what is the change in heat when 23.5g water vapor at 100C condenses to liquid 18.50C
To find the change in heat when water vapor condenses to a liquid, we need to calculate the heat lost by the water vapor.
The equation that relates heat transfer (q), mass (m), and change in temperature (ΔT) is:
q = m * c * ΔT
Where:
q is the change in heat
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 (final temperature - initial temperature)
First, we need to determine the initial and final temperatures of the water vapor.
Given:
Initial temperature (Ti) = 100°C
Final temperature (Tf) = 18.50°C
Next, we calculate the change in temperature (ΔT):
ΔT = Tf - Ti
ΔT = 18.50°C - 100°C
ΔT = -81.50°C
Now, let's determine the specific heat capacity of water vapor. The specific heat capacity of water vapor is approximately 1.996 J/g°C.
Using the equation, we can find the change in heat:
q = m * c * ΔT
q = 23.5g * 1.996 J/g°C * -81.50°C
Calculating the change in heat:
q ≈ -37,677.32 J
Therefore, the change in heat when 23.5g of water vapor at 100°C condenses to a liquid at 18.50°C is approximately -37,677.32 Joules. The negative sign indicates that heat is being lost by the water vapor during condensation.
q1 = heat released on condensation of steam at 100 C to liquid water at 100 C.
q1 = mass steam x heat vaporization.
q2 = heat released on cooling from liquid water from 100 C to 18.5 C.
q2 = mass water at 100 x specific heat H2O x (Tfinal-Tinitial)
Total Q = q1 + q2.