joules released when 125 g of steam at 100 ∘C condenses and cools to liquid at 20.0 ∘C
How much energy is released when 500 g of 373 K steam is condensed into liquid water?
124
Well, that depends. Are you a fan of steamy jokes?
To find the energy released when steam condenses and cools to liquid, you need to calculate the heat energy lost during the process. The heat energy lost can be calculated using the formula:
Q = m * ΔT * c
Where:
Q is the amount of heat energy lost
m is the mass
ΔT is the change in temperature
c is the specific heat capacity
In this case, the mass of the steam is given as 125 g. The change in temperature (ΔT) is the difference between the initial temperature (100 °C) and the final temperature (20 °C).
The specific heat capacity of water is approximately 4.18 J/g°C. However, during the process of phase change, the specific heat capacity is different. For water changing from steam to liquid at 100 °C, the specific heat capacity is called the heat of vaporization (ΔHvap). For water changing from liquid to solid at 0 °C, the specific heat capacity is called the heat of fusion (ΔHfus).
To calculate the heat energy lost during the process of steam condensing and cooling, we need to consider both the heat of vaporization and the heat of fusion. The heat of vaporization for water is approximately 2260 J/g, and the heat of fusion is approximately 334 J/g.
First, calculate the heat energy lost during the phase change from steam to liquid:
Q1 = m * ΔHvap
Q1 = 125 g * 2260 J/g
Next, calculate the heat energy lost during the cooling process from 100 °C to 20 °C:
Q2 = m * ΔT * c
Q2 = 125 g * (100 °C - 20 °C) * 4.18 J/g°C
Finally, add Q1 and Q2 together to get the total heat energy lost:
Q_total = Q1 + Q2
Keep in mind that this calculation assumes that there are no heat losses to the surroundings.
q1 = heat released on steam cooling from 125 C to steam at 100C.
q1 = mass steam x specific heat steam x (Tfinal - Tinitial) where Tf is 100 and Ti is 125.
q2 = heat released on condensation of steam at 100
C to liquid water at 100 C.
q2 = mass steam x heat of condensation.
q3 = heat released on cooling water from 100 C to water at 20 C.
q = mass H2O x specific heat liquid water x (Tfinal-Tinitial) where Tf is 20 and Ti is 100.
Total heat released is sum of individual qs.