I have revised this lab so that students are no longer required to find the heat capacity of the calorimeter itself. This was a tedious process which seldom resulted in anything like consistent numbers. By allowing students to assume the calorimeter is a perfect insulator we can make do with nothing more than what is hopefully a consistent systematic error. The old version is still available here.
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Calorimetry

Objective

In this lab students will measure the heat capacity of a metal sample and measure the molar heat of solution of potassium nitrate.


Materials


Background

Energy

The scientific concept of energy is a difficult one. The basic idea is that energy is the capacity to do work or to oppose natural forces such as gravity or electric fields. This seems natural enough. But energy is actually a difficult idea because it is a mathematical abstraction which by definition always has the same value. No matter what happens, the amount of energy involved in an event is always the same after the event as it was before the event. In fact, the amount of energy in the entire universe is understood to be constant. Formally, this idea is called the First Law of Thermodynamics. We will use this law in order to understand and interpret our work in the lab. Simply put, all of the energy lost by one object plus all of the energy gained by another equals zero.

A great discussion about the nature of energy comes from the first volume of the Feynman Lectures on Physics. I recommend you read Chapter 4 on this page: http://www.feynmanlectures.caltech.edu/I_04.html. I accessed the page on 03/26/2018.

The trouble with the First Law is that it says nothing about what forms energy may take. For example, we all know that heat is a form of energy. Formally, the definition of heat is a form of energy and when added to a system the temperature of the system rises as long as the system is not undergoing a chemical or physical change. Heat is involved in chemical and physical changes as the energy released when bonds form and the energy absorbed when bonds are broken. Light is also a form of energy: when you sit out in the sun you can feel your skin become warmer and heat is certainly a form of energy. Also, perhaps from other science classes, we know that energy can be associated with motion, so-called kinetic energy. The faster something is moving, the more energy it is understood to have. Finally, in another invention of scientists designed to keep track of this slippery abstraction, there is the concept of potential energy. One form of potential energy is called gravitational potential energy and can be understood through the following example. A 1 kg rock raised one meter above the floor will strike the floor at nearly 10 mph (4.4 m/s) when dropped. The same rock when dropped from 10 meters above the floor will strike at nearly 30 mph (14 m/s). Clearly the rock had more energy—potential energy—when raised to 10 m than when it was only 1 m above the floor.

Because energy is understood to have many forms it can be a difficult thing to measure and keep track of for purposes of making measurements of changes in energy. If energy should assume a form that you did not anticipate you may find that some energy is apparently lost. Fortunately in this lab activity the work is designed so that all energy will be in the form of heat. Even so, some careful thinking is involved in keeping track of the origins and destinations of the heat in this lab. We will use the First Law as follows: The heat lost by one object is gained by the others in contact with it. We will measure these exchanges of heat by way of measuring temperature. Heat itself is not directly measurable and there is no such thing as a ‘heatomometer’. Fortunately, there is such a thing as a thermometer.

Heat vs. Temperature

Though heat and temperature are not scientifically equivalent they seem to mean about the same thing to most people. The distinction between them is an important one, however, so we will explore it a bit here. In common speech heat and temperature are often interchangeable. “This cup of boiling water is hot,” means both that the temperature is high and that the water is full of heat. Although the former is true, the water does have a high temperature, the latter is not. A bathtub full of water heated to 37°C (body temperature) requires much more heat energy than a cup of water boiled for tea. Even cold water contains heat because if a colder material is placed in it, then heat will move out of the cold water and into the colder object. Heat does not have to be hot (that is, an object does not need to have a high temperature to have heat in it).

Temperature measures the average kinetic energy (which is related to the speed) of molecular motion. The higher the temperature, the faster the molecules in a sample of matter are moving. The lower the temperature, the slower the average speed of the molecules. When heat moves into a material it sometimes causes a change in temperature. Adding heat to a collection of molecules will cause their random motion to increase—the molecules will move faster on average than they did before heat was added. As long as no chemical reaction or phase change occurs this leads to a rise in temperature. When heat moves out of a system the average molecular speed is reduced. As long as there is no chemical reaction or phase change this leads to a drop in temperature. To be clear ‘cold’ is not a form of energy and cannot be added to something to cool it down. An object or the air in the room cools down as a result of the removal of heat energy and not because ‘coldness’ was added.

