Assume that the pressure is one atmosphere and determine the heat in joules required to produce 3.23 kg of water vapor at 100.0 °C, starting with (a)3.23 kg of water at 100.0 °C and (b)3.23 kg of liquid water at 0.0 °C

a) what exactly is the latent heat of vaporization?

b) Adding to a), what exactly is the specific heat content of substances.

Look those terms up.

To determine the heat required to produce water vapor, we need to use the concept of enthalpy. Enthalpy (H) is the heat content of a system at constant pressure, and it is typically measured in joules (J).

(a) To determine the heat required to produce water vapor starting with 3.23 kg of water at 100.0 °C:
1. First, we need to determine the enthalpy change from liquid water at 100.0 °C to water vapor at 100.0 °C.
- This can be done by using the formula: ΔH = m × Cp × ΔT, where ΔH is the enthalpy change, m is the mass, Cp is the specific heat capacity, and ΔT is the change in temperature.
- The specific heat capacity of water is approximately 4.18 J/g°C.
- The change in temperature is from 100.0 °C (liquid) to 100.0 °C (gas), which is 0 °C.
- The mass is given as 3.23 kg, so convert it to grams by multiplying by 1000.

2. Calculate the enthalpy change:
ΔH = (mass in grams) × Cp × ΔT
ΔH = (3.23 kg × 1000 g/kg) × (4.18 J/g°C) × (0 °C)

Note: Since the change in temperature is 0 °C, the enthalpy change in this case would be equal to zero. This is because there is no change in state (liquid to vapor) at the same temperature.

Therefore, the heat required to produce 3.23 kg of water vapor at 100.0 °C, starting with 3.23 kg of water at 100.0 °C is zero joules.

(b) To determine the heat required to produce water vapor starting with 3.23 kg of liquid water at 0.0 °C:
1. First, we need to determine the enthalpy change from liquid water at 0.0 °C to water vapor at 100.0 °C.
- We can use the enthalpy formula mentioned above: ΔH = m × Cp × ΔT.
- The specific heat capacity of water is the same as before.

2. Calculate the first enthalpy change from liquid water at 0.0 °C to liquid water at 100.0 °C:
ΔH1 = (mass in grams) × Cp × ΔT
ΔH1 = (3.23 kg × 1000 g/kg) × (4.18 J/g°C) × (100 °C - 0 °C)

3. Next, we need to determine the enthalpy change from liquid water at 100.0 °C to water vapor at 100.0 °C. This is known as the heat of vaporization (ΔHvap).
- The heat of vaporization for water at 100.0 °C is approximately 40.7 kJ/mol or 40.7 J/g.
- We need to convert the mass from grams to moles (since the heat of vaporization is given per mole) by dividing by the molar mass of water.

4. Calculate the enthalpy change from liquid water at 100.0 °C to water vapor at 100.0 °C:
ΔH2 = (mass in grams) / (molar mass) × (heat of vaporization)
ΔH2 = (3.23 kg × 1000 g/kg) / (18.015 g/mol) × (40.7 J/g)

5. Finally, calculate the total heat by summing up the enthalpy changes:
Total heat = ΔH1 + ΔH2

By following this process, you can determine the heat required to produce 3.23 kg of water vapor at 100.0 °C, starting with 3.23 kg of liquid water at 0.0 °C.