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?
To find the amount of energy required to heat the given amount of ice (H2O(s)) to steam (H2O(g)), we need to consider two steps:
Step 1: Heating the ice from -12.0 °C to 0 °C
Step 2: Melting the ice at 0 °C (at equilibrium)
Step 3: Heating the water from 0 °C to 100 °C
Step 4: Boiling the water from 100 °C to 127.0 °C (at equilibrium)
Step 5: Heating the steam from 100 °C to 127.0 °C
We will calculate the energy required for each of these steps and sum them up to get the total energy required.
Step 1: To heat the ice from -12.0 °C to 0 °C, we use the formula:
q1 = m * C1 * ΔT1
Where:
q1 is the heat energy,
m is the mass of the substance,
C1 is the specific heat capacity of the substance at that temperature range,
ΔT1 is the change in temperature.
Given:
m = 35.0 g
C1 (for ice) = 2.09 J/g°C (Specific heat capacity of ice in this temperature range is constant)
ΔT1 = 0 °C - (-12.0 °C) = 12.0 °C
q1 = 35.0 g * 2.09 J/g°C * 12.0 °C = 878.04 J
Step 2: To melt the ice at 0 °C, we use the formula:
q2 = m * ΔHf
Where:
q2 is the heat energy required for melting,
m is the mass of the substance,
ΔHf is the heat of fusion (specific latent heat) of the substance.
Given:
m = 35.0 g
ΔHf (for ice) = 334 J/g (Specific latent heat of fusion for ice at its melting point)
q2 = 35.0 g * 334 J/g = 11,690 J
Step 3: To heat the liquid water from 0 °C to 100 °C, we use the formula as in Step 1:
q3 = m * C2 * ΔT2
Where:
q3 is the heat energy,
m is the mass of the substance,
C2 (for liquid water) = 4.184 J/g°C (Specific heat capacity of water in this temperature range is constant)
ΔT2 = 100 °C - 0 °C = 100 °C
q3 = 35.0 g * 4.184 J/g°C * 100 °C = 14,684 J
Step 4: To boil the water from 100 °C to 127.0 °C, we use the formula:
q4 = m * ΔHv
Where:
q4 is the heat energy required for vaporization,
m is the mass of the substance,
ΔHv (for water) = 40.7 J/g (Specific latent heat of vaporization for water at its boiling point)
q4 = 35.0 g * 40.7 J/g = 1,424.5 J
Step 5: To heat the steam from 100 °C to 127.0 °C, we use the formula as in Step 1:
q5 = m * C3 * ΔT3
Where:
q5 is the heat energy,
m is the mass of the substance,
C3 (for steam) = 2.03 J/g°C (Specific heat capacity of steam in this temperature range is constant)
ΔT3 = 127.0 °C - 100 °C = 27.0 °C
q5 = 35.0 g * 2.03 J/g°C * 27.0 °C = 1,606.95 J
Now, we can sum up all the individual energies to get the total energy required:
Total energy = q1 + q2 + q3 + q4 + q5
Total energy = 878.04 J + 11,690 J + 14,684 J + 1,424.5 J + 1,606.95 J
Total energy = 30,283.49 J
Therefore, at 1 atm, approximately 30,283.49 J of energy is required to heat 35.0 g of H2O(s) at -12.0 °C to H2O(g) at 127.0 °C.
To determine the energy required to heat a substance from one temperature to another, you can use the equation:
q = m * c * ΔT
Where:
q = heat energy (in Joules)
m = mass of the substance (in grams)
c = specific heat capacity of the substance (in J/g °C)
ΔT = change in temperature (in °C)
In this case, you need to heat 35.0 g of water (H2O) from -12.0 °C to 127.0 °C at a constant pressure of 1 atm.
First, let's calculate the energy required to heat the ice (H2O(s)) from its initial temperature of -12.0 °C to its melting point of 0 °C.
ΔT1 = 0 °C - (-12.0 °C) = 12.0 °C
Next, we'll calculate the energy required to melt the ice at 0 °C and convert it to water (H2O(l)):
q1 = m * ΔHf
Where:
ΔHf = heat of fusion of water (in J/g)
The specific heat of fusion of water is 334 J/g. Thus,
q1 = 35.0 g * 334 J/g = 11,690 J
Now, let's calculate the energy required to heat the liquid water (H2O(l)) from 0 °C to its boiling point of 100 °C:
ΔT2 = 100 °C - 0 °C = 100 °C
Next, we'll calculate the energy required to vaporize the water (H2O(l)) at 100 °C and convert it to steam (H2O(g)):
q2 = m * ΔHv
Where:
ΔHv = heat of vaporization of water (in J/g)
The specific heat of vaporization of water is 2260 J/g. Thus,
q2 = 35.0 g * 2260 J/g = 79,100 J
Lastly, we need to calculate the energy required to heat the steam (H2O(g)) from 100 °C to 127.0 °C:
ΔT3 = 127.0 °C - 100 °C = 27.0 °C
Now, adding up all the individual energy values:
q_total = q1 + q2 + q3
q_total = 11,690 J + 79,100 J + (35.0 g * specific heat capacity of steam * ΔT3)
The specific heat capacity of steam is approximately 2.03 J/g °C.
q_total = 11,690 J + 79,100 J + (35.0 g * 2.03 J/g °C * 27.0 °C)
Calculating this expression will give you the total energy required to heat 35.0 g of H2O(s) at -12.0 °C to H2O(g) at 127.0 °C at a constant pressure of 1 atm.