"Calculate the enthalpy of vaporization of water given that, when 5.00 grams of steam (gaseous H2O) at 100°C was introduced into a beaker containing 500 grams of water at 20.0°C, the temperature rose to 26.2°C."
H2O (l) -> H2O (g)
ΔH = ΔH_vap
5g H2O (g) @ 100°C
500g H2O (l) @ 20°C
Final temperature is 26.2°C
Is ΔH_vap 40.7 kJ/mol have to be used or we're finding that?
http://www.jiskha.com/display.cgi?id=1473202007
To calculate the enthalpy of vaporization (ΔH_vap) of water, we first need to determine the amount of heat transferred during the process. The equation for heat transfer can be written as:
q = m * C * ΔT
Where:
q = heat transferred
m = mass of the substance
C = specific heat capacity of the substance
ΔT = change in temperature
In this case, the mass of the steam (gaseous H2O) is given as 5.00 grams and the change in temperature is from 20.0°C to 26.2°C, which is ΔT = 26.2°C - 20.0°C = 6.2°C.
Since the steam condenses and transfers heat to the water, the equation can be written as:
q = -q_water
Where q_water is the heat absorbed by the water.
Now, let's calculate q_water by rearranging the equation:
q_water = -q = -m * C * ΔT
The mass of water (liquid H2O) is given as 500 grams, and the specific heat capacity of water is approximately 4.18 J/g°C.
q_water = -500 g * 4.18 J/g°C * 6.2°C
Next, we need to convert the heat absorbed by the water from joules (J) to kilojoules (kJ). Since 1 kJ = 1000 J, we divide q_water by 1000:
q_water = -500 g * 4.18 J/g°C * 6.2°C / 1000
Finally, we can calculate the enthalpy of vaporization using the equation:
ΔH_vap = q_water / n
Where n is the number of moles of water. Since the molar mass of water (H2O) is approximately 18 g/mol, we can calculate the number of moles:
n = m / M = 5.00 g / 18 g/mol
Now, substitute the calculated values into the equation:
ΔH_vap = (-500 g * 4.18 J/g°C * 6.2°C / 1000) / (5.00 g / 18 g/mol)
Simplifying this expression will give you the enthalpy of vaporization (ΔH_vap) of water.