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.

**Problem #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.

Brief Description of the following problems:

a) 2 - 4: cold water + hot metal; asks for final temperature of mix (the most common problem type)

b) 5 & 6: hot water + cold metal; asks for final temperature of mix (the reverse of the most common problem type)

c) 7 & 8: cold water + hot metal; asks for final temperature of mix (incorporates use of a calorimeter constant)

d) 9 & 10: cold water + hot metal; one asks for final temp., one asks for mass (two different metals are involved)

e) 11 & 12: cold water + hot metal; asks for starting temperature of metal

f) 13: cold water + hot metal; asks for starting temperature of water

**Problem #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 sample problem #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.

**Problem #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.

**Problem #4:** A 12.48 g sample of an unknown metal is heated to 99.0 °C and then was plunged into 50.0 mL of 25.0 °C water. The temperature of the water rose to 28.1°C.

(a) How many joules of energy did the water absorb?

(b) How many joules of energy did the metal lose?

(c) What is the heat capacity of the metal?

(d) What is the specific heat capacity of the metal?

The definitions for heat capacity and specific heat capacity may be found here.

1) Solution to (a):

q = (50.0 g) (3.1 °C) (4.181 J g¯^{1}°C¯^{1}) = 648.52 J

I used 50.0 g because the density of water is 1.00 g/mL and I had 50.0 mL of water.

2) Solution to (b):

q = 648.52 J

We assume all heat absorbed by the water was lost by the metal. We assume no loss of heat energy to the outside during the transfer.

3) Solution to (c):

648.52 J / 74.0 °C = 8.76 J/°C (or 8.76 J/K)

4) Solution to (d):

(50.0 g) (3.1 °C) (4.181 J g¯^{1}°C¯^{1}) = (12.48 g) (74.0 °C) (x)Solve for x.

**Problem #5:** 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.

**Problem #6:** 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

**Problem #7:** A 610. g piece of copper tubing is heated to 95.3 °C and placed in an insulated vessel containing 45.0 g of water at 36.5 °C. Assuming no loss of water and heat capacity for the vessel of 10.0 J/K, what is the final temperature of the system (C_{p} of copper = 0.387 J/g-K)?

**Solution:**

1) This problem can be summarized thusly:

q_{lost by copper}= q_{gained by water}+ q_{gained by calorimeter}

2) Therefore:

(610. g) (95.3 °C - x) (0.387 J g^{-1}K^{-1}) = (45.0 g) (x - 36.5 °C) (4.184 J g^{-1}°C^{-1}) + [(10.0 J/K) (x - 36.5 °C)]Comment: The K and the °C cancel because the °C in this problem is a temperature difference (not one single specific value) and the "size" of one K = one °C.

3) Isn't algebra fun?

22497.471 - 236.07x = 198.28x - 7237.22424.35x = 29734.691

x = 70.1 °C

**Problem #8:** A 45.0 g sample of a substance at 55.0 °C (s = 1.66 cal/g °C) was placed into a coffee-cup calorimeter (c = 4.20 cal/°C) which contained 50.0 g of ethyl alcohol at 25.0 degrees c (s = 0.590 cal/g °C). What is the resulting temperature?

Comment prior to solution: note that this problem uses calories rather than Joules. This does not affect the solution technique.

**Solution:**

1) Set up the following equation:

(mass substance) (Δt substance) (C_{p}substance) = (mass alcohol) (Δt alcohol) (C_{p}alcohol) + (calorimeter constant) (Δt alcohol)There is an implicit assumption that the alcohol and the calorimeter start at the same temperature. This is a very safe assumption.

2) Insert appropriate values:

(45.0 g) (55.0 - x) (1.66 cal/g °C) = (50.0 g) (x - 25.0) (0.590 cal/g °C) + (4.20 cal/°C) (x - 25.0)

3) Algebra ensues:

4108.5 - 74.7x = 29.5x - 737.5 + 4.2x - 105x = 45.7 °C

**Problem #9:** A pure gold ring and pure silver ring have a total mass of 17.0 g. The two rings are heated to 65.4 °C and dropped into 12.4 mL of water at 22.3 °C. When equilibrium is reached, the temperature of the water is 24.7 °C. What is the mass of the gold ring?

