at 1 atm how much energy is required to heat 67 grams of H2O(s) at -22 degrees C to H2O(g) at 159 degrees C

q1 = heat to move ice at -22 to the melting point of zero C.

q1 = mass x specific heat ice x (Tfinal-Tinital)

q2 = heat to convert solid water (ice) at zero C to liquid water at zero C.
q2 = mass x heat fusion.

q3 = heat to move liquid water at zero C to the boiling point at 100 C.
q3 = mass x specific heat water x (Tfinal-Titial)

q4 = heat to convert liquid water at 100 C to steam at 100 C.
q4 = mass water x heat vaporization

q5 = heat to move steam at 100 C to steam at 158 C.
q5 = mass x specific heat steam x (Tfinal-Tinitial)

Total Q = q1 + q2 + q3 + q4 + q5.
Notice that you have only two formulas to memorize. One is for changing phase (at melting and boiling points); the other is with the same phase (from -22 to zero)(from zero to 100) (from 100 to 158). That should keep things simple for you.

To calculate the amount of energy required to heat a substance, we can use the equation:

Q = m * c * ΔT

Where:
Q is the amount of energy in Joules (J)
m is the mass of the substance in grams (g)
c is the specific heat capacity of the substance in J/g°C
ΔT is the change in temperature in degrees Celsius (°C)

First, let's calculate the energy required to heat the ice from -22°C to 0°C:

Q1 = m * c1 * ΔT1

Since we are converting H2O(s) (ice) to H2O(l) (liquid), we use the specific heat capacity of ice (c1 = 2.09 J/g°C).

Q1 = 67 g * 2.09 J/g°C * (0°C - (-22)°C)
= 67 g * 2.09 J/g°C * 22°C
= 3101.86 J

Next, let's calculate the energy required to heat the liquid water from 0°C to 100°C:

Q2 = m * c2 * ΔT2

For liquid water, we use the specific heat capacity of water (c2 = 4.18 J/g°C).

Q2 = 67 g * 4.18 J/g°C * (100°C - 0°C)
= 27994.6 J

Finally, let's calculate the energy required to convert the liquid water to water vapor:

Q3 = m * Hvap

Where Hvap is the heat of vaporization of water.
The heat of vaporization of water is approximately 40.7 kJ/mol or 40.7 J/g.

Q3 = 67 g * 40.7 J/g
= 2726.9 J

Now, let's add up the three amounts of energy to get the total energy required:

Q_total = Q1 + Q2 + Q3
= 3101.86 J + 27994.6 J + 2726.9 J
= 33823.36 J

Therefore, the amount of energy required to heat 67 grams of H2O(s) at -22°C to H2O(g) at 159°C is approximately 33,823.36 J (or 33.82336 kJ) at 1 atm of pressure.