When Two Samples of Water are Mixed, what Final Temperature Results?
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Sample Problem #1:
Determine the final temperature when 32.2 g of water at 14.9 °C mixes with 32.2 grams of water at 46.8
°C.
This is problem 8a from Worksheet #2.
First some discussion, then the solution. Forgive me if the points seem obvious:
1) The colder water will warm up (heat energy "flows" in to it). The warmer water will cool down (heat
energy "flows" out of it).
2) The whole mixture will wind up at the SAME temperature. This is very, very important.
3) The energy which "flowed" out (of the warmer water) equals the energy which "flowed" in (to the
colder water)
This problem type becomes slightly harder if a phase change is involved. For this example, no phase
change. What that means is that only the specific heat equation will be involved
Solution Key Number One: 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 t, but the FINAL temperature. This is what we are solving for.
The warmer water goes down from to 46.8 to x, so this means its t equals 46.8 minus x. The colder
water goes up in temperature, so its t 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 46.8 to x is 46.8 minus
x and the distance from x to 14.9 is x - 14.9.
Solution Key Number Two: 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:
(32.2) (46.8 - x)(4.184) = (32.2) (x - 14.9) (4.184)
Solve for x
Sample Problem #2:
Determine the final temperature when 10.0 g of steam at 100.0 °C mixes with 500.0 grams of water at
25.0 °C.
This problem is like 9 and 10 in Worksheet #2 with one difference. The sample problem has steam and
the worksheet problem have ice. The ONLY differences are the numbers used and the fact that ice is
associated with "cold water" side of the equation. The techniques are the same.
The "warm water" in this case is going to do two things:
a) as a gas, condense at 100.0 °C to liquid water
b) as a liquid, the temperature goes down some unknown amount
The colder water will receive all the energy for this to happen. Keeping in mind that the heat lost by the
warmer must equal the heat gained by the colder, we have this:
heat lost to cooler water by condensing steam + heat used to warm cooler water = heat gained by the
cooler water
Please note again that there are TWO sources of heat energy (steam condensing, then the warm water
cooling down). The total energy lost by both will equal the heat energy gained by the cooler water.
Here are the numbers:
[(40700)(10.0/18.0)] + [(10)(100 - x)(4.18)] = (500.0)(x - 25)(4.18)
Notice I used 40700 J rather than 40.7 kJ for the molar heat of vaporization.
Proceeding with the solution, I get:
2131.8x = 79041.11
and so, x = 37.1 °C.
Keep in mind that 'x' was identified with the final temperature, NOT the t.
Also, I did this problem with 4.18. Doing it with 4.184 gives slightly different numbers. Make sure you
check with your teacher as to the values of the various constants that he/she wishes for you to use.
Sample Problem #3:
Determine the final temperature when 18.0 g of ice at -10.0 °C mixes with 275.0 grams of water at 60.0
°C.
This is like problem 8e and several following in Worksheet #2.
The "cold water" in this case is going to do three things:
a) as a solid, warm up from -10 to zero
b) all 18 g will melt
c) as a liquid, the temperature goes up some unknown amount
The warmer water must provide all the energy for this to happen. Keeping in mind that the heat lost by the
warmer must equal the heat gained by the colder, we have this:
heat to warm ice + heat to melt ice + heat to warm cold water = heat lost by the warm water
Here are the numbers:
[(18)(10)(2.06)] + [(6020)(18.0/18.0)] + [(18)(x - 0)(4.184)] = (275.0)(60.0 - x)(4.184)
Notice I used 6020 J rather than 6.02 kJ for the molar heat of fusion.
Proceeding with the solution, I get:
1225.912x = 62945.2
and so, x = 51.3 °C.
Keep in mind that 'x' was identified with the final temperature, NOT the t.
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