In warm climates, drinking water is often cooled by evaporating water from the surfaces of clay pots or canvas bags. How many liters of water may be cooled from 37 ºC to 24 ºC by the evaporation of 100 g of water? The heat of vaporization of water is 2.4 kJ/g, and the specific heat of water is 4.18 J/g-K.
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To find out how many liters of water can be cooled by the evaporation of 100 g of water, we need to calculate the amount of heat energy required to cool the water from 37 ºC to 24 ºC.
First, let's calculate the heat energy required to cool 100 g of water from 37 ºC to 24 ºC using the specific heat equation:
Q = m * c * ΔT
Where:
Q = Heat energy
m = Mass of the water (100 g)
c = Specific heat of water (4.18 J/g-K)
ΔT = Change in temperature (24 ºC - 37 ºC = -13 ºC)
Q = 100 g * 4.18 J/g-K * (-13 ºC)
Q = -5414 J
Next, let's calculate the amount of water that can be cooled by the evaporation of 5414 J of heat energy.
Since the heat of vaporization of water is 2.4 kJ/g, we can convert 5414 J to kJ:
5414 J = 5414 J / 1000 = 5.414 kJ
Now, let's divide the heat energy by the heat of vaporization to find the amount of water that can be cooled:
Amount of water = 5.414 kJ / 2.4 kJ/g
Amount of water = 2.26 g
Since we know that the density of water is 1 g/mL, we can convert 2.26 g to liters:
Amount of water = 2.26 g * (1 mL / 1 g) * (1 L / 1000 mL)
Amount of water = 0.00226 L
Therefore, approximately 0.00226 liters of water can be cooled from 37 ºC to 24 ºC by the evaporation of 100 g of water.