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Determination of Specific Heat of Copper using Calorimetry, Lecture notes of Chemical Thermodynamics

An experiment to determine the specific heat of copper using the calorimetry method. The concept of specific heat, the historical definition of heat in terms of calories, and the conversion to joules. It also provides a detailed procedure for conducting the experiment, including the materials needed and the steps to take. The document concludes with the purpose of the experiment and the calculations to be performed.

What you will learn

  • What is the specific heat of copper?
  • What is the significance of the calorimetry method in determining the specific heat of materials?
  • How is the specific heat of copper determined experimentally?

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2021/2022

Uploaded on 09/12/2022

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SPECIFIC HEAT OF COPPER
Reminder – Goggles must be worn at all times in the lab
PRE-LAB DISCUSSION:
The amount of heat required to raise the temperature of a solid body depends on its change in temperature
(
Δ
T), its mass (m), and an intrinsic characteristic of the material forming the body called specific heat (cp).
The heat is calculated from the equation
Q = cp x m x
Δ
T
The unit for cp is thus heat per unit mass per unit temperature. The value of cp does depend on the
temperature. However, for the small temperature range we are interested in, it is a good approximation to
regard cp as temperature independent. Historically, heat (Q) was measured in terms of calories. The calorie
was defined as the amount of heat required to raise the temperature of 1 gram of water by 1 °C from 14.5 °C
to 15.5 °C at 1 atmosphere pressure. With this definition, the specific heat of water is 1.00 cal/(g⋅°C). The
use of the calorie began before it was established that heat is a form of energy and 1 calorie is equivalent to
4.18 J. The joule (J) has become the more favored unit in recent years. Thus, the units for cp that we will use
are J/(g⋅°C). The specific heat of water is then 4.18 J/(g⋅°C).
When two bodies in an isolated system, initially at different temperatures, are placed in intimate contact with
each other, in time they will come to equilibrium at some common intermediate temperature. Because of
energy conservation, the quantity of heat lost by the hot object is equal to that gained by the cold object
provided that no heat is lost to the surroundings. This is the basis for the method of calorimetry through
mixture: A metal sample whose specific heat is to be determined is heated in boiling water to 100 °C. It is
then quickly transferred to a Styrofoam calorimeter cup which contains a known volume of water of known
temperature. When the metal specimen and the calorimeter (including the water) come to equilibrium, the
final temperature is measured with a thermometer. It is assumed that the heat loss to the Styrofoam cup and
thermometer is negligible and if the heat exchange with the environment is kept small, then the heat lost by
the metal sample is equal to the total heat gained by the water.
PURPOSE:
To apply the experimental methods of calorimetry in the determination of the specific heat of a metal.
MATERIALS
Copper plumbing fixtures LabPro with temperature probe
Styrofoam cup Milligram balance
100 mL graduated cylinder Crucible tongs
Hotplate
PROCEDURE:
1. Get a hotplate and begin heating approximately 200 mL of water in a 400 mL glass beaker. Heat the
water to boiling.
2. Get a copper object from the counter. Determine its mass to three decimal places, and record the results
in the data section.
3. Lower the copper object into the water with your crucible tongs. Be careful not to drop the piece of
copper into the beaker (you might crack the beaker).
4. Obtain a Styrofoam cup “calorimeter” and add to it 75.0 mL of water. Remembering that the density of
water is 1 gram per mL, record the mass of the water in your data table.
5. Set up a LabPro CBL with a stainless steel temperature probe connected to Channel 1. Start the
DataMate program (APPS button). It will auto-recognize the type of probe. Data collection mode must be
set to “Time Graph,” to collect 1 data point every second for 180 seconds.
6. When the copper object has been in the boiling water for 3 minutes, begin the graphing of the water
temperature in the calorimeter, and then quickly move the piece of copper to the calorimeter using the
tongs. DO NOT try to handle the copper object with your hands. Do not allow the temperature probe to
come in direct contact with the piece of copper.
7. After three minutes are up, the calculator will display the final graph of the temperature change over the
three minutes. Scroll along the graph to determine the lowest temperature (initial, T1)
and the highest temperature (final, T2). Record these values in your data table.
8. Repeat the entire procedure two more times (three total) recording new values for change in
temperature.
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SPECIFIC HEAT OF COPPER

Reminder – Goggles must be worn at all times in the lab

PRE-LAB DISCUSSION:

The amount of heat required to raise the temperature of a solid body depends on its change in temperature

( Δ T ), its mass ( m ), and an intrinsic characteristic of the material forming the body called specific heat ( cp ).

