### When Two Samples of Water are Mixed, what Final Temperature Results?Problems 1 - 10

Problem #1: 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.

Solution:

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 Δt.

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.

Problem #2: 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.

Solution:

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 (qlost = qgain), we have this:

heat to warm ice 10 degrees + heat to melt ice + heat to warm cold water by unknown amount = 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. That is because the other two parts of the left-hand side of the equation will give Joules as their answer. I used 6020 so that all three parts would be in Joules. If I had used 6.02, then that middle part woud have been in units of kJ.

Notice the use of x − 0 and 60.0 − x. This time, visualize (or write out) the number line used above. The zero is to the left, the 60.0 to the right and the x is in between the 0 and the 60.0.

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 Δt.

Problem #3: 55.0 mL of ethanol (d = 0.789 g/mL) at 8.0 °C is mixed with 55.0 mL of water at 28.2 °C. Assuming no heat is lost, what is the final temperature of the mixture?

Solution:

Two things must be done first: (1) determine how many grams of ethanol are present and (2) determine the specific heat for ethanol.

55.0 mL times the density (0.789 g/mL) gives the mass of ethanol present. This value is 43.395 g. I will use this value and round off at the end of the calculation.

The specific heat value needs to be looked up on the Internet. The most common value found in a search is 2.44 J g¯1 °C¯1, so that is what we will use.

Letting 'x' equal the final temperature, we have this:

(43.395) (x − 8.0) (2.44) = (55.0) (28.2 − x) (4.184)

You may finish this problem.

Problem #4: 900.0 L of palm oil at 47.0 °C is mixed with 200.0 L of palm oil at 76.0 °C. What is the final temperature? Assume no heat is lost to the surroundings.

Solution:

Notice that no specific heat for palm oil is provided. That is because none is needed. Also, a bit of Internet sleuthing shows that the density of palm oil is about 0.9 g/cm3.

1) Let us convert 900.0 L and 200.0 L to grams using the density:

(900,000 mL) (0.9 g/cm3) = 8.10 x 105 g

(200,000 mL) (0.9 g/cm3) = 1.80 x 105 g

Remember that 1 mL equals 1 cm3.

2) We will let 'x' equal the final temperature, so the two temperature changes are:

the 900 L goes up in temperature, so use 'x − 47.0'

the 200 L goes down in temperature, so use '76.0 − x'

3) Keeping in mind that the heat lost by the 200 L equals the heat gained by the 900 L, we write:

(1.80 x 105) (76.0 − x) = (8.10 x 105) (x − 47.0)

x = 52.3 °C

I did not write the specific heat values because they would just cancel out. This is because we have palm oil on both sides of the equation. If the substances were different (see examples above), the two specific heats would have to be included.

Also, when I did this problem on the calculator, I dropped the 105 portion of the mass. In addition, if you were to ignore the use of the density and use, say 200 and 900, you'd get 51.0 °C for the answer.

Problem #5: 364 g of water at 34.0 °C is added to ice at 0.0 °C. If the final temperature of the system (which you can assume is isolated) is 0.0 °C, determine how much ice melted. The specific heat of water is 4186 J/kg ⋅ °C. The latent heat of fusion for water is 335,000 J/kg.

Solution:

1) Determine energy lost by warm water:

q = (mass) (Δt) (Cp)

q = (0.364 kg) (34.0 °C) (4186 J/kg ⋅ °C)

q = 51805.936 J

2) Determine how much ice is melted by 51805.936 J:

51805.936 J / 335,000 J/kg = 0.155 kg = 155 g (to three sig figs)

Problem #6: How much ice (in grams) would have to melt to lower the temperature of 353.0 mL of water from 26.0 °C to 6.0 °C?

Solution:

1) Determine heat lost by cooling water from 26.0 to 6.0:

q = (353.0 g) (20.0 °C) (4.184 J g¯1 °C¯1)

q = 29539.04 J

Note the silent conversion from volume of water to mass of water using the density of 1.00 g/mL.

2) The heat lost by the warm water does two things:

a) melt an unknown mass of ice
b) take the same mass of melted ice from zero Celsius to 6 °C

3) We can express it thusly:

29539.04 J = (6020 J/mol) (x / 18.015 g/mol) + (x) (6.0 °C) (4.184 J g¯1 °C¯1)

29539.04 J = 334.166x + 25.104x

x = 82.2 g

Problem #7: An unknown volume of water at 18.2 °C is added to 27.8 mL of water at 33.6 °C. If the final temperature is 23.5 °C, what was the unknown volume? (Assume that no heat is lost to the surroundings; density of water is 1.00 g/mL.)

Solution:

Note the our 'x' in this problem will be an unknown mass of water. Since we know the starting temperatures and the final temperature, we can calculate the Δt values.