When two materials are placed in contact they will exchange heat only if they are at different temperatures. Heat naturally moves from the object with a higher temperature into the object with a lower temperature. Anyone can verify this who has held a cup of hot coffee: the heat from the cup moves into the hand holding it, causing it to become warmer—in other words, the temperature of the hand rises as the temperature of the cup falls. If two objects placed in contact have the same temperature then they will not exchange heat energy. If a piece of metal is placed in boiling water it will quickly have the same temperature as the water. When that same piece of hot metal is taken out of the boiling water and placed into cooler water it will lose a quantity of heat and cause the water to warm up. Formally, we can write an equation for this process. The equation below is only true if the exchange occurs in a closed system which is insulated against the loss or gain of heat by exchange with the surroundings.

qmetal + qcold = 0
qmetal = heat lost by the hot metal
qcold = heat gained by the cold water

Heat and temperature can be particularly confusing when chemical reactions or phase changes come into play. Heat added to a system can cause molecular motion to increase in speed but it can also be used to break bonds holding one molecule to its neighbor. Water molecules are held together in the solid and liquid form with a strong form of inter-molecular bond called a hydrogen bond. To break these bonds requires an input of energy and when heat is added to an ice cube it causes the water molecules to shake more violently and some of the heat energy is consumed in breaking hydrogen bonds. Under circumstances where the number of bonds broken equals the number of new bonds that form (a special condition called equilibrium) the temperature can remain the same even while heat enters the system. The heat is consumed in causing the phase change from a solid to a liquid: it breaks bonds instead of speeding up the molecules.

Even though the system’s temperature may remain the same, the temperature of the surroundings may drop. If the heat that drives the melting is coming from the counter-top in a kitchen then that counter-top will experience a drop in temperature. An endothermic process is one in which heat enters a system and is consumed. In the case of melting ice heat enters the system and its temperature remains constant while the temperature of the surroundings falls.

Heat can leave a system when new bonds form between molecules. As steam condenses the water molecules must slow down in order to approach one another without immediately bouncing off. Heat must leave the system for the temperature to fall enough so that this can happen. At a specific temperature it is possible to set up an equilibrium where just as many molecules condense as evaporate. In this case the temperature of the system will remain constant inside the system. When the molecules do manage to stick together they do so with a release of heat into the surroundings. This is similar to the way that a stretched rubber band releases energy when it is allowed to snap: bond formation releases energy. For example, if the water molecules stick together on the surface of a piece of glass then the glass will gain the heat energy and its temperature will rise. Heat exits the system but the system can remain the same temperature while the surroundings absorb the heat and experience a rise in temperature. An exothermic process is one in which heat exits a system.

Exothermic vs. Endothermic

In the study of the exchange of energy (a.k.a., Thermodynamics) a useful distinction is made between processes that release energy and those that absorb energy. When the system releases energy into the surroundings the process is called exothermic. When the system absorbs energy from the surroundings the process is called endothermic. Physical changes that are exothermic include the condensation of water and the freezing of water. In both cases heat must exit the system in order for the molecules to slow down enough for the forces that attract the molecules to one another to be able to act. As the bonds between molecules form, this also releases heat. That the freezing of a substance is exothermic may seem counterintuitive at first because we think of it happening in a cold box on top of the refrigerator. The reality is that the heat removed from the inside of that box is released at the back of the refrigerator: the coils on the back are noticeably warm to the touch. They are warm with the heat removed from the food and ice cubes inside. Physical changes that are endothermic include melting ice and boiling water. In both cases heat must be supplied from the outside. In the case of ice this may be by passive transmission of heat into the ice cube from a warm glass of water. In the case of boiling water heat must be constantly supplied by the stove burner in order for the boiling to continue. In both melting and boiling molecules must speed up, which requires an input of heat, and break bonds holding molecules together, which also requires an input of heat. Another endothermic physical change is the formation of a solution of certain salts. When these salts dissolve in water they absorb heat and the temperature of the solution drops compared to the original temperature of the water. This shows that there is a net absorption of energy in the process and more energy is used breaking bonds than is released in the formation of new bonds. One example of a salt which has an endothermic heat of solution is potassium nitrate (KNO3). Other salts may release energy when mixed with water, notably calcium chloride (CaCl2).

