Go to calculating final temperature when mixing metal and water: problems 1 - 15

Go to calculating final temperature when mixing two samples of water

These problems are exactly like mixing two amounts of water, with one small exception: the specific heat values on the two sides of the equation will be different. The water specific heat will remain at 4.184, but the value for the metal will be different. These values are tabulated and lists of selected values are in most textbooks.

**Example #1:** Determine the final temperature when a 25.0 g piece of iron at 85.0 °C is placed into 75.0 grams of water at 20.0 °C.

First some discussion, then the solution. Forgive me if the points seem obvious:

a) The colder water will warm up (heat energy "flows" in to it). The warmer metal will cool down (heat energy "flows" out of it).

b) The whole mixture will wind up at theSAMEtemperature. This is very, very important.

c) The energy which "flowed" out (of the warmer water) equals the energy which "flowed" in (to the colder water)

**Solution Key Number One:** We start by calling the final, ending temperature 'x.' Keep in mind that BOTH the iron and the water 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 iron 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, so its Δt equals x minus 20.0.

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 to 20.0 is x minus 20.0

**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:

q_{lost}= q_{gain}

So, by substitution, we then have:

(25.0) (85.0 − x)(0.45) = (75.0) (x − 20.0) (4.184)

Solve for x

Please note the use of the specific heat value for iron. It is 0.45 J per gram degree Celsius.

Noting that 75/25 = 3, we arrive at:

38.25 − 0.45x = 12.552x − 251.04

then

13.002x = 289.29

The answer is

22.25 °C

if you aren't too fussy about significant figures.

Note that the iron drops quite a bit in temperature, while the water moves only a very few (2.25 in this case) degrees. This is the typical situation in this type of problem.

**Example #2:** Determine the final temperature when 10.0 g of aluminum at 130.0 °C mixes with 200.0 grams of water at 25.0 °C.

There is no difference in calculational technique from Example #1. Please note the starting temperature of the metal is above the boiling point of water. In reality, the sample may vaporize a tiny amount of water, but we will assume it does not for the purposes of the calculation.

**Solution:**

1) Set up the numbers:

q_{aluminum}= q_{water}(10) (130 − x) (0.901) = (200.0 )(x − 25) (4.18)

2) Noting that 200/10 = 20, I get:

117.13 − 0.901x = 83.6x − 2090x = 26.12 °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 a slightly different answer. Make sure you check with your teacher as to the values of the various constants that he/she wishes for you to use.

**Example #3:** Determine the final temperature when 20.0 g of mercury at 165.0 °C mixes with 200.0 grams of water at 60.0 °C. (C_{p} for Hg = 0.14 J per gram degree Celsius.)

We will ignore the fact that mercury is liquid. it does not dissolve in water.

**Solution:**

(20.0) (165.0 − x) (0.14) = (200.0) (x − 60.0) (4.18)Noting that 200/20 = 10, I get:

23.1 - 0.14x = 41.8x − 2508

41.94x = 2531.1

x = 60.35 °C

Note that the water moves only 0.35 of one degree. Keep in mind that there is a large amount of water compared to the mercury AND that it takes a great deal more energy to move water one degree as compared to the same amount of mercury moving one degree.

**Example #4:** 10.0 g of water is at 59.0 °C. If 3.00 g of gold at 15.2 °C is placed in the calorimeter, what is the final temperature of the water in the calorimeter? (The specific heat of gold is 0.128 J/g °C.)

**Solution:**

1) Set up the following:

q_{water}= q_{gold}(10.0) (59.0 − x) (4.184) = (3.00) (x − 15.2) (0.128)

2) Algebra:

2468.56 − 41.84x = 0.384x − 5.836842.224x = 2474.3968

x = 58.6 °C

Note that, in this case, the water cools down and the gold heats up. This is opposite to the most common problem of this type, but the solution technique is the same.

