Hot to Find Grams of Water in Solution
When Two Samples of Water are Mixed, what Final Temperature Results?
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Worksheet #2
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Example #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" into 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 Δt, 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 Δt equals 46.8 − x. The colder water goes up in temperature, so its Δt equals x − 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 − x and the distance from x to 14.9 is x − 14.9.
These two distances on the number line represent our two Δt values:
a) the Δt of the warmer water is 46.8 minus x
b) the Δt of the cooler water is x minus 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
However:
q = (mass) (Δt) (Cp)
So:
(mass) (Δt) (Cp) = (mass) (Δt) (Cp)
With qlost on the left side and qgain on the right side.
Substituting values into the above, we then have:
(32.2) (46.8 − x)(4.184) = (32.2) (x − 14.9) (4.184)
Solve for x
Example #2: 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 − 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
Example #3: 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.
Solution:
(60.0) (28.2 − x)(4.184) = (30.0) (x − 8.00) (4.184)
Example #4: A 29.5 g sample of methanol at 208.9 K is mixed with 54.3 g of methanol at 302.3 K. Calculate the final temperature of the mixture assuming no heat is lost to the containers and surroundings. The specific heat of methanol is 2.53 J g¯1 K¯1
Solution:
Let the final temperature be 'x.' Therefore, the Δt for the warmer methanol will be '302.3 − x' and for the colder methanol, it is 'x − 208.9.' Remember, 'x' is the final temperature and it is lower than the warmer methanol and higher than the colder methanol.
Remember:
(1) (mass) (Δt) (Cp) = (mass) (Δt) (Cp)(2) qlost on the left; qgain on the right.
Substituting and solving, we have:
(29.5) (x − 208.9) (2.53) = (54.3) (302.3 − x) (2.53)29.5x − 6162.55 = 16414.89 − 54.3x
83.8x = 22577.44
x = 269.4 K
In case, you're not sure what happened to the 2.53, I simply divided both sides by 2.53 first.
Example #5: A sheet of nickel weighing 10.0 g and at a temperature of 18.0 °C is placed flat on a sheet of iron weighing 20.0 g and at a temperature of 55.6 °C. What is the final temperature of the combined metals? Assume that no heat is lost to the surroundings.
Solution:
This problem requires us to find the specific heats for nickel and iron. To do this, we will use this site. The values given are respectively, 0.54 J g¯1 °C¯1 and 0.46 J g¯1 °C¯1
Notice that the units on the site are kJ kg¯1 K¯1. In addition, notice that I wrote J g¯1 °C¯1. Also, notice that there is no numerical difference when using either specific heat unit (the kJ one or the J one). In other words:
one kJ kg¯1 K¯1 = one J g¯1 °C¯1
The left-hand unit is the IUPAC-approved one; the one on the right-hand side is the one in most common use.
On to the solution:
qlost = qgain
Therefore:
(20.0) (55.6 − x) (0.46) = (10.0) (x − 18.0) (0.54)9.2 (55.6 − x) = 5.4 (x − 18)
511.52 − 9.2x = 5.4x − 97.2
14.6x = 608.72
x = 41.7 °C
Example #6: 10.0 g of steam at 100. °C is mixed with 50.0 g of ice. What is the final temperature of the 60.0 g of liquid water?
Solution:
1) Before launching into the numbers, consider what is going on:
Energy is being liberated when:steam is condensing
hot water is cooling downEnergy is being absorbed when:
ice is melting
cold water is heating upThose two amounts of energy equal each other:
(steam is condensing) + (hot water is cooling down) = (ice is melting) + (cold water is heating up)Each of those four parts will have a calculation associated with it.
2) Here they are:
steam is condensing (10.0 g) (2259 J/g) hot water is cooling down (10.0 g) (100 − x) (4.184 J/g °C) ice is melting (50.0 g) (334 J/g) cold water is heating up (50.0 g) (x − 0) (4.184 J/g °C)
3) The set up to be solved:
[(10.0 g) (2259 J/g)] + [(10.0 g) (100 − x) (4.184 J/g °C)] = [(50.0 g) (334 J/g)] + [(50.0 g) (x − 0) (4.184 J/g °C)]22590 + 4184 − 41.84x = 16700 + 209.2x
251.04x = 10074
x = 40.1 °C
Example #7: How many grams of ice at −17.0 °C must be added to 741 grams of water that is initially at a temperature of 70.0 °C to produce water at a final temperature of 12.0 °C?
Assume that no heat is lost to the surroundings and that the container has negligible mass. The specific heat of liquid water is 4184 J/kg °C and of ice is 2000. J/kg °C. For water the normal melting point is 0.0 °C and the heat of fusion is 334 x 103 J/kg.
