specific heat capacity experiment data

Required Practical: Investigating Specific Heat Capacity ( AQA GCSE Physics )

Revision note.

Required Practical 1: Investigating Specific Heat Capacity

Aims of the experiment.

• The aim of the experiment is to determine the specific heat capacity of a substance, by linking the amount of energy transferred to the substance with the rise in temperature of the substance
• Independent variable = Time,  t
• Dependent variable = Temperature, θ
• Material of the block
• Current supplied,  I
• Potential difference supplied,  V

Equipment List

• Thermometer = 1 °C
• Stopwatch = 0.01 s
• Voltmeter = 0.1 V
• Ammeter = 0.01 A

Apparatus to investigate the specific heat capacity of the aluminium block

• Start by assembling the apparatus, placing the heater into the top of the block
• Measure the initial temperature of the aluminium block from the thermometer
• Turn on the power supply and start the stopwatch
• Whilst the power supply is on, the heater will heat up the block. Take several periodic measurements, eg. every 1 minute of the voltage and current from the voltmeter and ammeter respectively, calculating an average for each at the end of the experiment up to 10 minutes
• Switch off the power supply, stop the stopwatch and leave the apparatus for about a minute. The temperature will still rise before it cools
• Monitor the thermometer and record the final temperature reached for the block
• An example table of results might look like this:

Analysis of Results

• The thermal energy supplied to the block can be calculated using the equations:
• E = thermal energy, in joules (J)
• Q = Charge, in coulombs (C)
• I = current, in amperes (A)
• V = potential difference, in volts (V)
• t = time, in seconds (s)
• Rearrange to make Q the subject
• Substitute into the Q = It equation
• Rearrange to make E the subject
• The change in thermal energy is defined by the equation:
• Δ E = change in energy, in joules (J)
• m =  mass, in kilograms (kg)
• c =  specific heat capacity, in joules per kilogram per degree Celsius (J/kg °C)
• Δ θ =  change in temperature, in degrees Celsius (°C)
• Rearranging for the specific heat capacity, c :
• To calculate Δθ:
• To calculate Δ E:
• I = average current, in amperes (A)
• V =  average potential difference (V)
• θ f = final time, in seconds (s)
• θ i = initial time, in seconds (s)
• These values are then substituted into the specific heat capacity equation to calculate the specific heat capacity of the aluminium block

Evaluating the Experiment

Systematic Errors:

• Make sure the voltmeter and ammeter are initially set to zero, to avoid zero error

Random Errors:

• This means the measured value of the specific heat capacity is likely to be higher than what it actually is
• To reduce this effect, make sure the block is fully insulated
• This would eliminate errors from the voltmeter, ammeter and the stopwatch
• Make sure the temperature value is read at eye level from the thermometer, to avoid parallax error
• The experiment can also be repeated with a beaker of water of equal mass, the water should heat up slower than the aluminium block

Safety Considerations

• Run any burns immediately under cold running water for at least 5 minutes
• Allow time for all the equipment, including the heater, wire and block to cool before packing away the equipment
• Keep water away from all electrical equipment
• Wear eye protection if using a beaker of hot water

Teacher tip

From my experience of teaching this practical investigation, it is usually done as a teacher demonstration because of the specialist equipment used. Therefore, students find it more difficult to engage with than a hands-on practical, and so can often find it quite confusing. This Required Practical does come up in exams very frequently, so you do need to make the extra effort to understand the equipment, the set-up and how the measurements are taken.

The main idea is that the measurements of current and potential difference allow you to calculate the energy transferred to the metal block, and from there you can use the change in temperature and the energy supplied to calculate the specific heat capacity of the metal.

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Author: Leander

Leander graduated with First-class honours in Science and Education from Sheffield Hallam University. She won the prestigious Lord Robert Winston Solomon Lipson Prize in recognition of her dedication to science and teaching excellence. After teaching and tutoring both science and maths students, Leander now brings this passion for helping young people reach their potential to her work at SME.

14 Calorimetry for Determination of Specific Heat Capacity

Calorimetry for determination of specific heat capacity.

This lab is designed to align with AAOT science outcome #1: Gather, comprehend, and communicate scientific and technical information in order to explore ideas, models, and solutions and generate further questions.