**Example #5:** 105.0 mL of H_{2}O is initially at room temperature (22.0 °C). A chilled steel rod (2.00 °C) is placed in the water. If the final temperature of the system is 21.5 °C, what is the mass of the steel bar? (specific heat of water = 4.184 J/g °C; specific heat of steel = 0.452 J/g °C)

**Solution:**

(105.0 g) (0.5 °C) (4.184 J °C^{-1}g^{-1}) = (x) (19.5 °C) (0.452 J °C^{-1}g^{-1})x = 24.9 g

**Example #6:** A pure gold ring and pure silver ring have a total mass of 15.0 g. The two rings are heated to 62.4 °C and dropped into a 13.6 mL of water at 22.1 °C. When equilibrium is reached, the temperature of the water is 23.9 °C. (Assume a density of 0.998 g/mL for water.)

(a) What is the mass of the gold ring?

(b) What is the mass of the silver ring?

Comment: specific heat values are available in many places on the Internet and in textbooks. Here is an example.

**Solution:**

1) The basic equation to be used is this:

Heat gained or lost = (mass) (temperature change) (specific heat)In more compact form: q = (m) (Δt) (C

_{p})There will be three of them used below.

2) The two masses associated with the gold and the silver rings:

Let y = grams gold

Therefore, 15.0 − y = grams silver

3) The mass of water:

13.6 mL x 0.998 g/mL = 13.5728 g

4) Heat gained by water:

q = (13.5728 g) x (1.8 °C) (4.184 J g¯^{1}°C¯^{1})q = 102.2195 J

The 1.8 is arrived at thusly: 23.9 − 22.1

5) As the gold ring and the silver ring cool down, they liberate energy that sums to 102.2195 J. The sum can be expressed thusly:

(y) (38.5 °C) (0.129 J g¯^{1}°C¯^{1}) + (15 − y) (38.5 °C) (0.235 J g¯^{1}°C¯^{1}) = 102.2195 JRemember, a change of 1 °C equals a change of 1 K. That means 0.129 J g¯

^{1}°C¯^{1}is the same thing as 0.129 J g¯^{1}K¯^{1}The 38.5 was arrived at in the same manner as the 1.8 just above.

6) Algebra!

4.9665y + (15 − y) (9.0475) = 102.21954.9665y + 135.7125 – 9.0475y = 102.2195

-4.081y = -33.493

y = 8.21 g Au

15.0 − 8.21 = 6.79 g Ag

**Example #7:** A ring has a mass of 8.352 grams and is made of gold and silver. When the ring has been heated to 94.52 °C and then dropped into 13.40 g water at 20.00 °C, the temperature of the water after thermal equilibrium was reached was 22.00 °C. What is the percent by mass of gold and silver in the ring?

**Solution:**

The heat given off by the silver plus the heat given off by the gold equals the heat absorbed by the water.Set the mass of silver to be 'x.' That means that the mass of the gold is 8.352 minus x

(x) (72.52 °C) (0.235 J/g °C) + (8.352 − x) (72.52 °C) (0.129 J/g °C) = (13.40 g) (2.00 °C) (4.184 J/g °C)

The 72.52 comes from 94.52 minus 22.00

17.0422x + (8.352 − x) (9.35508) = 112.1312

17.0422x + 78.13362816 − 9.35508x = 112.1312

7.68712x = 33.99757184

x = 4.422667 g

mass percent of gold: (4.422667 / 8.352) * 100 = 52.95%

mass percent of silver: 100.00 − 52.95 = 47.05%

**Example #8:** A 74.0 g cube of ice at −12.0 °C is placed on a 10.5 kg block of copper at 23.0 °C, and the entire system is isolated from its surroundings. After a few minutes, the ice has melted and the temperature of the system has reached equilibrium. Calculate the final temperature of the system.

Comment: none of the appropriate constants are supplied. You would have to look up the proper values, if you faced a problem like this. If you examine your sources of information, you may find they differ slightly from the values I use. This is common. Many of the values used have been determined experimentally and different sources will often contain slightly different values.

**Solution:**

1) Ice goes from −12 to 0:

q = (74.0 g) (12.0 °C) (2.06 J/g °C) = 1829.28 J

2) Ice melts:

q = (74.0 g / 18.0 g/mol) (6.02) = 24.7489 kJ

3) Liquid water goes through an unknown temperature increase to the final value of x

q = (74.0 g) (x − 0) (4.184 J/g °C) = 309.616x

4) The copper loses heat and drops in temperature to the final value of x:

q = (10500 g) (23.0 − x) (0.385 J/g °C)

5) The amount of heat lost by the copper equals the heat gained by the water:

(10500) (23.0 − x) (0.385) = 1829.28 + 24748.9 + 309.616xNotice how the kJ from the ice melting is used as J rather than kJ. The copper mass is expressed in grams rather than kg.