Solution:
1) How much energy is lost by the 70.0 °C as it cools to 12.0 °C?
q = (4184 J/kg °C) (0.741 kg) (58.0 °C)q = 173619.264 J
2) The ice absorbing the energy will do three things:
(a) warm up from −17 to 0
(b) melt
(c) warm up (as a liquid) from 0 to 12
3) Each of those three changes has a calculation associated with it:
(a) q = (x) (17.0 °C) (2000. J/kg °C)
(b) q = (334 x 103 J/kg) (x)
(c) q = (x) (12.0 °C) (4184. J/kg °C)
4) The sum of those three calculations is 173619.264 J:
173619.264 J = [(x) (17.0 °C) (2000. J/kg °C)] + [334 x 103 J/kg) (x)] + [(x) (12.0 °C) (4184. J/kg °C)]173619.264 J = [(34000 J/kg) (x)] + [(334000 J/kg) (x)] + [(50208 J/kg) (x)]
(418208 J/kg) (x) = 173619.264 J
x = 173619.264 J / (418208 J/kg)
x = 0.415 kg = 415 g
Example #8: Suppose 45.0 grams of water at 85.0 °C is added to 105.0 grams of ice at 0.0 °C. The molar heat of fusion of water is 6.02 kJ/mol and the specific heat of water is 4.184 J g¯1 °C¯1. Based on these data:
(a) What will be the final temperature of the mixture?
(b) How many grams of ice will melt?
Solution:
1) Determine how much energy is lost by the 45.0 grams of water as it cools down to zero Celsius:
q = (45.0 g) (85.0 °C) (4.184 J g¯1 °C¯1)q = 16003.8 J
2) Determine energy required to melt the entire 105.0 grams of ice:
q = (105.0 g / 18.015 g/mole) (6020 J/mole)q = 35087.43 J
3) The warm water does not provide sufficient energy to melt all the ice. Let us determine how much ice will be melted by the 16003.8 J:
16003.8 J = (x / 18.015 g/mole) (6020 J/mole)x = 47.9 g <--- the answer to (b)
4) Since ice remains in contact with the liquid water, the final temperature of the mixture will be zero degrees Celsius.
Example #9: Suppose 145.0 grams of water at 85.0 °C is added to 105.0 grams of ice at 0.0 °C. The molar heat of fusion of water is 6.02 kJ/mol and the specific heat of water is 4.184 J g¯1 °C¯1. Based on these data:
(a) What will be the final temperature of the mixture?
(b) How many grams of ice will melt?
Solution:
1) Determine how much energy is lost by the 145.0 grams of water as it cools down to zero Celsius:
q = (145.0 g) (85.0 °C) (4.184 J g¯1 °C¯1)q = 51567.8 J
2) Determine energy required to melt the entire 105.0 grams of ice:
q = (105.0 g / 18.015 g/mole) (6020 J/mole)q = 35087.43 J
3) The warm water provides more than enough energy to melt all the ice (there's the answer to part b). How much energy is left over:
51567.8 J − 35087.43 J = 16480.37 J
4) We now have 250.0 g (from 145.0 + 105.0) of liquid water at zero Celsius and we are going to add 16480.37 J. What temperature results?
16480.37 J = (250.0 g) (Δt) (4.184 J g¯1 °C¯1)Δt = 15.8 °C (to three sig figs)
Since the water started at zero, 15.8 °C is the temperature of the entire amount of water at the end. It's the answer to part a.
5) This problem can also be solved with one large equation:
heat used to melt ice + heat use to raise temp = heat lost by warm water[(105.0 g / 18.015 g/mole) (6020 J/mole)] + [(105.0 g) (x − 0 °C) (4.184 J g¯1 °C¯1)] = [(145.0 g) (85.0 °C − x) (4.184 J g¯1 °C¯1)]
35087.43 J + [(439.32 J °C¯1) (x)] = [(606.68 J °C¯1) (85.0 °C − x)]
35087.43 J + [(439.32 J °C¯1) (x)] = 51567.8 J − [(606.68 J °C¯1) (x)]
(1046 J °C¯1) (x) = 16480.37 J
x = 15.8 °C
Example #10: 40.0 grams of ice at −11.0 °C are placed into 295 g of water at 25.0 °C. Assuming no energy is transferred to or from the surroundings, calculate the final temperature of the water after all the ice melts.
