A quantity of steam (750 g) at 122°C is condensed, and the resulting water is frozen into ice at 0°C. How much heat was removed?
q1 = heat removed in changing T from steam at 122 to steam at 100.
q1 = mass steam x specific heat steam x (Tfinal-Tinitial)
q2 = heat removed on condensation at 100 C to water at 100 C.
q2 = mass steam x delat H(heat condensation)
q3 = heat removed in changing T of liquid water from 100 C to 0 C.
q3 = mass H2O x specific heat H2O x (Tfinal-Tinitial)
q4 = heat removed on freezing liquid water at 0C to solid ice at 0C.
q4 = mass H2O x delta H fusion.
Total is q1 + q2 + q3 + q4.
To determine the amount of heat removed in this process, there are two steps to consider:
1. Heat removed during the condensation of steam
2. Heat removed during the freezing of water into ice
Step 1: Heat removed during the condensation of steam
To calculate the heat removed during the condensation, we need to use the formula:
q = m × ΔHv
Where:
q is the heat removed
m is the mass of the substance (750 g)
ΔHv is the enthalpy of vaporization
The enthalpy of vaporization of water is approximately 2,260 J/g.
Calculating:
q = 750 g × 2,260 J/g
q = 1,695,000 J (or 1.695 × 10^6 J)
So, the heat removed during the condensation of steam is 1,695,000 J.
Step 2: Heat removed during the freezing of water into ice
To calculate the heat removed during the freezing, we need to use the formula:
q = m × ΔHf
Where:
q is the heat removed
m is the mass of the substance (750 g)
ΔHf is the enthalpy of fusion
The enthalpy of fusion of water is approximately 334 J/g.
Calculating:
q = 750 g × 334 J/g
q = 250,500 J (or 2.505 × 10^5 J)
So, the heat removed during the freezing of water into ice is 250,500 J.
Therefore, the total heat removed in this process is the sum of the heat removed during condensation and freezing:
Total heat removed = Heat removed during condensation + Heat removed during freezing
Total heat removed = 1,695,000 J + 250,500 J
Total heat removed = 1,945,500 J (or 1.945 × 10^6 J)
Therefore, the total amount of heat removed is 1,945,500 J.
To determine the amount of heat removed, we need to calculate the energy change during the process. The conversion of steam to water involves the change of state from gas to liquid, and then the conversion of water to ice involves the change of state from liquid to solid.
1. First, let's calculate the heat required to convert steam to water at 122°C, which involves the process of condensation. To do this, we need to use the latent heat of vaporization for water.
The latent heat of vaporization for water is 2.26 × 10^6 J/kg.
Given:
Mass of steam (m) = 750 g = 0.75 kg
Initial temperature (Ti) = 122°C = 122 + 273 = 395 K (Kelvin)
Heat required for condensation (Q1) can be calculated using the formula:
Q1 = m × Lv
where Lv is the latent heat of vaporization.
Q1 = 0.75 kg × 2.26 × 10^6 J/kg
Q1 ≈ 1.695 × 10^6 J
2. Next, let's determine the heat required to convert water at 0°C to ice at 0°C. This involves the process of freezing. To do this, we'll use the latent heat of fusion for water.
The latent heat of fusion for water is 3.34 × 10^5 J/kg.
Given:
Mass of water (m) = 0.75 kg (since the amount of water is the same as the initial amount of steam)
Initial temperature (Ti) = 0°C = 0 + 273 = 273 K (Kelvin)
Heat required for freezing (Q2) can be calculated using the formula:
Q2 = m × Lf
where Lf is the latent heat of fusion.
Q2 = 0.75 kg × 3.34 × 10^5 J/kg
Q2 ≈ 2.505 × 10^5 J
3. Finally, the total heat removed (Qtotal) is the sum of Q1 and Q2:
Qtotal = Q1 + Q2
Qtotal ≈ 1.695 × 10^6 J + 2.505 × 10^5 J
Qtotal ≈ 1.945 × 10^6 J
Therefore, approximately 1.945 × 10^6 Joules of heat were removed during the process.