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 value will remain at 4.18, 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:
1) The colder water will warm up (heat energy "flows" in to it). The warmer metal 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)
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:
qlost = qgain
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.
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.
Here are the numbers:
(10)(130 - x)(0.901) = (200.0)(x - 25)(4.18)
Noting that 200/10 = 20, I get:
117.13 - 0.901x = 83.6x - 2090
and so, x = 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. (Cp for Hg = 0.14 J per gram degree Celsius.)
[By the way, we will ignore the fact that mercury is liquid. It so happened that iron, aluminum and mercury were the three specific heats given in the textbook I used. So there!]
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.
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.
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.
Solution to (c):
648.52 J / 74.0 °C = 8.76 J/°C (or 8.76 J/K)
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: 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:
(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)
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 #6: 105.0 mL of H2O 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 (Cp of copper = 0.387 J/g-K)?
Solution:
1) This problem can be summarized thusly:
qlost by copper = qgained by water + qgained 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