• digital device with spreadsheet program
• digital device with internet access
• Model the contents of a calorimeter as an isolated system to determine the unknown specific heat value of an unidentified metal.
• Identify the metal by comparing the determined specific heat value to tabulated values.
• Apply an improved model that includes calorimeter components in the isolated system and compare the results with the model that excludes the calorimeter from the system.
• Estimate the uncertainty in the specific heat value caused by cooling of the hot metal during the transfer to the calorimeter.

Experimental Methods

The video below shows the methods used to warm a piece of metal in a hot water bath and then transfer the metal to a Styrofoam cup filled with water inside a calorimeter. The Styrofoam cup provides insulation, and the cup itself is suspended within an aluminum cylinder with a layer of air between to provide additional insulation. The shiny metal cylinder also reduces radiative exchange of thermal energy. The cylinder itself is then suspended within a larger metal shell, once again separated by a layer of insulating air. A lid prevents significant convection or evaporation, but does have a hole to allow insertion of of a thermometer to monitor the contents of the calorimeter.  Due to the hold in the lid, the calorimeter is an open system. If we assume the hole is sufficiently small that there is negligible exchange of particles in/out of the calorimeter then we can treat the contents as a ___________ system. If we also assume that the insulation is essentially perfect and the heat transfer in/out of the calorimeter is negligible then we can treat the contents as an _____________ system.

Analysis Methods

2) Enter the values from the video into your model for the metal specific heat and calculate a value. Show you work.

Conclusions

3) Look at this table of specific heat values and compare your value to those listed to identify the metal in this experiment. List the value of the metal you chose from the list and calculate a % difference from the value found in this experiment.

Further Questions

4)  List some possible sources of uncertainty in our determination of the specific heat value that were related to the experimental methods and measurement tools, but not caused by mistakes in calculations.

5) Our model neglected any temperature change by the Styrofam cup in the calorimeter. Now include the Styrofoam cup in the model and recalculate the specific heat value of the metal.  Assume the initial and final temperatures of the Styrofoam were the same as the water it contained. Look up a value for Styrofoam specific heat capacity from a reliable source and provide a full URL for your source. Show all work.

6) How much did the determined specific heat value change when accounting for heating of the Styrofoam cup, and did it increase or decrease? Calculate a % difference between the two values produced by the two models.

7) Did accounting for this effect bring the specific heat value closer to a different metal the the one you identified previously? Explain.  Does it appear to be necessary to include the Styrofoam cup in the analysis for this type of experiment? Explain your reasoning.

Our analysis methods neglected any cooling of the metal during the transfer from the hot water bath to the calorimeter. The following infrared images indicate that the metal is radiating and that air surrounding the metal is undergoing natural convection (which implies that the air is first warmed by conduction from the metal).

To estimate how much cooling may have occurred, another experiment was performed. The metal was once again removed from the hot water bath and it’s temperature was measured and recorded by an infrared thermometer as seen in the following video.

8) How much did the temperature of the metal change during three seconds in the air just after being removed from the hot water bath?

9) Assume the same temperature change occurred for the metal in the original calorimetry experiment and that the metal was actually that much colder than you originally thought. Recalculate the metal specific heat with the expected lower initial temperature value. Show all work.

10) How much did the determined specific heat value change when accounting for cooling of the metal during transfer, and did it increase or decrease? Calculate a % difference between the two values produced by the two models.

11) Did accounting for this effect bring the specific heat value closer to a different metal than the one you identified previously? Explain. Does it appear necessary to be concerned about the heat loss during transfer to the calorimeter for this type of experiment? Explain your reasoning. (Consider the error in determining specific heat that you discovered was caused by these effects and compared to the difference in specific heat between metals).

12) The metal used in the experiment was actually zink. Was that the metal you identified?

13) Calculate a % difference between the final experimental specific heat capacity of the metal and the tabulated value for zink (look it up and provide a citation for your source).

General Physics Remote Lab Manual Copyright © by Lawrence Davis is licensed under a Creative Commons Attribution-NonCommercial 4.0 International License , except where otherwise noted.