(23.0 − x) (4042.5) = 26578.18 + 309.616x

92977.5 − 4042.5x = 26578.18 + 309.616x

66399.32 = 4352.116x

x = 15.2568 °C

x = 15.2 °C (to three sig figs, I followed the rule for rounding with 5)

**Example #9:** How many grams of water can be heated form 25.0 °C to 35.0 °C by the heat released from 85.0 g of iron that cools from 85.0 °C to 35.0 °C? The specific heat of iron is 0.450 J/g °C

**Solution:**

1) How much heat is lost by the iron?

temp change ---> 85.0 °C − 35.0 0°C = 50.0 °Cq = (mass) (temp. change) (specific heat)

q = (85.0 g) (50.0 °C) (0.450 J/g °C)

q = 1912.5 J

2) Assume all 1912.5 J go to heat water:

q = (mass) (temp. change) (specific heat)1912.5 J = (x) (10.0 °C) (4.184 J/g °C)

x = 45.7 g

Note that the specific heat for liquid water is not provided in the text of the problem.

3) This problem could have been solved by setting the two equations equal and solving for 'x.':

(85.0 g) (50.0 °C) (0.450 J/g °C) = (x) (10.0 °C) (4.184 J/g °C)

**Example #10:** Find the mass of liquid H_{2}O at 100.0 °C that can be boiled into gaseous H_{2}O at 100.0 °C by a 130.0 g Al block at temp 402.0 °C? Assume the aluminum is capable of boiling the water until its temperature drops below 100.0 °C. The heat capacity of aluminum is 0.900 J g¯^{1} °C¯^{1} and the heat of vaporization of water at 100 °C is 40.65 kJ mol¯^{1}

**Solution:**

1) Heat that Al can lose in going from its initial to its final temperature:

q = m Δt C_{p}q = (130.) g (302.0 °C) (0.900 J g¯

^{1}°C¯^{1}) = 35334 J = 35.334 kJ35.334 kJ of heat are available to vaporize water.

2) Use 35.334 kJ and the heat of vaporization of water to calculate moles and then mass of water vaporized:

q = (ΔH) (moles)35.334 kJ = (40.65 kJ/mol) (x)

x = 0.869225 moles H

_{2}Omass H

_{2}O = (0.869225 mol) (18.015 g/mol) = 15.659 gTo three sig figs, this is 15.6 g

**Bonus Example:** A 250. gram sample of metal is heated to a temperature of 98.0 °C. It is placed in 100. grams of water in a brass calorimeter cup with a brass stirrer. The total mass of the cup and the stirrer is 50.0 grams. The initial teperature of the water, stirrer, and calorimeter is 20.0 °C. The final equilibrium temperature of the system is 30.0 °C. What is the specific heat of the metal sample? (The specific heat of brass is 0.0920 cal g¯^{1} C¯^{1}.)

**Solution:**

1) The amount of heat given off by the sample of metal is absorbed by (a) the water and (b) the brass calorimeter & stirrer. How much heat was trapped by the water?

q = m Δt C_{p}q = (100. g) (10.0 °C) (1.00 g cal g¯

^{1}C¯^{1})q = 1000 cal

Note: 1.00 g cal g¯

^{1}C¯^{1}is the specific heat for liquid water.

2) How much heat was absorbed by the brass calorimeter and stirrer?

q = m Δt C_{p}q = (50.0 g) (10.0 °C) (0.092 cal g¯

^{1}C¯^{1})q = 54.28 cal

3) Total heat given off by the metal sample:

1000 cal + 54.28 cal = 1054.28 cal

4) Determine the specific heat of the metal sample:

q = m Δt C_{p}1054.28 cal = (250. g) (68.0 °C) (C

_{p})C

_{p}= 0.062 g cal g¯^{1}C¯^{1}

Go to calculating final temperature when mixing metal and water problems 1 - 15

Go to calculating final temperature when mixing two samples of water