Chemical changes are either exothermic or endothermic based on whether more energy is released due to bond formation than is consumed in breaking old bonds or vice versa. Chemical changes that are exothermic include burning hydrocarbons and mixing certain acids and bases. The heat released by these reactions comes from potential energy stored in the arrangement of the atoms and molecules: when new bonds form this releases heat. Endothermic chemical changes include the reaction between baking soda and vinegar. Mix a little of these two chemicals together in a glass held in the hand and it is clear that it becomes cool to the touch. Energy is absorbed from the surroundings in order to break bonds. The newly formed bonds do not release as much energy compared with the energy used to break bonds. As a result, the mixture’s temperature drops.




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Calculating Heat and Specific Heat

Heat energy will be measured in joules (J) in this lab because J is the SI unit of energy. An older unit of energy is the calorie and it originates in the idea of specific heat since in units of calories the specific heat of water is 1 calorie/g·°C. This calorie is 1,000 times smaller than the Calories you are meant to fear as displayed on the Nutrition Facts labels on your food. Those Calories are really kilocalories. Heat is calculated mathematically based on the change in temperature of a substance, the mass of the substance, and a quantity called the specific heat of the substance. The specific heat (also called heat capacity) is a constant that quantifies the amount of heat required to raise the temperature of one gram of a substance by 1 degree (°C or K). The specific heat is different at different temperatures but for purposes of this lab we will assume that it is in fact constant over the temperature range we will encounter. The specific heat of water is 4.184 J/g·°C. Specific heat will be denoted as a lower-case letter ‘s’. By comparison with water, many other substances have much lower specific heats. Copper, for example, has a specific heat of 0.385 J/g·°C. In effect, much less energy is required to raise the temperature a certain mass of copper than to raise the temperature of an equal mass of water.

There is a simple equation that we will use to calculate the amount of heat involved in a change in temperature. It is the product of the mass of the substance, the specific heat, and the change in temperature. The change in temperature is usually written as ΔT and is defined as the final temperature minus the initial temperature: ΔT = Tfinal – Tinitial. So heat is calculated as follows for 1.00 × 102 g of water that experiences a rise in temperature from 25.0°C to 45.0°C:

q = msΔT
q = (100 g)(4.184 J/g·°C)(45.0°C - 25.0°C)
q = 8.37 × 103 J or 8.37 kJ

This shows the amount of heat required to raise the temperature of 100 g of water by 20°C. If the substance had been copper instead the same temperature rise would have required only 770 J. Alternatively, if the 8.37 kJ had been used to heat 100 g of copper the temperature change would have been much greater: ΔT = 227°C! What follows is a pair of examples for the types of calculations you will carry out as part of analyzing your lab data.

There is another page on my site to help students to do calculations for calorimetry. That material is found here.

Heat Capacity of a Metal

A calorimeter is made using two nested styrofoam cups and filled with 175.0 g of water at an initial temperature of 25.0°C. A quantity of water is heated to 100.0°C and 25.0 g of aluminum is heated in the boiling water for five minutes. Once the aluminum has the same temperature as the boiling water it is removed with a pair of tongs and placed into the water in the calorimeter. What is the heat capacity of the piece of aluminum if the final temperature is 27.2°C?

For a calorimeter that we assume is perfect (it neither absorbs nor releases any heat) we can set up an equation showing that the heat lost by the metal plus the heat gained by the water equals zero:


qAl + qH2O = 0
qAl = (25.0 g)(s)(27.2°C – 100.0°C)
qH2O = (175.0 g)(4.184 J/g·°C)(27.2°C – 25.0°C)

Setting qH2O equal to –qAl and solving for s gives:
s = 0.897 J/g·°C

This example was written so that the answer gives the accepted reference value for the specific heat of aluminum metal. Your work in the lab will involve a similar measurement but may not give a result that exactly matches the reference value. Differences may be due to heat lost in heating the calorimeter or due to the fact that a little hot water may have still been clinging to the metal when it was dropped in the calorimeter. The thermometer itself may be a source of heat or absorb some heat and play a role in determining the final temperature. Finally, inaccuracies in thermometer readings cannot be ruled out. Each of these will have a different effect on the outcome of your measurement, making it higher or lower.

