Thermochemistry
determine the final temperature if 45.67 kJ of heat energy is removed from 18.5 g of H2O (g) at 122 degrees Celsius
useful information
sp. heat H2O (s) = 2.03 J/g(degree C)
sp heat H2O (l) = 4.18 J/g(degree C)
Sp heat H2O (g) = 2.01 J/g(degree C)
enthalpy H Fus = 334 J/g
enthalpy H vap = 2260 J/g
I would do this in steps.
First, heat released in cooling from 122 to 100.
q = mass x specific heat steam x (deltaT)
q = 18.5 x 2.01 x 22= about 818 approximately. You should confirm ALL of these numbers.
q = heat released on condensing the steam.
q = mass x heat vap = 18.5 x 2260 = about 42000,
42000 + 818 = about 43,000 J.
That's close to what you have at 45,670 J. The difference is 45,670-43000 = about 3,000 (again, redo these numbers)
Then
-3000 = mass H2O x specific heat liquid water x (Tfinal-Tinitial) and solve for Tfinal.
Thank You
for my final temperature I got around
83.1 degrees celsius
How did you know that the temperature dropped down to 100 degrees C? I have the same exact problem, except the heat is being added, not taken away. And the initial temperature is 73.5 degrees C.
To determine the final temperature when heat energy is removed from a substance, we can use the equation:
q = m × C × ΔT
where:
q is the heat energy transferred (in Joules),
m is the mass of the substance (in grams),
C is the specific heat capacity of the substance (in J/g·°C), and
ΔT is the change in temperature (in °C).
In this case, we are given the heat energy (45.67 kJ), the mass of H2O gas (18.5 g), and the initial temperature (122 °C). We need to find the final temperature (Tf).
First, we need to convert the given heat energy into Joules:
45.67 kJ × 1000 J/kJ = 45670 J
Since we are considering H2O in the gas phase, we will use the specific heat capacity for H2O (g), which is 2.01 J/g·°C.
Using the equation q = m × C × ΔT, we can rearrange it to solve for ΔT:
ΔT = q / (m × C)
Substituting the given values:
ΔT = 45670 J / (18.5 g × 2.01 J/g·°C) ≈ 1266 °C
Now we can calculate the final temperature:
Tf = Initial temperature - ΔT
= 122 °C - 1266 °C
= -1144 °C
Therefore, the final temperature would be approximately -1144 °C. Note that this result seems unusual since water would typically be in a different phase at such a low temperature, so it is important to double-check the problem setup and assumptions.