1) Calculate the two Δt values:

cooler water ⇒ 23.5 − 18.2 = 5.2
warmer water ⇒ 33.6 − 23.5 = 10.1

2) qlost = qgain:

(x) (5.3) (4.184) = (27.8) (10.1) (4.184)

5.3x = 280.78

x = 53.0 g

This is 53.0 mL of water.

Problem #8: A student mixed 6.00 g of ice at −3.4 °C with 1.00 g of steam at 103.0 °C. What is the final temperature of this mixture?

Solution:

heat gained in warming up = heat lost in cooling down

heat gained by ice + heat to melt ice + heat to raise water temp = heat lost by steam + heat lost as steam condenses + heat lost as water cools

(6.00 g) (3.4 °C) (2.06 J/g °C) + (6.00 g / 18.0 g/mol) (6020 J/mol) + (6.00 g) (x − 0 °C) (4.184 J/g °C) = (1.00 g) (3.0 °C) (2.02 J/g °C) + (1.00 g / 18.0 g/mol) (40700 J/mol) + (1.00 g) (100 − x °C) (4.184 J/g °C)

42.024 + 2006.667 + 25.104x = 6.06 + 2261.11 + 418.4 − 4.184x

29.288x = 636.879

x = 21.7 °C

Problem #9: A cube of ice is taken from the freezer at −5.5 °C and placed in a 98.0 g aluminum calorimeter filled with 326.0 g of water at room temperature of 20.0 °C. The final situation is observed to be all water at 15.0 °C. What was the mass of the ice cube?

Solution:

1) The ice does three things:

(a) heat up from −5.5 to 0
(b) melt at 0
(c) heat up from 0 to 15

2) The three calculations are:

qa = (x) (5.5 °C) (2.02 J / g °C)
qb = (x / 18.015 g/mol) (6.02 kJ/mol)
qc = (x) (15.0 °C) (4.184 J / g °C)

3) The water and calorimeter do this:

(d) water ---> cool down from 20 to 15
(e) calorimeter ---> cool down from 20 to 15

3) The two calculations are:

qd = (326.0 g) (5.0 °C) (4.184 J / g °C)
qe = (98.0 g) (5.0 °C) (0.900 J / g °C)

4) The amount of heat absorbed by the ice is equal to the heat lost by the water and Al calorimeter:

qa + qb + qc = qd + qe

However, something important must be done and it's associated with qb.

(x) (5.5 °C) (2.02 J / g °C) + (x / 18.015 g/mol) (6020 J/mol) + (x) (15.0 °C) (4.184 J / g °C) = (326.0 g) (5.0 °C) (4.184 J / g °C) + (98.0 g) (5.0 °C) (0.900 J / g °C)

I changed 6.02 kJ/mol to 6020 J/mol.

5) Finishing:

11.11x + 334.166x + 62.76x = 6819.92 + 441

408.036x = 7260.92

x = 17.8 g

Problem #10: 20.0 g of steam at 100.0 °C is bubbled into a mixture of 50.0 g of water and 200.0 g of ice at exactly 0 °C. All of the steam condenses to water. What is the composition of the system at the end?

Latent heat of fusion of ice = 334.16 J/g
Latent heat of vaporization of water = 2259.2 J/g

Solution:

1) Let's condense the 20.0 g of steam and then cool it to zero:

(20.0 g) (2259.2 J/) = 45184 J
(20.0 g) (100 °C) (4.184 J/g °C) = 8368 J

45184 + 8368 = 53552 J

2) We now have a system of 70.0 g of water at 0 °C and 200.0 g of ice with 53552 J "floating" about. (The 70.0 comes from the 50.0 already there plus the 20.0 of the steam that condensed.) Let's see how much ice will be melted by the 53552 J:

53552 J / 334.16 J/g = 160.26 g

3) How much ice is left?

200.0 g − 160.3 g = 39.7 g <--- part 1 of the answer

4) How much liquid is present?

70.0 g + 160.3 g = 230.3 g <--- part 2 of the answer

Problem #11: Suppose that 35.46 g of ice at −6.8 °C is placed in 69.12 g of water at 91.0 °C in a perfectly insulated vessel. Calculate the final temperature. (The molar heat capacity for ice is 37.5 J K¯1 mol¯1 and that for liquid water is 75.3 J K¯1 mol¯1. The molar enthalpy of fusion for ice is 6.01 kJ/mol.)

