How much heat is needed to convert 2.00 kg of liquid ammonia at -40.0 oC to its vapor at -11.0 oC? The specific heat of liquid ammonia, NH3 (l), is 4.70 J/g-K and that of the vapor NH3 (g), is 2.20 J/g-K. The enthalpy of vaporization of ammonia is 23.3 kJ/mol and its boiling point is -33.3 oC. Please help!!

Add the following:

Heat to go from -40C to -33C
Heat to vaporize at -33C
Heat to go from -33 to-11C as a vapor.

To solve this problem, we need to consider the different stages of the process:

1. Heating the liquid ammonia from -40.0 oC to its boiling point at -33.3 oC.
2. Vaporizing the liquid ammonia at its boiling point.
3. Heating the vapor from the boiling point to the final temperature of -11.0 oC.

Step 1: Heating the liquid from -40.0 oC to -33.3 oC
To determine the amount of heat required to raise the temperature of a substance, we use the formula:

Q = m * C * ΔT,

where Q is the heat required, m is the mass of the substance, C is the specific heat capacity, and ΔT is the change in temperature.

Given:
- Mass (m) = 2.00 kg
- Specific heat capacity (C) of liquid ammonia (NH3(l)) = 4.70 J/g-K
- Temperature change (ΔT) = (-33.3 - (-40.0)) oC = 6.7 oC

First, we need to convert the mass from kilograms to grams:
- Mass (m) = 2.00 kg * 1000 g/kg = 2000 g

Now we can calculate the heat required for the first step:
Q1 = m * C * ΔT
= 2000 g * 4.70 J/g-K * 6.7 oC

Step 2: Vaporizing the liquid ammonia
To vaporize the liquid ammonia at its boiling point, we need to use the enthalpy of vaporization (ΔHvap). The enthalpy of vaporization represents the amount of heat required to convert one mole of a substance from its liquid state to its gas state.

Given:
- Enthalpy of vaporization (ΔHvap) of ammonia = 23.3 kJ/mol

Since the given enthalpy of vaporization is in kilojoules per mole and we have the mass of ammonia in grams, we need to convert the mass to moles:

- Molar mass of ammonia (NH3) = 17.03 g/mol

The number of moles (n) can be calculated using the formula:

n = m / M,

where n is the number of moles, m is the mass in grams, and M is the molar mass.

n = 2000 g / 17.03 g/mol

Now we can calculate the heat required for the second step:
Q2 = n * ΔHvap
= (2000 g / 17.03 g/mol) * 23.3 kJ/mol * 1000 J/kJ

Step 3: Heating the vapor from the boiling point to the final temperature of -11.0 oC
For this step, we use the same formula as in step 1.

Given:
- Specific heat capacity (C) of ammonia vapor (NH3(g)) = 2.20 J/g-K
- Temperature change (ΔT) = (-11.0 - (-33.3)) oC = 22.3 oC

We can calculate the heat required for the third step:
Q3 = m * C * ΔT
= 2000 g * 2.20 J/g-K * 22.3 oC

Finally, we can find the total heat required by adding up the heat for each step:
Total heat required = Q1 + Q2 + Q3

Calculate each step and add up the values to find the answer.