In lesson 1-6,
you were introduced to the concept of heat. That lesson described the energy of
chemical reactions, and explained the difference between endothermic and exothermic
reactions. Chemists measure the heat given off or taken in during a chemical
reaction to determine the energy of a specific chemical or physical reaction. In
this lesson, you will learn to calculate the amount of heat transferred during a physical
or chemical change.
Scientists use a device called a calorimeter
to measure the transfer of heat during a physical or chemical change. You will be
using a device like this when you conduct the Calorimetry Laboratory. It is
essential that you understand the calculations required in that activity, so that you can
benefit from the learning opportunity that the laboratory makes available to you.
As you should know by now, the formula for
heat transfer calculations is:
amount of heat transferred = mass x change
in temperature x specific heat
I will remind you that the symbols
that are used for the formula will vary from textbook to textbook, but the values
that they represent never change. One way to write the heat transfer formula is
shown below:
q = m(DT)Cp
Where q = heat transferred, DT = the change in temperature and Cp
= the specific heat.
The SI units for heat transferred are joules,
however calories are still often used for problems involving water. You should
memorize the conversion factor; 4.18 J = 1 cal. The units for specific heat
are joules/grams x degrees Celsius (J/g x oC) or Calories/grams x degrees
Celsius (cal/g x oC). Temperature is usually given in degrees
Celsius.
You will solve these problems logically and
algebraically. Logically, meaning you will strive to understand the logic of
performing each step, and that you will check to make sure that your answer makes
sense. As in any algebra problem, you will only have one unknown. The rest of
the information will be provided for you. We will work through one example of each
of the possible types of heat transfer problems that you will be responsible for.
Then you can move on to try the worksheets and the online quiz programs.
Type 1. Heat Transferred (q) is the unknown:
Ex. Aluminum has a specific heat of 0.902 J/g x oC.
How much heat is lost when a piece of aluminum with a mass of 23.984 g cools from a
temperature of 415.0 oC to a temperature of 22.0 oC?
Step 1: First read the question and try to understand what
they are asking you. Can you picture a piece of
aluminum foil that is taken out of an oven. Imagine the aluminum losing heat to its
surroundings until the temperature goes from 415.0 oC to 22.0 oC.
Step 2: Write the original formula.
q = m(DT)Cp
Step 3: List the known and unknown factors.
Looking at the units in the word problem will help you determine which is which.
q = ?
m = 23.984 g
DT = (415.0 oC - 22.0 oC) = 393.0 oC
(remember, they asked for the change in temperature)
Cp = 0.902 J/g x oC
Step 4. Substitute your values into the formula
q = ?
m = 23.984 g
DT = (415.0 oC - 22.0 oC) = 393.0 oC
Cp = 0.902 J/g x oC
q = m(DT)Cp
q = 23.984 g x 393.0 oC
x 0.902 J/g x oC
Step 5. Cross out units where possible, and
solve for unknown.
q = 23.984 g x 393.0
oC x 0.902 J/g x oC
q = 8501.992224 J
Step 6. Round to the correct number of
significant digits and check to see that you answer makes sense.
q = 8.50 x 103 J
Our answer makes sense because joules (J) are
acceptable units for q, and the value should be positive based on the wording of the
question.
Type 2. mass (m) is the
unknown:
Ex. The temperature of a sample of water increases
by 69.5 oC when 24 500 J are applied. The specific heat of liquid water
is 4.18 J/g x oC. What is the mass of the sample of water?
Step 1: First read the question and try to
understand what they are asking you. Energy is being
used to change the temperature of a sample of water by 69.5 oC. What size
sample of water would require 24 500 J to make that change?
Step 2: Write the original formula, and then modify it
isolate the unknown.
q = m(DT)Cp
q =
m(DT)Cp
--- -------------
(DT)Cp (DT)Cp
m = q/(DT)Cp
Step 3: List the known and unknown factors.
Looking at the units in the word problem will help you determine which is which.
q = 24 500 J
m = ?
