How long would it take for 1.50mol of water at 100.0 degrees to be converted completely into steam if heat were added at a constant rate of 23.0J/s ? the constants are: Specific heat of ice: 2.09J(g/degrees celcius)
Specific heat of liquid water: 4.18 J (g/degrees celcius)
Enthalpy of fusion: 334J/g
Enthalpy of vaporization: 2250 J/g
Sam, Alisha, et al. This question was answered below by Bob Pusley.
To determine the time it would take for 1.50 mol of water to be completely converted into steam, we need to calculate the amount of heat required for the phase changes.
First, we calculate the heat required to raise the temperature of the water from 100.0 degrees Celsius to its boiling point (100.0 degrees Celsius):
q1 = (m * c * ΔT)
Where:
m = mass of water (in grams)
c = specific heat of liquid water (in J/(g·°C))
ΔT = change in temperature (in °C)
Since we have 1.50 mol of water and the molar mass of water (H2O) is approximately 18.0 g/mol, the mass of water is calculated as:
mass = (1.50 mol) * (18.0 g/mol) = 27.0 g
Using the specific heat of liquid water (4.18 J/(g·°C)), the change in temperature (ΔT) is:
ΔT = (100.0 °C - 0 °C) = 100.0 °C
Plugging in the values, we can calculate q1:
q1 = (27.0 g) * (4.18 J/(g·°C)) * (100.0 °C) = 112,554 J
Next, we calculate the heat required to melt the ice at its boiling temperature (100.0 degrees Celsius) into water:
q2 = (m * ΔHf)
Where:
m = mass of water (in grams)
ΔHf = enthalpy of fusion (in J/g)
Again, the mass of water is 27.0 g, and the enthalpy of fusion (ΔHf) is 334 J/g. Plugging in the values, we can calculate q2:
q2 = (27.0 g) * (334 J/g) = 9,018 J
Then, we calculate the heat required to raise the temperature of the water from its boiling point (100.0 degrees Celsius) to its boiling point (100.0 degrees Celsius):
q3 = (m * c * ΔT)
The mass of water is still 27.0 g, and the specific heat of water (c) is 4.18 J/(g·°C). Again, the change in temperature (ΔT) is:
ΔT = (100.0 °C - 100.0 °C) = 0 °C
Plugging in the values, we can calculate q3:
q3 = (27.0 g) * (4.18 J/(g·°C)) * (0 °C) = 0 J
Finally, we calculate the heat required to vaporize the water into steam:
q4 = (m * ΔHv)
Where:
m = mass of water (in grams)
ΔHv = enthalpy of vaporization (in J/g)
Again, the mass of water is 27.0 g, and the enthalpy of vaporization (ΔHv) is 2250 J/g. Plugging in the values, we can calculate q4:
q4 = (27.0 g) * (2250 J/g) = 60,750 J
To calculate the total heat required for the entire process, we sum up the individual heat values:
Q_total = q1 + q2 + q3 + q4 = 112,554 J + 9,018 J + 0 J + 60,750 J = 182,322 J
Finally, we can determine the time it would take for the water to be completely converted into steam by dividing the total heat required (Q_total) by the rate of heat addition (23.0 J/s):
t = Q_total / rate
Plugging in the values, we can calculate t:
t = 182,322 J / 23.0 J/s ≈ 7,930 s ≈ 2.20 hours
Therefore, it would take approximately 2.20 hours for 1.50 mol of water at 100.0 degrees Celsius to be converted completely into steam if heat were added at a constant rate of 23.0 J/s.