The heat capacity of liquid water is 4.18 J/g•C and the heat of vaporization is 40.7 kJ/mol. How many kilojoules of heat must be provided to convert 1.00 g of liquid water at 67C into 1.00 g of steam at 100C?
A. 22.7 kJ
B. 40.8 kJ
C. 2.2 kJ
D. 2,400 J
E. 40.8 J
How much energy does it take to convert 12.5 grams of water to water vapor?
To answer this question, we need to calculate the amount of heat required to raise the temperature of liquid water from 67°C to its boiling point (100°C), and then the amount of heat required to convert it into steam at 100°C.
1. Calculate the heat required to increase the temperature of 1.00 g of liquid water from 67°C to 100°C:
Heat = mass * heat capacity * change in temperature
Heat = 1.00 g * 4.18 J/g·°C * (100°C - 67°C)
Heat = 1.00 g * 4.18 J/g·°C * 33°C
Heat = 138.18 J
2. Calculate the heat required to convert 1.00 g of liquid water at 100°C into steam at 100°C:
Heat = moles * heat of vaporization
Moles = mass / molar mass
Moles = 1.00 g / 18.015 g/mol (molar mass of water)
Moles = 0.05546 mol
Heat = 0.05546 mol * 40.7 kJ/mol
Heat = 2.256122 kJ
3. Add the two calculated heats together to get the total heat required:
Total Heat = 138.18 J + 2.256122 kJ
Total Heat = 2.394122 kJ
Therefore, the correct answer is C. 2.2 kJ.
To solve this problem, we need to consider two steps:
1. Heating the liquid water from 67°C to 100°C.
2. Vaporizing the liquid water at 100°C into steam.
Step 1: Heating the liquid water from 67°C to 100°C.
The heat required to raise the temperature of a substance can be calculated using the formula: q = m * C * ΔT, where q is the heat in joules (J), m is the mass in grams (g), C is the specific heat capacity in J/g•°C, and ΔT is the change in temperature in °C.
For this step, we need to calculate the heat required to raise the temperature of 1.00 g of liquid water from 67°C to 100°C. The equation would be:
q1 = 1.00 g * 4.18 J/g•°C * (100°C - 67°C)
q1 = 1.00 g * 4.18 J/g•°C * 33°C
q1 = 138.54 J
Step 2: Vaporizing the liquid water at 100°C into steam.
The heat required to convert a substance from one phase to another can be calculated using the formula: q = n * ΔHvap, where q is the heat in joules (J), n is the number of moles, and ΔHvap is the heat of vaporization in J/mol.
To calculate the number of moles of water, we can use the molar mass of water (H2O), which is approximately 18.02 g/mol. The equation would be:
n = 1.00 g / 18.02 g/mol
n ≈ 0.0555 mol
For this step, we need to calculate the heat required to vaporize 0.0555 mol of water. The equation would be:
q2 = 0.0555 mol * 40.7 kJ/mol * 1000 J/kJ
q2 = 2257.85 J
q2 ≈ 2.26 kJ
Finally, we need to sum up the heat from both steps to obtain the total heat required to convert 1.00 g of liquid water at 67°C into 1.00 g of steam at 100°C:
q_total = q1 + q2
q_total = 138.54 J + 2257.85 J
q_total ≈ 2396.39 J
q_total ≈ 2.40 kJ
Therefore, the correct answer is D. 2,400 J.