If 1.20g of steam at 100.0 celcius condenses into 38.5 g of water, initially at 27 celcius what is final temperature of water
Note the correct spelling of celsius.
On condensation, the steam loses
q = mass steam x heat condensation.
That water will be at 100 C.
Then
[mass steam water x specific heat H2O x (Tfinal-Tinitial)] + [mass cool water x specific heat H2O x (Tfinal-Tinitial)] - q from condensation.
Substitute the numbers and solve for Tf. I estimated the Tf as about 45 or so.
To find the final temperature of the water, we can use the principle of conservation of energy.
First, let's calculate the heat gained by the water when it condenses the steam. We can use the equation:
q = m × c × ΔT
where:
q is the heat gained or lost by the substance,
m is the mass of the substance,
c is the specific heat capacity of the substance, and
ΔT is the change in temperature.
For the water, the mass is 38.5 g, the specific heat capacity is approximately 4.18 J/g°C (specific heat capacity of water), and the change in temperature is the final temperature (unknown) minus the initial temperature (27°C).
So, the equation becomes:
q = 38.5 g × 4.18 J/g°C × (final temperature - 27°C)
Next, let's calculate the heat lost by the steam when it condenses into water. The amount of heat lost by the steam is equal to the amount of heat gained by the water. The equation remains the same:
q = m × c × ΔT
For the steam, the mass is 1.20 g, the specific heat capacity is approximately 2.03 J/g°C (specific heat capacity of steam), and the change in temperature is 100°C (the boiling point of water at atmospheric pressure) minus the final temperature.
So, the equation becomes:
q = 1.20 g × 2.03 J/g°C × (100°C - final temperature)
Since the heat gained by the water is equal to the heat lost by the steam, we can set the two equations equal to each other:
38.5 g × 4.18 J/g°C × (final temperature - 27°C) = 1.20 g × 2.03 J/g°C × (100°C - final temperature)
Now, we can solve this equation to find the final temperature of the water.