Molar Heat of Solution

Another example of an endothermic process is the formation of a solution of certain inorganic salts. Potassium nitrate (KNO3) has an endothermic heat of solution. This means that heat is absorbed from the surroundings to enable the KNO3 to dissolve. This leads to a reduction in temperature of the water and the vessel in which the solution is made. Heat has been used up and taken away from the water to create the solution.

Find the molar heat of solution for KNO3. First, dissolve 1.01 g of KNO3 in 25.0 g of water at 25.0°C. When this is done in a calorimeter which we assume neither absorbs nor releases heat the final temperature is 21.8°C

qKNO3 + qwater = 0
qKNO3 + (4.18 J/°C·g)(26.01 g)(21.8 - 25.0°C) = 0
Note the mass for qwater in this expression is given as 26.01 g and not 25.0 g. This is intentional: the mass of material in the calorimeter is in fact the sum of the water plus the KNO3. The calculation assumes that this mass has the same specific heat as plain water. The qKNO3 term represents the quantity of energy consumed in dissolving the KNO3.
Solving for qKNO3 gives:
qKNO3 = 347.9 J
This amount of heat is for 1.01 g of KNO3 so the molar heat of solution is:
 347.9 J      101.1 g      1 kJ
-------- * --------- * ------- = 34.8 kJ/mol
 1.01 g      1 mol      1000 J

Note that the value for the heat of solution of KNO3 is positive. If you calculated the heat for the water it would prove to have a negative value. When solved for qKNO3 the equation gives a positive number. From the system’s point of view an exothermic process has a negative value for its calculated heat and an endothermic process has a positive value for its calculated heat. This comes about naturally from the basic equation for heat: q = sm(Tf – Ti). When Tf is greater than Ti then temperature increased (ΔT is positive). To raise temperature requires an input of heat. If heat is an input then a process is endothermic. The reverse is true when Ti is greater than Tf.


Pre-lab Problems

The following problems will help you to understand the concepts involved in this lab and to be able to do the math required for the analysis of your lab results.

  1. Describe one chemical change and two physical changes which are exothermic. Justify your answer by stating where the energy came from which was released as heat.
  2. Describe one chemical change and two physical changes which are endothermic. Justify your answer by stating how the heat is absorbed.
  3. Identify each of the following as exothermic or endothermic:
    1. boiling water
    2. the cooling effect of perspiration
    3. condensing water
    4. freezing ice
    5. melting ice
    1. burning natural gas
    2. neutralizing a base with acid
    3. the chemical reaction in a cold pack
    4. the chemical reaction in a hot pack
  1. Under certain circumstances orange growers will spray their oranges with water on winter nights when there is likely to be a ‘freeze’. How does the fact that water freezes to the surface of the fruit help to protect the fruit itself from becoming frozen? (Try an internet search of protecting oranges from freezing. See also http://extension.arizona.edu/sites/extension.arizona.edu/files/pubs/az1222.pdf)
  2. The specific heat of water is 4.184 J/g·°C and that of aluminum is 0.897 J/g·°C. Which substance will show a larger increase in temperature upon the addition of 100 J of heat to 100 g, the aluminum or the water? Justify your answer with a calculation.
  3. What is the difference between heat and temperature?
  4. A few iron nails with a total mass of 22.4 g are heated in a water bath for five minutes. The water bath was boiling with a constant temperature of 101°C according to the thermometer. When the nails were carefully transferred into 182.0 g of water in a coffee cup calorimeter the temperature of the water rose from 25.0°C to 26.0°C. What is the heat capacity of the iron?
  5. When 100.0 mL of a solution of an acid with the general formula HA is mixed with 100.0 mL of a solution of a base with the general formula BOH in a calorimeter the temperature increases from 21.7°C to 35.8°C. The calorimeter may be assumed to neither absorb nor release heat. The specific heat of solutions with low concentrations are considered to be identical with the specific heat of water. The density of such solutions is also taken to be identical with the density of water.
    1. Did the mixture in the calorimeter give up or absorb heat? Justify your answer.
    2. Was the reaction between the two chemicals exothermic or endothermic? Justify your answer.
    3. How much heat energy was released in the chemical reaction?
  6. Hydrogen chloride gas is highly soluble in water. When dissolved in water the solution is called hydrochloric acid. When 10.0 g of hydrogen chloride is dissolved in 100.0 g of water in a calorimeter at 25.0°C the temperature rises to 69.5°C.
    1. Is this process exothermic or endothermic? Justify your answer.
    2. How much heat was given off by the creation of this particular solution? Assume that the specific heat of the HCl solution is the same as that of water and that the density is still about 1 g/mL. Also, remember to use the total mass of the contents of the calorimeter in calculating the heat of the solution.
    3. What is the molar heat of solution for hydrogen chloride?



