At 1 atm, how much energy is required to heat 79.0 g of H2O(s) at –20.0 °C to H2O(g) at 121.0 °C?
figure the heat to take ice to OC
mc deltatemp
figure the heat to melt the ice at OC
Heatfusion*mass
figure the heat to warm it from 0C to 100C
mc deltatemap
figure the heat to vaporize the water at 100C
Heatvaporizatioin*mass
Figure the heat to heat the steam from 100C to 121C.
mc deltaTemp
add them together.
remember, ice, water, steam all have different specific heat capacities c.
q1 = heat to raise T solid H2O from -20 to ice at zero C.
q1 = mass H2O x specific heat ice x (Tf-Ti) where Tf = 0 and Ti = -20
q2 = heat to melt ice.
q2 = mass ice x heat fusion.
q3 = heat to raise T of liquid H2O from zero C to 100 C.
q3 = mass H2O x specific heat liquid H2O x (Tfinal-Tinitial) where Tf = 100 and Ti = 0.
q4 = heat to change liquid H2O at 100 C to steam at 100 C.
q4 = mass H2O x heat vaporization.
q5 = heat to raise steam from 100 C to 121 C.
q5 = mass steam x specific heat steam x (Tfinal-Tinitial) where Tf is 121 C and Ti is 100 C.
Total q = q1 + q2 + q3 + q4 + q5
To determine the amount of energy required to heat the given mass of water from -20.0 °C to 121.0 °C, we need to calculate the energy at each step and then add them together.
First, we need to calculate the energy required to heat the ice from -20.0 °C to its melting point, which is 0 °C. We can use the specific heat capacity of ice (c = 2.09 J/g·°C) for this calculation:
Q1 = m × c × ΔT1
= 79.0 g × 2.09 J/g·°C × (0 °C - (-20.0 °C))
= 79.0 g × 2.09 J/g·°C × 20.0 °C
= 3,315.6 J
Next, we calculate the energy required to melt the ice to liquid water at 0 °C. The heat of fusion (Hf) for water is 333.55 J/g:
Q2 = m × Hf
= 79.0 g × 333.55 J/g
= 26,336.45 J
Finally, we calculate the energy required to heat the liquid water from 0 °C to 121.0 °C using the specific heat capacity of water (c = 4.18 J/g·°C):
Q3 = m × c × ΔT2
= 79.0 g × 4.18 J/g·°C × (121.0 °C - 0 °C)
= 79.0 g × 4.18 J/g·°C × 121.0 °C
= 40,008.02 J
To get the total energy required, we add up the three values:
Total Energy = Q1 + Q2 + Q3
= 3,315.6 J + 26,336.45 J + 40,008.02 J
= 69,660.07 J
Therefore, at 1 atm, the amount of energy required to heat 79.0 g of H2O(s) at -20.0 °C to H2O(g) at 121.0 °C is approximately 69,660.07 J.