Solution:

1) The ice will absorb energy from the warm water. That energy gained makes the ice do three things:

1) heat from −6.8 °C to 0 °C
2) melt at 0 °C
3) heat from 0 °C to the final temperature

2) Here the equations for the three energy-gaining behaviors of the ice:

q1 = (35.46 g / 18.0 g/mol) (6.8 °C) (37.5 J K¯1 mol¯1) <--- ice heating up
q2 = (35.46 g / 18.0 g/mol) (6010 J/mol) <--- ice melting
q3 = (35.46 g / 18.0 g/mol) (x − 0 °C) (75.3 J K¯1 mol¯1) <--- melted ice heating up to final temp

3) The warm water will cool down (it loses energy). Here is the equation for the warm water cooling down:

q4 = (69.12 g / 18.0 g/mol) (91.0 °C − x) (75.3 J K¯1 mol¯1)

4) You solve this equation:

q1 + q2 + q3 = q4

Note that only one variable has been used. The 'x' is the final temperature.

5) Substitute into the above and do a bit of algebra.

[(35.46/18.0) (6.8) (37.5)] + [(35.46/18.0) (6010)] + [(35.46/18.0) (x) (75.3)] = (69.12/18.0) (91.0 °C − x) (75.3)

[(1.97) (6.8) (37.5)] + [(1.97) (6010)] + [(1.97) (x) (75.3)] = (3.84) (91.0 − x) (75.3)

502.35 + 11839.7 + 148.341x = 26312.832 − 289.152x

437.493x = 13970.782

x = 31.9 °C

6) Two comments:

1) I changed 6.01 kJ/mol to 6010 J/mol so as to align all the units to Joules.
2) I ignored the °C and K issues on the temps. That's because all the temps involved are changes in temp, not a single, fixed temperature. Since 1 °C is the same size as 1 K, everything cancels.

Problem #12: A 40.0 g ice cube at −19.0 °C are placed into 275 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.

Needed information:

Heat capacity of H2O(s) = 37.7 J/(mol K)
Heat capacity of H2O(ℓ) = 75.3 J/(mol K)
Enthalpy of fusion of H2O = 6.02 kJ/mol

Solution:

1) The ice cube does three things:

(a) heat from −19 to zero C
(b) melt at zero C
(c) heat up from zero to the final temp

2) The liquid water does one thing. It cools down from 25.0 °C to the final temp.

3) The heat gained by the ice equals the heat lost by the 25.0 °C of water.

4) The energy involved in the ice cube:

qa = (40.0 g / 18.0 g/mol) (19 K) (37.7 J/(mol K))
qb = (40.0 g / 18.0 g/mol) (6020 J/mol) <--- note how I used J instead of kJ
qc = (40.0 g / 18.0 g/mol) (x − 0) (75.3 J/(mol K))

5) The 25.0 °C water does this:

qd = (275 g / 18.0 g/mol) (25 − x) (75.3 J/(mol K))

This term:

(25 − x)

is the change in temp from 25 down to the final temp. I put x − 0 in the other one to be explicit that it's a change to x from a starting temp of 0.

6) We want this set up:

qa + qb + qc = qd

(40.0 g / 18.0 g/mol) (19 K) (37.7 J/(mol K)) + (40.0 g / 18.0 g/mol) (6020 J/mol) + (40.0 g / 18.0 g/mol) (x − 0) (75.3 J/(mol K)) = (275 g / 18.0 g/mol) (25 − x) (75.3 J/(mol K))

1591.78 + 13377.78 + 167.33x = 28760.42 − 1150.42x

1317.75x = 13790.86

x = 10.5°C

Problem #13: A 200. gram liquid sample of Alcohol X is prepared at −10.0 °C. The sample is then added to 300. g of water at 20.0 °C in a sealed styrofoam container. When thermal equilibrium is reached, the temperature of the alcohol-water solution is 10.0 °C. What is the specific heat capacity of the alcohol? Assume the sealed container is an isolated system. The specific heat capacity of water is 4.184 kJ kg¯1 °C¯1.

Solution:

1) The key assumption is that all of the heat lost by the warmer water is gained by the cooler alcohol. In an equation:

qwater = qalcohol

or

(mass of water) (temp change of water) (specific heat of water) = (mass of alcohol) (temp change of alcohol) (specific heat of alcohol)

2) Energy lost by the water:

(0.300 kg) (10.0 °C) (4.184 kJ kg¯1 °C¯1) = 12.552 kJ

3) 12.552 kJ will be gained by the alcohol:

12.552 kJ = (0.200 kg) (20.0 °C) (x)

x = 3.1375 kJ kg¯1 °C¯1

To three significant figures, the answer is 3.14 kJ kg¯1 °C¯1

4) Often you will find steps 2 and 3 above combined as follows:

(0.300 kg) (10.0 °C) (4.184 kJ kg¯1 °C¯1) = (0.200 kg) (20.0 °C) (x)

Problem #14: A 41.58 g sample of water at 73.3 °C is added to a sample of water at 23.6 °C in a constant-pressure calorimeter. If the final temperature of the combined water is 38.9 °C and the heat capacity of the calorimeter is 26.3 J °C¯1, calculate the mass of the water originally in the calorimeter.

