Return to Mixing Water tutorial
Sample Problem #1:
Determine the final temperature when 45.0 g of water at 20.0 °C mixes with 22.3 grams of water at 85.0 °C.
Solution:
We start by calling the final, ending temperature 'x.' Keep in mind that BOTH water samples will wind up at the temperature we are calling 'x.' Also, make sure you understand that the 'x' we are using IS NOT the Δt, but the FINAL temperature. This is what we are solving for.
The warmer water goes down from to 85.0 to x, so this means its Δt equals 85.0 minus x. The colder water goes up in temperature (from 20.0 to the ending temperature), so its Δt equals x minus 14.9.
That last paragraph may be a bit confusing, so let's compare it to a number line:
To compute the absolute distance, it's the larger value minus the smaller value, so 85.0 to x is 85.0 minus x and the distance from x (the larger value) to 20.0 (the smaller value) is x - 20.0.
The energy amount going out of the warm water is equal to the energy amount going into the cool water. This means:
qlost = qgain
So, by substitution, we then have:
(22.3) (85.0 - x)(4.184) = (45.0) (x - 20.0) (4.184)
Solve for x
Sample Problem #2:
Determine the final temperature when 30.0 g of water at 8.00 °C mixes with 60.0 grams of water at 28.2 °C.
(60.0) (28.2 - x)(4.184) = (30.0) (x - 8.00) (4.184)