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Lab Procedure

Safety

The following list does not cover all possible hazards, just the ones that can be anticipated. Move slowly and carefully in the lab: haste and impatience have caused more than one accident.

Procedure

Calorimeter Apparatus (9K)

This lab has two main parts. First you will use the calorimeter to determine the specific heat of a sample of metal. In Part II you will measure the heat of solution of potassium nitrate. Ordinary table salt shows no change in temperature upon dissolving in water. Potassium nitrate on the other hand has a positive heat of solution, drawing heat from the water and reducing its temperature.


Part I

In this part of the lab you will use a simple constant-pressure calorimeter. Next you will heat a sample of metal in boiling water, which has a constant temperature. By placing the hot metal into the calorimeter and measuring the change in temperature you will be able to calculate specific heat of the metal.

  1. The data you will need to complete this part of the lab are as follows:
    1. The mass of water in the calorimeter
    2. The mass of the metal sample
    3. The initial temperature of the metal
    4. The initial temperature of the water in the calorimeter
    5. The final temperature of the water and metal
  2. If applicable, set up your temperature probe and computer interface. When ready to collect data, click the ‘Collect’ button. By default the system records temperature every 0.5 sec and graphs temperature vs. time. The temperature will be displayed on the screen at all times. If necessary, adjust the displayed thermometer so that it will show temperatures up to or slightly above 100°C.
  3. Find the mass of your calorimeter when it’s empty; include a magnetic stir bar if you are using a magnetic stirrer. Next, fill the calorimeter with about enough water to completely submerge your metal sample. Measure the exact amount of water you add using a lab balance by subtracting the mass of the calorimeter (plus stir bar). Record the mass to the maximum number of significant figures.
  4. Measure the mass of your metal sample to the maximum number of significant figures.
  5. Set up the stir bar and thermometer probe in such a way that the stir bar does not hit the thermometer and so that there is room to add your metal sample.
  6. Set up the ring, stand, and wire gauze in preparation for heating water over the burner. Set it up in such a way that it is not too easy to knock it down once the burner is on and the water is hot. Draw approx. 300 mL of water from the tap into a 400-mL beaker and start heating it over the burner. You will heat the water to the boiling point to serve as a constant-temperature water bath for your metal sample.
  7. Add the metal sample to the water bath using a pair of crucible tongs. Once the water is boiling and has had a constant temperature for a minute or so start timing. Heat the metal in boiling water for at least five minutes.
  8. Once the water is boiling, measure the temperature of the hot water bath. This temperature should remain constant. Record the temperature of the water bath as the initial temperature of the metal. Remove the thermometer (or probe) and cool it off under running tap water. Once cool, place it in the hole in the calorimeter cover and submerge the end of it in the water. Stir and wait until the temperature stabilizes and then record the temperature as the initial temperature of the water. Begin collecting temperature data if using the computerized probe.
  9. Without tipping over the beaker of boiling water, carefully remove your metal sample. It will have remained at the temperature you last measured as long as you did not move or turn off the burner. Tap the metal on the counter-top once to remove drops of hot water and then carefully place it into the calorimeter. Make sure that it is completely submerged and then place the cover on it and stir the water with the thermometer or by using a magentic stirrer.
  10. Watch the thermometer (or the display) to find out when the temperature reaches a maximum value. Since the calorimeter is insulated this should remain constant for a time before it begins to fall again. The highest temperature that is reached is the final temperature for the experiment. Record this and data collection for this trial is complete.
  11. As time allows, and considering there is another part to this lab, do additional trials to get 2 – 4 total measurements using the same sample. Simply keep the hot water bath hot and don’t forget to measure the exact mass of water in the calorimeter each time.
  12. Alternatively: Find the heat capacity of one or more other kinds of metal. There are at least five metals for you to test.