Solution:

1) The energy from the warm water will do two things:

(a) heat up the cooler water from 23.6 °C to 38.9 °C
(b) heat up the calorimeter from 23.6 °C to 38.9 °C

2) In so doing, the temperature of the warmer water will drop from 73.3 °C to 38.9 °C. Since the warmer water is the only source of heat energy, let up determine how much heat is lost:

q = (mass) (temp. change) (specific heat)

q = (41.58 g) (34.4 °C) (4.184 J g¯1 °C¯1) = 5984.59 J

3) From step 1 above, here are the set-ups for (a) and (b):

qa = (m) (15.3 °C) (4.184 J g¯1 °C¯1)
qb = (26.3 J °C¯1) (15.3 °C) = 402.39 J

4) The sum of (a) and (b) equals 5984.59 J:

5984.59 J = (m) (15.3 °C) (4.184 J g¯1 °C¯1) + 402.39 J

64.0152m = 5582.2

m = 87.2 g (to three sig figs)

Problem #15: Equal masses of hot water and ice are mixed together. All of the ice melts and the final temperature of the mixture is 0 °C. If the ice was originally at 0 °C, what was the initial temperature of the hot water?

Solution:

1) Let us assume we have 18.0 g of ice and 18.0 g of hot water present.

2) The key is to realize that the only thing the ice did is melt, it did not change its temperature. So, let us calculate the amount of heat needed to melt our 18.0 g (or, 1.00 mole) of ice:

q = (6.02 kJ/mol) (1.00 mol) = 6.02 kJ

3) The only source of heat is the hot water, which provides 6020 J (I converted the 6.02 kJ to J.) of heat. Let us calculate the temperature change of 18.0 g of hot water as it loses 6020 J of heat:

6020 J = (18.0 g) (x) (4.184 J g¯1 °C¯1)

x = 79.9 °C

The hot water was at an initial temperature of 79.9 °C and went to 0 °C as the ice melted (and stayed at 0 °C the entire time).

Problem #16: In real calorimeters, most of the heat released by the bomb is absorbed by water, but a certain amount is also absorbed by the metal and insulation surrounding the water tank. A certain calorimeter absorbs 24 J/°C. If 50.0 g of 52.7 °C water is mixed with the calorimeter's original 50.0 g of 22.3 °C water, what will be the final temperature of the mixture?

Solution:

1) We have some warm water that will lose heat. We can set up a calculation to determine the amount:

(50.0 g) (4.184 J/g °C) (52.7 °C − T)

The T is the final temp and (52.7 °C − T) is the temp change.

2) All the heat that is lost goes two places:

(a) the cool water
(b) the calorimeter itself

3) Here's the set up for (a):

(50.0 g) (4.184 J/g °C) (T − 22.3 °C)

Remember, the T (final temp) is higher than 22.3 °C so the temp change starts at 22.3 °C and goes up to T.

4) Here's the set up for (b):

(24 J/°C) (T − 22.3 °C)

The calorimeter is at the same starting temp (22.3 °C) as the cool water and it winds up at the same T (final temp) so it goes through the same change in temp (called the ΔT)

5) Now, we set up an equation . . . :

(50.0 g) (4.184 J/g °C) (52.7 − T) = (50.0 g) (4.184 J/g °C) (T − 22.3) + (24 J/°C) (T − 22.3)

The left side is the heat coming out of the warm water and the right side is the heat going into two destinations (the calorimeter itself and the cool water).

6) . . . and solve it:

(50.0) (4.184) (52.7 − T) = (50.0) (4.184) (T − 22.3) + (24) (T − 22.3)

11024.84 − 209.2T = (209.2T − 4665.16) + (24T − 535.2)

11024.84 − 209.2T = 233.2T − 5200.36

442.4T = 16225.2

T = 36.7 °C (to three sig figs)

Bonus Problem: Water at 0 °C was placed in a dish inside a vessel maintained at low pressure by a vacuum pump. After a quantity of water had evaporated, the reminder froze. If 9.31 g of ice at 0 °C was obtained, how much liquid water must have evaporated? (At 0 °C, ΔHfusion = 6.02 kJ/mol; ΔHvap = 45.054 kJ/mol)

Solution:

1) The water that froze (9.31 g) released:

(9.31 g) (6.02 kJ/mole) (1 mole / 18.015 g) = 3.11108 kJ

2) This energy is given up by the water vaporizing:

energy = (moles of water that evaporated) (heat of vaporization)

3.1108 kJ = (mass of water that evaporates) (45.054 kJ/mole)

0.069046 moles of water evaporated

(0.069046 mol) (18.015 g/mol) = 1.24 g (to three sig figs)

Note the use of ΔHvap for 0 °C. At 100 °C, it equals 40.7 kJ/mol. (Here is the source I used for the 45.054 value.)