**Solution:**

1) Set up the following equation:

(mass gold) (Δt gold) (C_{p}gold) + (mass silver) (Δt silver) (C_{p}silver) = (mass water) (Δt water) (C_{p}water)

2) Insert appropriate values:

(x) (40.7 °C) (0.129 J g^{-1}°C^{-1}) + (17.0 g - x) (40.7 °C) (0.237 J g^{-1}°C^{-1}) = (12.4 g) (2.4 °C) (4.184 J g^{-1}°C^{-1})

3) Algebra:

x = 8.98 g

4) Comments:

a) I looked up the values for the specific heats of gold and silver online. By the way, you should have memorized the sprecific heat value for liquid water by now.

b) The terms for the masses of the gold and silver rings comes from the fact that their sum is 17.0 g. We assign 'x' to be the mass of the gold ring, therefore the mass of the silver ring is 17.0 minus x.

**Problem #10:** A 5.00 g sample of aluminum (specific heat capacity = 0.89 J °C^{-1} g^{-1}) and a 10.00 g sample of iron (specific heat capacity = 0.45 J °C^{-1} g^{-1}) are heated to 100.0 °C. The mixture of hot iron and aluminum is then dropped into 91.9 g of water at 23.7 °C. Calculate the final temperature of the metal and water mixture, assuming no heat loss to the surroundings.

**Solution:**

1) Set this up:

q_{Al}+ q_{Fe}= q_{water}(5.00 g) (100 °C - x) (0.89 J °C

^{-1}g^{-1}) + (10.00 g) (100 °C - x) (0.45 J °C^{-1}g^{-1}) = (91.9 g) (x - 23.7 °C) (4.184 J °C^{-1}g^{-1})

2) Let's do some algebra (and drop all the units):

(445 - 4.45x) + (450 - 4.5x) = 384.5096x - 9112.87752393.4596x = 10007.87752

x = 25.4 °C

**Problem #11:** A 50.6 g sample of iron metal is heated and put into 104.0 g of water at 19.7 °C in a calorimeter. If the final temperature of the iron sample and the water is 24.3 °C, what was the temperature of the iron sample when it was placed in the water?

**Solution:**

1) heat lost by iron = heat gained by water:

(mass iron) (Δt iron) (C_{p}iron) = (mass water) (Δt water) (C_{p}water)(50.6 g) (x - 24.3 °C) (0.450 J/g °C) = (104.0 g) (4.6 °C) (4.184 J/g °C)

The 4.6 came from 24.3 minus 19.7.

The x - 24.3 °C is the Δt of the iron. It went from a hot temperature 'x' to a cooler temperature of 24.3 °C.

2) Solve for x:

(50.6x - 1229.58) (0.450) = 2001.625622.77x - 553.311 = 2001.6256

22.77x = 2554.9366

x = 112.2 °C

**Problem #12:** A 505.0 g piece of copper tubing is heated to 99.9 °C and placed in an insulated vessel containing 59.8 g of water at 24.8 °C. Assuming no loss of water and a heat capacity for the vessel of 10.0 J/K, what is the final temperature of the system? (C_{p} of copper = 0.387 J/g K)

**Solution:**

See problem #11 for the pattern to follow for the solution. No answer is provided.

Please note that K and °C will cancel in the problem. This is because (a) the temperatures in the actual calclation are differences between two temperature values and (b) the "size" of 1 K equals the "size" of 1 °C.

**Problem #13:** What volume of 18.5 °C water must be added, together with a 1.23-kg piece of iron (C_{p} = 0.449 J/g degrees C) at 68.5 °C in an insultated container, so that the final temperature of the water/metal mix remains constant at 25.6 °C?

**Solution:**

heat lost by the metal = heat gained by the water(1230 g) (42.9 °C) (0.449 J/g °C) = (mass) (4.184 J/g °C) (7.1 °C)

mass = 797.562 grams

rounding to 3 sig figs seems reasonable

798 mL