The heat is calculated from the equation

Q = c p x m x Δ T

The unit for c (^) p is thus heat per unit mass per unit temperature. The value of c (^) p does depend on the temperature. However, for the small temperature range we are interested in, it is a good approximation to regard c (^) p as temperature independent. Historically, heat (Q) was measured in terms of calories. The calorie

was defined as the amount of heat required to raise the temperature of 1 gram of water by 1 °C from 14.5 °C to 15.5 °C at 1 atmosphere pressure. With this definition, the specific heat of water is 1.00 cal/(g⋅°C). The use of the calorie began before it was established that heat is a form of energy and 1 calorie is equivalent to 4.18 J. The joule (J) has become the more favored unit in recent years. Thus, the units for c (^) p that we will use

are J/(g⋅°C). The specific heat of water is then 4.18 J/(g⋅°C).

When two bodies in an isolated system, initially at different temperatures, are placed in intimate contact with each other, in time they will come to equilibrium at some common intermediate temperature. Because of energy conservation, the quantity of heat lost by the hot object is equal to that gained by the cold object provided that no heat is lost to the surroundings. This is the basis for the method of calorimetry through mixture: A metal sample whose specific heat is to be determined is heated in boiling water to 100 °C. It is then quickly transferred to a Styrofoam calorimeter cup which contains a known volume of water of known temperature. When the metal specimen and the calorimeter (including the water) come to equilibrium, the final temperature is measured with a thermometer. It is assumed that the heat loss to the Styrofoam cup and thermometer is negligible and if the heat exchange with the environment is kept small, then the heat lost by the metal sample is equal to the total heat gained by the water.

PURPOSE:

To apply the experimental methods of calorimetry in the determination of the specific heat of a metal.

MATERIALS

Copper plumbing fixtures LabPro with temperature probe Styrofoam cup Milligram balance 100 mL graduated cylinder Crucible tongs Hotplate

PROCEDURE:

  1. Get a hotplate and begin heating approximately 200 mL of water in a 400 mL glass beaker. Heat the water to boiling.
  2. Get a copper object from the counter. Determine its mass to three decimal places, and record the results in the data section.
  3. Lower the copper object into the water with your crucible tongs. Be careful not to drop the piece of copper into the beaker (you might crack the beaker).
  4. Obtain a Styrofoam cup “calorimeter” and add to it 75.0 mL of water. Remembering that the density of water is 1 gram per mL, record the mass of the water in your data table.
  5. Set up a LabPro CBL with a stainless steel temperature probe connected to Channel 1. Start the DataMate program (APPS button). It will auto-recognize the type of probe. Data collection mode must be set to “Time Graph,” to collect 1 data point every second for 180 seconds.
  6. When the copper object has been in the boiling water for 3 minutes, begin the graphing of the water temperature in the calorimeter, and then quickly move the piece of copper to the calorimeter using the tongs. DO NOT try to handle the copper object with your hands. Do not allow the temperature probe to come in direct contact with the piece of copper.
  7. After three minutes are up, the calculator will display the final graph of the temperature change over the three minutes. Scroll along the graph to determine the lowest temperature (initial, T 1 ) and the highest temperature (final, T 2 ). Record these values in your data table.
  8. Repeat the entire procedure two more times (three total) recording new values for change in temperature.
  1. When done, return the copper and the Styrofoam cup to the counter. Turn off the hotplate. Do not touch the hot water beaker or the hotplate until they have cooled enough for handling. This will usually be at least 30 minutes!

OBSERVATIONS AND DATA:

Trial #1 Trial #2 Trial #

  1. Mass of the copper object g g g
  2. Volume of water in the calorimeter mL mL mL
  3. Initial temperature of water in calorimeter(T 1 ) (^) °C °C °C
  4. Final (highest) temperature of calorimeter(T 2 ) (^) °C °C °C

CALCULATIONS: Show your work!

Trial #1 Trial #2 Trial #3 Average

Δ Twater ( T 2 - T 1 )

Δ copper (T 2 - 100 °C)

1. Copy the table above, and calculate the temperature change of the water, Δ Twater , T 2 - T 1 for each trial.

Then calculate the average of these values for water.

2. In the same table, calculate the temperature change of the copper, Δ Tcopper , T 2 - 100 °C (Yes your

answer, will be negative). Then calculate the average of these values for copper.

  1. Calculate the amount of energy absorbed by the water. In order to do this, we use the specific heat ( c (^) p )

of pure water, 4.18 J/(g⋅°C), and we assume the density of the water was 1.00 gram/mL. Express your final answer in joules.

Q water = c p water x mwater x Δ T water

  1. Next, we assume that the energy absorbed by the water is equal to the energy lost by the copper. -Q water = Q copper On this assumption, we can calculate the specific heat of copper using the mass of the copper (m), the

value of Q calculated above, and Δ Tcopper.

c (^) p copper = ___Q copper ___

mcopper ⋅ Δ T copper

  1. Using the known value of the specific heat of copper, 0.385 J/(g⋅°C), calculate the absolute error in your averaged experimental result.
  2. Calculate your relative error (PERCENT ERROR). Remember that:

x 100

AcceptedValue

AbsoluteError

% Error =