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• CBSE Class 11
• CBSE Class 11 Physics Practical
• To Determine Specific Heat Capacity Of A Given Solid By Method Of Mixture

To Determine Specific Heat Capacity of a Given Solid by Method of Mixture

Specific heat capacity is defined as the amount of heat required for one gram of substance at 1℃. To understand how to determine the specific heat capacity of a given solid by the method of mixtures, please read the below article.

To determine the specific heat capacity of a given solid by the method of mixtures.

Materials Required

• A hypsometer
• Calorimeter
• A lid and outer jacket
• Solid in small pieces
• Two half-degree thermometers
• Clamp stand

The calorimeter is a device used to measure the heat flow of a chemical or physical reaction. Calorimetry is the process of measuring this heat. It consists of a metal container to hold water above the combustion chamber and a thermometer to measure the temperature change. Hypsometer is an instrument used to determine the boiling point of water at a given altitude.

• Put thermometers A and B in a beaker containing water and note their reading. Let thermometer A be a standard to find the correction that is to be applied to thermometer B.
• Put the thermometer B into the copper tube of a hypsometer containing the powder of the given solid. Before placing the hypsometer on the burner, add a sufficient amount of water.
• Record the weight of the calorimeter with a stirrer and lid over it.
• Add water (temperature between 5 to 8℃) to the calorimeter at half-length and weigh it again.
• Heat the hypsometer till the temperature of the solid is steady.
• Note the temperature of water in calorimetry. Now slowly stir and add the solid powder from the hypsometer to the calorimeter and record the final temperature of the mixture.
• Remove thermometer A from the calorimeter.
• Note the weight of the calorimeter with the contents and lid.

Observations

Reading of thermometer A = T A = ……… ℃

Reading of thermometer B = T B = ……… ℃

Correction applied in B with respect to A (T A – T B ) = …….. ℃

Mass of calorimeter and stirrer m = …….. g

Water equivalent of calorimeter ω = m × 0.095 = …….. g

Specific heat of copper calorimeter = 0.095 cal/g

Mass of calorimeter + stirrer + lid = m 1 = ……. g

Mass of calorimeter + stirrer + lid + cold water = m 2 = …… g

The steady temperature of hot solid = T s = …….. ℃

Corrected temperature of hot solid T =T s – (T A – T B ) = ……… ℃

The temperature of cold water = t =……… ℃

The temperature of the mixture = Ө =…….. ℃

Mass of calorimeter, stirrer, lid, cold water, and solid = m 3 =…… g

Calculations

Mass of cold water = m 2 – m 1 =…….. g

Mass of hot solid = m 3 – m 2 =…….. g

Rise of the temperature of cold water and calorimeter = Ө – t =………. ℃

Fall in temperature of solid = T – Ө =……… ℃

Heat gain by calorimeter, cold water and stirrer = [ω + ( m 2 – m 1 )(Ө – t)] =……. (a)

Heat lost by solid = (m 3 – m 2 ) × C × (T – Ө) =……. (a)

Here, C is the specific heat of solid to be calculated.

According to the principle of calorimeter, heat lost = heat gained

Specific heat of given solid by the method of the mixture is ………. cal g -1 ℃ -1 .

Precautions

• A sufficient amount of solid powder should be used to cover the tip of the thermometer.
• A sufficient amount of water should be taken in a hypsometer.
• Dropping of solid should be quick and gentle.
• To avoid excess radiation, the calorimeter should be polished from the outside.
• The temperature of cold water should not cross the dew point.

Sources of Error

• There might be heat loss while transferring solid into the calorimeter.
• During conduction, convection, and radiation, there might be heat loss.
• The bulbs of the thermometer might not be completely inside the solid.

Note: To determine the specific heat of a given liquid by the method of mixtures, the liquid is taken in place of cold water to determine the specific heat. Proceed with the same procedure as in the experiment.

Q1. What is the heat?

Ans: Heat is defined as the quality of being hot at high temperatures.

Q2. Define the specific heat of the substance.

Ans: It is defined as the amount of heat energy required to raise the temperature of 1 gram of a substance by 1℃.

Q3. State the principle of calorimetry.

Ans: The principle of calorimetry is heat lost is equal to the heat gained.

Q4. Why is calorimeter made of copper is used in the experiment?

