At 1 atm, how much energy is required to heat 35.0 g of H2O(s) at –12.0 °C to H2O(g) at 127.0 °C?
Do this in steps and add each step to find total q.
1. Within a phase the formula is
q = mass x specfic heat x (Tfinal-Tinitial).
For example to heat 20g liquid water at zero C to 100 C (all in the same phase)
q = 20 x specific heat x (100-0) = ?
2. At the phase change (solid to liquid or liquid to vapor) it is
q = mass x heat fusion at freezing point or
q = mass x heat vaporization at the boiling point.
Then add the q values together to find the total.
To determine the amount of energy required to heat 35.0 g of H2O(s) at -12.0 °C to H2O(g) at 127.0 °C, we will use the formula:
q = m * ΔT * C
where:
q is the amount of energy (in Joules),
m is the mass of the substance (in grams),
ΔT is the change in temperature (in Celsius), and
C is the specific heat capacity of the substance (in J/g°C).
First, we need to calculate the energy required to heat the solid ice from -12.0 °C to its melting point at 0 °C using the formula:
q1 = m * ΔT1 * C1
where:
m is the mass of H2O(s) (35.0 g),
ΔT1 is the change in temperature (-12.0 °C - 0 °C = -12.0 °C), and
C1 is the specific heat capacity of ice, which is 2.09 J/g°C.
Using the given values, we can calculate q1:
q1 = (35.0 g) * (-12.0 °C) * (2.09 J/g°C) = -878.4 J
Next, we need to calculate the energy required to convert the solid ice at 0 °C to liquid water at 0 °C. This process is known as the heat of fusion and can be calculated using the formula:
q2 = m * Hf
where:
m is the mass of H2O (35.0 g),
Hf is the heat of fusion for water, which is 334 J/g.
Calculating q2:
q2 = (35.0 g) * (334 J/g) = 11,690 J
Then, we need to calculate the energy required to heat the liquid water from 0 °C to the boiling point at 100 °C using the formula:
q3 = m * ΔT3 * C3
where:
m is the mass of H2O (35.0 g),
ΔT3 is the change in temperature (100 °C - 0 °C = 100 °C), and
C3 is the specific heat capacity of liquid water, which is 4.18 J/g°C.
Calculating q3:
q3 = (35.0 g) * (100 °C) * (4.18 J/g°C) = 14,630 J
Lastly, we need to calculate the energy required to convert the liquid water at 100 °C to water vapor at 100 °C. This process is known as the heat of vaporization and can be calculated using the formula:
q4 = m * Hv
where:
m is the mass of H2O (35.0 g),
Hv is the heat of vaporization for water at 100 °C, which is 2260 J/g.
Calculating q4:
q4 = (35.0 g) * (2260 J/g) = 79,100 J
Now, we can calculate the total energy required by summing up the values obtained from q1, q2, q3, and q4:
Total energy = q1 + q2 + q3 + q4
Total energy = -878.4 J + 11,690 J + 14,630 J + 79,100 J
Total energy = 104,542 J
Therefore, at 1 atm, it would require approximately 104,542 J of energy to heat 35.0 g of H2O(s) at -12.0 °C to H2O(g) at 127.0 °C.