DT = 69.5 oC
Cp = 4.18 J/g x oC
Step 4. Substitute your values into the
formula.
q = 24 500 J
m = ?
DT = 69.5 oC
Cp = 4.18 J/g x oC
m = q/(DT)Cp
m = 24 500 J/69.5 oC x
4.18 J/g x oC
Step 5. Cross out units where possible, and
solve for unknown.
m = 24 500 J/69.5 oC
x 4.18 J/g x oC
m = 84.3344463184 g
Step 6. Round to the correct number of
significant digits and check to see that you answer makes sense.
m = 84.3 g
Our answer makes sense because grams are the correct
units for mass, and the value should be positive.
Type 3. change in temperature (DT) is the unknown:
Ex. 850 calories of heat are applied to a 250
g sample of liquid water with an initial temperature of 13.0 oC. Find a)
the change in temperature and b) the final temperature. (remember, the specific heat
of liquid water, in calories, is 1.00 cal/g x oC.)
Step 1: First read the question and try to
understand what they are asking you. Here they are
heating up a sample of water. They want to know how many degrees increase will
result from 850 calories of heat. Further, they want to know the final temperature
of the water, which will simply be equal to the initial temperature + the change in
temperature.
Step 2: Write the original formula, and then
modify it isolate the unknown.
q = m(DT)Cp
q = m(DT)Cp
---- ---------------
m Cp m Cp
DT = q/m x
Cp
Step 3: List the known and unknown factors.
Looking at the units in the word problem will help you determine which is which.
q = 850 cal
m = 250 g
DT = ?
Cp = 1.00 cal/g x oC
Step 4. Substitute your values into the
formula
q = 850 cal
m = 250 g
DT = ?
Cp = 1.00 cal/g x oC
DT = q/m x
Cp
DT = 850
cal/250 g x 1.00 cal/g x oC
Step 5. Cross out units where possible, and
solve for unknown.
DT = 850 cal/250
g x 1.00 cal/g x oC
Answer to step a) DT
= 3.4 oC
Answer to step b) final temperature =
13.0 oC + 3.4 oC = 16.4 oC
Step 6. Round to the correct number of
significant digits and check to see that you answer makes sense.
Answers are already rounded correctly. They
make sense because they show the correct units for temperature and because the final
temperature is higher than the initial temperature, as it should be.
Type 4. Specific Heat (Cp) is the
unknown:
Ex. When 34 700 J of heat are applied to
a 350 g sample of an unknown material the temperature rises from 22.0 oC
to 173.0 oC. What must be the specific heat of this material?
Step 1: First read the question and try to
understand what they are asking you. Specific heat is a
concept that some students struggle with. The question is about finding the specific
heat by seeing how much the temperature changes when a certain amount of heat is applied.
Metal heats up faster than water because it has a low specific heat. If a
material has a low specific heat, the temperature change will be greater for a given
amount of heat, when all other things are equal.
Step 2: Write the original formula, and then
modify it to isolate the unknown.
q = m(DT)Cp
q = m(DT)Cp
--- -------------
m(DT) m(DT)
Cp = q/m(DT)
Step 3: List the known and unknown factors.
Looking at the units in the word problem will help you determine which is which.
q = 34 700 J
m = 350 g
DT = (173.0oC - 22.0oC) = 151.0 oC
Cp = ?
Step 4. Substitute your values into the
formula
q = 34 700 J
m = 350 g
DT = (173.0oC - 22.0oC) = 151.0 oC
Cp = ?
Cp = q/m(DT)
Cp = 34 700 J/350 g x 151.0 oC
Step 5. Cross out units where possible, and
solve for unknown.
Cp = 34 700 J/350 g x 151.0 oC
Cp = 0.65657521286 J/g x oC
Step 6. Round to the correct number of
significant digits and check to see that you answer makes sense.
Cp = 0.66 J/g x oC
Our answer is logical, and the units are correct.