Ans: The specific heat of copper is very low. The absorption and liberation of heat will be less during heat transfer.

Q5. Is heat gained always equal to the heat lost?

Ans: If there is no chemical reaction taking place between the components only then heat loss will be always equal to the heat gain.

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1.0 Introduction

When performing an experiment and carefully following the proper procedure methods, the results obtained should be relatively accurate. Many experiments require multiple trials and sometimes never result in a complete conclusion.

The specific heat test described in this article required many modifications and repetitions to display conclusive results. The results obtained were relatively accurate even taking into consideration the modification of using a calorimeter made of Styrofoam cups. Often results from experiments using modifications can be inaccurate or inconclusive, especially when there is only a single recorded data point. In this article we will go through proper data acquisition and analysis.

When using proper procedure for this in-home experiment, it is unlikely that 100% accuracy will be achieved due to the unstoppable nature of thermodynamics. Following this method can give an accuracy of approximately 80%. This is a respectable degree of accuracy to be achieved from a simple DIY application.

2.0 Procedure

The procedure for this experiment is thoroughly covered in the coinciding Specific Heat Test article. A brief summary of the procedure is outlined below.

A mass of water was measured than poured into the calorimeter, the water remained there until it reached room temperature. For the most accurate and readable results, there should only be enough water in the calorimeter to completely cover the sample.

Figure 1: Calorimeter made of two Styrofoam cups and a thermometer

A beaker filled with approximately 300mL of water was placed on a hot plate. A sample of the stainless-steel variety used for the purpose of this experiment was then placed in a test tube and set in a stand so that the majority of the tube was submerged in the beaker. The sample tube was placed vertically to ensure that it did not come into contact with the bottom or sides of the beaker. Once the sample tube was placed properly; the hot plate was turned on.

Figure 2: Beaker on a hot plate with a test tube containing the sample submerged in boiling water

During the initial trial of this experiment a solid 25.22-gram cylindrical stainless-steel-316 sample was used. Using a test tube to hold the sample in the water (as shown above), did not lead to favorable results. This poor experimental outcome is likely due to the sample’s radius being smaller than the radius of the test tube. This additional space around the sample added a layer of insulating air in the test tube. This error could be averted if the sample were a powder or composed of smaller pieces.

However, if changes were made to the sample at this point in the experiment, the sample would be completely destroyed. For the remainder of the experiment the sample was held in place with a pair of insulated tongs instead of the test tube. Altering this element of the procedure for the remainder of the trials dramatically improved the results.

Figure 3: Beaker on a hot plate with a sample submerged and hovered above the bottom by tongs

Once the water in the beaker was boiling, it was recommended to wait approximately 10 minutes to ensure the sample is evenly heated. Note that to achieve a boiling condition the sample needs to be heated to 100 degrees Celsius.

2.3 Material Transfer

After 10 minutes the test tube was detached from the stand and the sample was poured into the calorimeter. The sample must be poured safely but quickly into the calorimeter, so that a minimal amount of heat is lost to the surrounding air. It is also crucial not to transfer any water from the beaker into the calorimeter. If water from the beaker entered the calorimeter in would add heat that is unrelated to the sample and alter the calculated mass of water.

A Styrofoam cup calorimeter was used for the first two trials of this experiment. For the third and fourth trials a metal thermos replaced the Styrofoam cups to determine the difference a conductive material had on the results.

Figure 4: A Calorimeter made from a thermos and a digital thermometer

2.4 Record Data

Once the sample was stabilized in the enclosed calorimeter, data was collected from the thermometer. For the purposes of this experiment, data was recorded at every 30 second interval. However, this experiment is not time sensitive so depending on the conditions, data can be reordered at multiple different intervals. Data was continuously recorded until the temperature of the substance started to drop indicating that the highest possible temperature was achieved.

Recording temperatures at intervals is the recommended option compared to just waiting for the temperature peak. Having a consistent recording time ensures the most accurate results and minimalizes chances for error.

3.0 Analyzing your Data

3.1 a preliminary look.

The data obtained from the temperature analysis was entered into Microsoft excel (any program would suffice). Using Microsoft excel the data was placed into a graph format for easy comparison between trials. The graphs produced from the data would indicate any errors that could have possibly occurred during the experimental trials. If errors are observed the experiment would have to be repeated before continuing further analysis.

