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Determining the Specific Heat of a Metal: An Experiment in Thermodynamics, Lecture notes of Thermodynamics of Materials

International University Institute of Luxembourg (IUIL)Thermodynamics of Materials

An experiment designed to determine the specific heat of a metal using a calorimeter. The concept of specific heat, the equation used to calculate it, and the safety precautions and procedure for conducting the experiment. Students will learn about the importance of accurate temperature measurements and minimizing errors in the experiment.

Typology: Lecture notes

2021/2022

Uploaded on 08/05/2022

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Experiment 7 - Finding the Specific Heat of a Metal
Heat is needed to warm up anything, but the amount of heat is not the same for all
things. Therefore it is desirable to know the specific heat of a substance. The amount of
heat needed to raise the temperature of one gram of a substance by one Celsius degree is
call the “specific heat” or “specific heat capacity” of that substance. For pure water, that
amount is one calorie, which is equivalent to 4.184 joules. Almost all other substances
have lower specific heats than water does.
Specific heat is defined in terms of one gram of substance and a one degree
temperature rise, but of course it is applied to other-sized samples and other temperature
changes. So, in general terms:
Heat that must be put = (temperature change) × (grams sample) × (specific heat)
into a sample
or Q =s•m•∆T, where Q is the amount of heat, s is the specific heat, m is the mass of the
sample, and ∆T is the temperature change. This equation can be used to calculate the
amount of heat that must be involved when the other three values are known or measured.
If a substance is cooling off, rather than warming up, the term Q represents the heat that
is given off by the sample (instead of the heat that must be put into the sample). This
same equation can also be used to calculate a specific heat, when the other three terms are
known or measured. That is what you will be doing in the present experiment.
The heat that is taken from a hot sample into cooler surrounding material, such as
water, can be measured in an insulated container called a calorimeter. A weighed
amount of the substance being studied is heated to some known (measured) temperature
and is then poured quickly into the calorimeter that already contains a measured amount
of water at a known, measured temperature. Heat flows from the hot substance into the
cooler water, until the sample temperature and the water temperature become equal. (This
final temperature will be somewhere in between the initial temperatures of the two
substances.)
When two objects at different temperatures are placed in contact with each other,
heat always flows from the hotter to the cooler object. Heat will flow until the two reach
thermal equilibrium, when they are at the same temperature. It will always be true that
the number of calories (or joules) of heat lost by the originally hot object will equal the
number of calories (or joules) gained by the originally cold object. In other words, the
amount of heat lost is equal to the amount of heat gained (Qlost = Qgained). In this
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Experiment 7 - Finding the Specific Heat of a Metal

Heat is needed to warm up anything, but the amount of heat is not the same for all things. Therefore it is desirable to know the specific heat of a substance. The amount of heat needed to raise the temperature of one gram of a substance by one Celsius degree is call the “specific heat” or “specific heat capacity” of that substance. For pure water, that amount is one calorie , which is equivalent to 4.184 joules. Almost all other substances have lower specific heats than water does. Specific heat is defined in terms of one gram of substance and a one degree temperature rise, but of course it is applied to other-sized samples and other temperature changes. So, in general terms: Heat that must be put = (temperature change) × (grams sample) × (specific heat) into a sample or Q =s•m•∆T, where Q is the amount of heat, s is the specific heat, m is the mass of the sample, and ∆T is the temperature change. This equation can be used to calculate the amount of heat that must be involved when the other three values are known or measured. If a substance is cooling off, rather than warming up, the term Q represents the heat that is given off by the sample (instead of the heat that must be put into the sample). This same equation can also be used to calculate a specific heat, when the other three terms are known or measured. That is what you will be doing in the present experiment. The heat that is taken from a hot sample into cooler surrounding material, such as water, can be measured in an insulated container called a calorimeter. A weighed amount of the substance being studied is heated to some known (measured) temperature and is then poured quickly into the calorimeter that already contains a measured amount of water at a known, measured temperature. Heat flows from the hot substance into the cooler water, until the sample temperature and the water temperature become equal. (This final temperature will be somewhere in between the initial temperatures of the two substances.) When two objects at different temperatures are placed in contact with each other, heat always flows from the hotter to the cooler object. Heat will flow until the two reach thermal equilibrium , when they are at the same temperature. It will always be true that the number of calories (or joules) of heat lost by the originally hot object will equal the number of calories (or joules) gained by the originally cold object. In other words, the amount of heat lost is equal to the amount of heat gained (Qlost = Qgained). In this

experiment, the amount of heat that is lost by a sample of metal as it cools is equal to the amount of heat gained by the water in the calorimeter. This assumes that no heat is lost from the calorimeter to its surroundings (the room), and that the amount of heat that is absorbed by the calorimeter itself is so small we can ignore it. Thus: Heat lost by the metal = heat gained by the water, or Qmetal = (∆Tm)(mm)(sm) = Qwater = (∆Tw)(mw)(sw) where the subscripts m and w identify the metal and the water. In this equation, you will know both ∆T values because you will measure initial and final temperatures. You will know all of the water values, and all but one of the metal values. You will then be able to solve for the “unknown” value, the specific heat of the metal. (Note: in the above equation, we are assuming that we will be using the absolute value of ∆T. If you are not using the absolute values of ∆T, then Qmetal = - Q (^) water.) Safety Precautions:

  • Wear your safety goggles.
  • Wash your hands after handling the metals. Waste Disposal:
  • There is no waste for this experiment. Return the metal samples so that they can be re-used.

Procedure

  1. Obtain a calorimeter (2 nested Styrofoam cups), a thermometer, and a sample of metal in a large test tube.
  2. Weigh the sample of metal in the test tube. Then, carefully transfer the metal sample to a clean dry beaker while you weigh the empty test tube, so you can determine the actual weight of the sample alone. NOTE: If it is awkward to weigh the tube and sample on the balance pan, you can support it in a small beaker. Since the beaker will be part of the weight when the empty tube is weighed also, its own weight will cancel out.
  3. Replace the metal in the test tube and put the test tube in a beaker of water. The beaker should contain enough water so that the top of the metal is below the surface of the water. Heat the water to boiling and allow it to boil for a few minutes to allow the metal to attain the temperature of the boiling water (100°C).

Questions

  1. A large error can be caused in this experiment by allowing some hot water to enter the calorimeter along with the hot metal. If this happens, would it make the specific heat value obtained for the metal higher than it really is or lower? Explain.
  2. Another error is caused by the fact that the calorimeter itself absorbs some of the heat from the metal. Does this error make the specific heat value obtained for the metal higher than it really is or lower? Explain.