximately 38 grams of water can be cooled from 44 °C to 15 °C by the evaporation of 40 g of water.