At 1 atm, how much energy is required to heat 49.0 g of H2O(s) at –12.0 °C to H2O(g) at 169.0 °C?
q1 = heat to raise T of solid to zero C.
q1 = mass ice x specific heat ice x delta T.
q2 = heat to melt ice at zero C to water at zero C.
q2 = mass ice x heat fusion
q3 = heat to raise T of liquid water from zero C to liquid water at 100 C.
q3 = mass water x specific heat H2O x delta T.
q4 = heat to boil water at 100 C to steam at 100 C.
q4 = mass water x heat vaporization.
q5 = heat to raise T of steam from 100 C to 169 C.
q5 = mass steam x specific heat steam x delta T.
Total = q1 + q2 + q3 + q4 + q5.
To elaborate on DrBob's answer (nearly a decade later):
There are five things to calculate here:
Q1: the energy needed to raise the ice to 0°C
Q2: the energy needed to melt the ice
Q3: the energy needed to raise the water to 100°C
Q4: the energy needed to boil the water
Q5: the energy needed to raise the steam to 169°C
Whatever homework you're using should give you the heat transfer constants so I'll be using the following but adjust your calculations accordingly if your constants are different:
Enthalpy of fusion at 0°C: 333.6J/g
Enthalpy of vaporization of at 100°C: 2257 J/g
Specific heat of solid H2O (ice): 2.087 J/(g*°C)
Specific heat of liquid H2O (water): 4.184 J/(g*°C)
Specific heat of gas H2O (steam): 2.000 J/(g*°C)
Q1 [heating the ice]: (49g)(2.087 J/g*°C)(12°C) = 1227.156 J
Q2 [melting the ice]: (49g)(333.6 J/g) = 16346.4 J
Q3 [heating the water to 100°C]: (49g)(4.184 J/g*°C)(100°C) = 20501.6 J
Q4 [boiling the water]: (49g)(2257 J/g) = 110593 J
Q5 [heating the steam to 169°C]: (49g)(2 J/g*°C)(69) = 6762
Q1 + Q2 + Q3 + Q4 + Q5 = 155430.156 J or 155.4 kJ
To calculate the energy required to heat a substance, you need to use the formula:
q = m * C * ΔT
where:
q is the amount of energy required (in Joules)
m is the mass of the substance (in grams)
C is the specific heat capacity of the substance (in J/g°C)
ΔT is the change in temperature (in °C)
To calculate the specific heat capacity of water, we'll use the following values:
C(sol) = 4.184 J/g°C (specific heat capacity for liquid water)
C(vap) = 2.03 J/g°C (specific heat capacity for water vapor)
Now let's break down the heating process into two steps: heating the ice from -12.0 °C to 0 °C, and then converting it to steam at 100 °C, and finally heating the steam from 100 °C to 169 °C.
First step: Heating the ice to 0 °C:
q1 = m * C(sol) * (0 - (-12.0))
Second step: Melting the ice at 0 °C to liquid water at 100 °C:
q2 = m * ΔH(fusion)
Third step: Heating the liquid water from 100 °C to steam at 100 °C:
q3 = m * C(vap) * (100 - 100)
Fourth step: Heating the steam to 169 °C:
q4 = m * C(vap) * (169 - 100)
Finally, sum up all the individual heat values:
q_total = q1 + q2 + q3 + q4
Let's calculate it step by step:
Step 1:
q1 = 49.0 g * 4.184 J/g°C * (0 - (-12.0)) = 2437.376 J
Step 2: (to calculate ΔH(fusion), we'll use the value 6.01 kJ/mol)
First, convert the mass of water to moles:
n = m/M
where M is the molar mass of water (18.0153 g/mol)
n = 49.0 g / 18.0153 g/mol ≈ 2.718 mol
Then, calculate the heat required to melt the ice:
q2 = n * ΔH(fusion)
q2 = 2.718 mol * 6.01 kJ/mol = 16.34638 kJ = 16.34638 * 1000 J = 16346.38 J
Step 3:
q3 = 49.0 g * 2.03 J/g°C * (100 - 100) = 0 J
Step 4:
q4 = 49.0 g * 2.03 J/g°C * (169 - 100) = 3341.03 J
Finally, add up all the heat values:
q_total = q1 + q2 + q3 + q4
q_total = 2437.376 J + 16346.38 J + 0 J + 3341.03 J = 22124.786 J
Therefore, the energy required to heat 49.0 g of H2O(s) at –12.0 °C to H2O(g) at 169.0 °C at 1 atm is approximately 22124.786 Joules.