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 67°C into 1.00 g of steam at 100°C?

q1 = heat to raise T of liquid H2O from 67 C to 100 C.

q1 = mass x specific heat H2O x (Tfinal-Tinitial)

q2 = heat to convert liquid water @ 100 C to steam @ 100 C.
q2 = mass x delta Hvap.

The specific heat capacity of liquid water is 4.184 J/goC

Calculate the energy (in kJ) required to heat 25 g of liquid water from 25oC to 100 oC

To calculate the amount of heat required, we need to consider two steps:

Step 1: Heating the liquid water from 67°C to 100°C.
Step 2: Vaporizing the water at 100°C to steam.

Step 1: Heating the liquid water from 67°C to 100°C
The heat capacity of liquid water is given as 4.18 J/g·°C. We have 1.00 g of liquid water, and the temperature needs to increase by (100 - 67) = 33°C.

The amount of heat required for step 1 can be calculated using the formula:
q = m * C * ΔT

Where:
q is the heat required
m is the mass of the substance (in grams)
C is the heat capacity (in J/g·°C)
ΔT is the change in temperature (in °C)

Plugging in the values:
q1 = 1.00 g * 4.18 J/g·°C * 33°C
q1 = 138.18 J

Step 2: Vaporizing the water at 100°C to steam
The heat of vaporization for water is given as 40.7 kJ/mol. Since we have 1.00 g of water, we need to convert grams to moles before calculating the heat.

The molar mass of water (H2O) is 18.015 g/mol.

The number of moles (n) can be calculated using the formula:
n = mass / molar mass

Plugging in the values:
n = 1.00 g / 18.015 g/mol
n ≈ 0.055 mol

The amount of heat required for step 2 can be calculated using the formula:
q2 = n * ΔHv

Where:
q2 is the heat required
n is the number of moles
ΔHv is the heat of vaporization (in J/mol)

Plugging in the values:
q2 = 0.055 mol * 40.7 kJ/mol
q2 = 2.2385 kJ or 2238.5 J

Now we can calculate the total amount of heat required:
q_total = q1 + q2
q_total = 138.18 J + 2238.5 J
q_total ≈ 2376.68 J or 2.377 kJ

Therefore, approximately 2.377 kJ of heat must be provided to convert 1.00 g of liquid water at 67°C into 1.00 g of steam at 100°C.

To determine the amount of heat required to convert 1.00 g of liquid water at 67°C to steam at 100°C, we need to consider two processes: raising the temperature of the water to its boiling point and then vaporizing it.

First, let's calculate the heat required to raise the temperature of the water from 67°C to its boiling point, which is 100°C. We can use the formula:

Q = m * C * ΔT

Where:
Q is the heat required (in Joules),
m is the mass of the water (in grams),
C is the heat capacity of water (in J/g·°C),
ΔT is the change in temperature (in °C).

In this case, the mass of water is 1.00 g, the heat capacity of water is 4.18 J/g·°C, and the change in temperature is 100°C - 67°C = 33°C.

Q = 1.00 g * 4.18 J/g·°C * 33°C
Q = 137.94 J

Now, we need to calculate the heat required for the phase change from liquid to steam. This is given by the equation:

Q = n * ΔH

Where:
Q is the heat required (in Joules),
n is the number of moles of water,
ΔH is the heat of vaporization (in J/mol).

To determine the number of moles, we use the molar mass of water, which is approximately 18.02 g/mol.

n = m / M
n = 1.00 g / 18.02 g/mol
n = 0.055 mol

Now we can calculate the heat required for vaporization:

Q = 0.055 mol * 40.7 kJ/mol * (1000 J/1 kJ)
Q = 2.2435 kJ

Finally, we add up the heat required for temperature change and vaporization:

Total heat required = 137.94 J + 2.2435 kJ
Total heat required = 2.3814 kJ

Therefore, it requires approximately 2.3814 kilojoules of heat to convert 1.00 g of liquid water at 67°C into 1.00 g of steam at 100°C.