Heat capacity of H2O(s) = 37.3 J/(mol K)
Heat capacity of H2O(ℓ) = 75.3 J/(mol K)
Enthalpy of fusion of H2O(s) = 6.02 kJ/mol
Solution:
1) Here's what the ice does:
(a) heats up from −11 to zero (37.3 J/(mol K) is involved here)
(b) it melts while staying at zero (6.02 kJ/mol is involved here)
(c) it heats from zero to some unknown temperature (75.3 J/(mol K) is involved here)
2) The set ups for the three above:
qa = (40 g / 18.0 g/mol) (11 °C) (37.3 J/(mol K)) = 911.78 J3) The 295 g of water will cool down from 25 to the final temp, which is the unknown 'x'qb = (40 g / 18.0 g/mol) (6.02 kJ/mol) = 13.378 kJ = 13378 J
qc = (40 g / 18.0 g/mol) (x) (75.3 J/(mol K) <--- can't solve yet
qd = (295 g / 18.0 g/mol) (25 − x) (75.3J/(mol K)
4) Set qa + qb + qc equal to qd and solve for x:
911.78 J + 13378 J + (40 g / 18.0 g/mol) (x) (75.3 J/(mol K) = (295 g / 18.0 g/mol) (25 − x) (75.3 J/(mol K)14289.78 + 167.33x = 30852.08 − 1234.08x
1401.41x = 16562.3
x = 11.8 °C <--- that's our final temperature
5) See that 11 °C up above? It cancels with all the K values. That's because it's an eleven degree difference and the size of one Celsius is equal to the size of 1 Kelvin. Don't add 273 to all the different K's that are in the problem.
Bonus Example #1: 100.0 mL of water were initially at a temperature of 60.1 °C. After adding ice, the final temperature was 1.9 °C and the final volume 171.0 mL. Calculate the molar enthalpy of fusion for ice.
Solution:
1) The warm water has given up some energy. Let us calculate that amount:
q = (100 g) (58.2 °C) (4.184 J g¯1 °C¯1)q = 24350.88 J
2) That energy did two things:
1) melted 70 g of ice2) raised 70 g of liquid water from 0 to 1.9
3) I'm going to calculate how much energy is involved in the second thing:
q = (71 g) (1.9 °C) (4.184 J g¯1 °C¯1)q = 564.4216 J
4) That energy did not melt ice, so let's get rid of it:
24350.88 − 564.4216 = 23786.4584 J
5) Now, for the molar enthalpy:
23.7864584 kJ / (71 g / 18.015 g/mol) = 6.04 kJ/mol (to three sig figs)
Bonus Example #2: 50.0 g of methanol (CH3OH) at 42.0 °C is mixed with 375 g of water at 10.0 °C. What is the final temperature of the mixture?
Solution:
1) We look up the boiling point of methanol and find it to be 64.7 °C. Since both the methanol and the water remain as liquids, only the specific heats for liquid will be involved in the calculation:
methanol ---> 79.9 J/(mol K)
water ---> 4.184 J/(g K)Note that I deliberately gave the specific heats in different units.
2) The units on all values MUST be the same. I'll change water:
4.184 J 18.015 g ––––––– x ––––––– = 75.37476 J / (mol K) g K mol
3) The heat lost by the warmer methanol goes entirely to heat the cooler water, with no loss to the surroundings:
qmethanol = qwater(moles) (temp. change) (specific heat) = (moles) (temp. change) (specific heat)
(50.0 g / 32.04 g/mol) (42.0 − x) (79.9 J/(mol K)) = (375 g / 18.015 g/mol) (x − 10) (75.37476 J / (mol K))
5236.89369 − 124.687945x = 1569x − 1569
20926.89369 = 1693.687945x
x = 12.4 °C
4) Suppose I had changed the methanol value:
79.9 J 1 mol ––––––– x ––––––– = 2.49376 J / (g K) mol K 32.04 g
5) And solve:
(mass) (temp. change) (specific heat) = (mass) (temp. change) (specific heat)(50.0 g) (42.0 − x) (2.49376 J / (g K)) = (375 g) (x − 10) (4.184 J / (g K))
5236.896 − 124.688x = 1569x − 15690
Which is the same result as above in step 3.
6) In the above calculations, the units on the temperature are Celsius whereas Kelvin is involved in the specific heats. These units will cancel because they are temperature changes and not measurements of a specified temperature.
The end result is that you will get a Kelvin value corresponding to 12.4 °C if you do the calculation with Kelvins:
(50.0 g) (315 − x) (2.49376 J / (g K)) = (375 g) (x − 283) (4.184 J / (g K))39276.72 − 124.688x = 1569x − 444027
483303.72 = 1693.688x
x = 285.3558 K
Which is 12.4 °C when changed to Celsius and rounded off.
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Worksheet #2
Hot to Find Grams of Water in Solution
Source: https://www.chemteam.info/Thermochem/MixingWater.html