Part II

In this part of the lab you will add a carefully measured mass (~3 g) of potassium nitrate (KNO3) to a calorimeter containing approximately 1 cm depth of water at a known temperature. Because this substance has a positive enthalpy of solution the temperature will drop as heat is removed from the water and calorimeter.

  1. The data you will need to complete this part of the lab are as follows:
    1. The mass of the water in the calorimeter.
    2. The initial temperature of the cold water.
    3. The exact mass of the KNO3.
    4. The final temperature of the mixed KNO3 and cold water.
  2. Obtain a small amount, a bit more than 3 g, of KNO3. Grind it thoroughly into a very fine powder using the mortar and pestle.
  3. Using the analytical balance, measure approx. 3 g of the finely ground KNO3 into a weighing boat. Do not try to get the mass to be exactly 3.000 g because you will just divide by this number, whatever it is, to get the molar heat of solution.
  4. Prepare the dry calorimeter by measuring approx. water into it until it is about 1 cm deep. Record the mass of water exactly (it should be near 75 g). Place the thermometer or probe into the calorimeter and record the temperature once you have determined it is not changing. It is your initial temperature (Ti).
  5. When you are ready, add the KNO3 to the water in the calorimeter.
  6. Cover the calorimeter and stir slowly and constantly with the temperature probe (or using a magnetic stirrer). Collect data on the computer and watch for a temperature minimum. On the graph you will see the temperature vs. time plot drop and then flatten out when the temperature becomes constant. This constant temperature is the final temperature (Tf). If a computerized temperature probe is not available: Carefully observe the temperature and record it every 10 seconds until the temperature is constant for three consecutive readings.
  7. Repeat this procedure one or two more times, as time allows. Add water to the calorimeter and measure its mass exactly. Ensure that all temperatures are equilibrium temperatures by stirring the water and waiting for the temperature to stop changing before recording the temperature.
  8. Calculate the heat of solution of your sample of KNO3 and the heat of solution per mole of KNO3 before you leave the lab. By doing so you may be able to find and correct errors which you cannot do once you put away your materials and clean up.



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Analysis

No formal lab report will be required. Instead make a data table for relevant data using a spreadsheet or word processing program. Then answer the following questions in a typed document. For calculations, show only the set-up and result. Do not show all intermediate algebraic steps.

  1. What is the specific heat of your metal sample(s)? Include a data table with relevant data. Show the setup and conclusion of your calculation.
  2. Identify your metal and compare your average result with a reference value. Calculate a percent difference between your result and the reference value. If your value does not match the reference value exactly then explain what factors may have led to your result being too high or too low.
  3. What is the molar heat (or enthalpy) of solution of KNO3 according to your data? Include a data table with relevant data. Show the setup and conclusion of your calculation.
  4. The accepted value of the enthalpy of solution for KNO3 is +34.9 kJ/mol. Calculate a percent difference between your result and this accepted value.
  5. How reliable is your determination of the molar heat of solution of KNO3? Consider both precision (how close are your two measurements?) and accuracy (how close is your result to the accepted value?). What possible sources of error are there in making this measurement? Identify at least two and work out their consequences as far as how they will affect your measurement.
  6. What change in temperature would you expect from your calorimeter if it contains 60.0 g of water at 25.0°C and you add 50.0 g of silver that has been heated to 100°C? (Use your textbook or other resource to find the specific heat of silver metal).
  7. Which is lower in chemical potential energy: a mixture of methane and oxygen gas? Or a mixture of carbon dioxide gas and water vapor? How do you know? (Hint: write a chemical reaction relating these four substances using as a model one of the basic types of chemical reactions you have learned about).
  8. Consider the process of freezing water. When liquid water becomes ice, is this process one in which heat is released or absorbed? In other words, is freezing exothermic or endothermic? Explain.
  9. Why is the temperature of the boiling water constant even as the bunsen burner below it pours heat into the water?

Grading

Answer the questions in the analysis section of this lab handout (above) in a typed document.




Last updated: Nov 18, 2020
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