Acceptable data will display a noticeable curve in the graph indicating the relationship between temperature and time. Additional modifications that could be implemented to improve results could be breaking the sample into smaller pieces, having a larger mass of sample or having less water in the calorimeter. All of these would impact the deviation in the temperature difference displaying more drastic changes in temperature. Increased variation would make the results more noticeable and the data calculations would also be more accurate.

3.2 A Deeper Look

The time that the peak temperature occurred is displayed noticeably in the graphs of the experimental data. Depending on the compatibility of the data with the software being used for the analysis a line of best fit can be produced from the peak temperature. Errors bars are optional depending on the purpose of the experiment and the results obtained. The majority of the error in the results would be due to the unpreventable loss of heat to the air and the calorimeter.

3.3 Results

Figure 5: Graph displaying the relationship between Temperature (°C) and Time (minutes) during the first trial using a Styrofoam cup calorimeter and heated sample test tube

Figure 7: Graph displaying the relationship between Temperature (°C) and Time (minutes) during the third trial using a thermos as the calorimeter, tongs held the sample while heated.

Figure 6: Graph displaying the relationship between Temperature (°C) and Time (minutes) during the second trial using a Styrofoam cup, tongs held the sample while heated.

Figure 8: Graph displaying the relationship between Temperature (°C) and Time (minutes) during the fourth trial using a thermos at room temperature as the calorimeter, tongs held the sample while heated.

(Note that the ideal form of a graph of temperature change versus time due to a gradient would be logarithmic.)

3.4 Calculations

The first law of thermal dynamics states that all energy in the universe is conserved. This law can be applied to the results from the experiment. When the sample cools down, energy is lost in the form of heat. As the law states, this energy doesn’t just disappear, in the case of this experiment that energy is absorbed by the water. This increase in energy will cause the water’s temperature to rise. Measuring this change determines the amount of heat absorbed by the water which leads to determining the heat lost by the stainless-steel sample.

Q w = -Q s    Where Q is the change in heat.

Using specific heat of water and the mass of water being used, we can express the above more informatively.

Q w = c w m w (T peak – T initial ) w = -Q s = c s m s (T peak – T initial ) s

The formula can then be rearranged for determining the specific heat of the sample.

c s = c w m w ΔT w / m s ΔT s

3.5 Discussing Results

For the first trial of the experiment, Styrofoam cups were used as the calorimeter along with a test tube for holding the sample. The results from this trial were inconclusive and had a large amount of error. In the second trial a pair of insulated tongs replaced the test tube for holding the sample. The results from this trial displayed substantially less error. Modifications were made for the second trial in an attempt to produce viable results.

In the second trial, the water in the calorimeter rested longer at room temperature before the data was collected. This modification produced a result of 305.4 Joules per kilogram kelvin, which was 67.9% accurate to the theoretical value of grade 316 stainless steel, 470.

Another modification was used for the third and fourth trial to determine the change in accuracy when using a conductive calorimeter. A standard beverage thermos was used as the calorimeter for these trials. The thermometer used to measure temperature was upgraded to a digital one with long wire probes that enabled the thermos to have a tight seal. The first trial using this modification produced a result of 553.29 Joules per kilogram kelvin, which is 84.9% accurate, however, it was above the theoretical value. Accuracy above theoretical value indicated that an error was made.

Using a thermos as the calorimeter means that the water would take longer to reach room temperature . It is debatable if the water inside the thermos not being at room temperature would affect the results of the experiment. However, after further analysis the size of the thermos was determined as the cause of error in accuracy values. Due to the large size of the thermos, additional spaced remained inside the container for room temperature air to be trapped when the lid was placed on. This large volume of air was in direct contact with the water inside the thermos and affected the water’s temperature.

A further modification was made for the fourth trial of this experiment to limit error caused by the room temperature air. Only enough water to cover the sample was added and it was stressed that this water was at room temperature. This modification received results of 367.56 Joules per kilogram kelvin, which is 78.2% accurate.

4.0 Conclusion

An accurate result was not the goal of these experiments. Science is a slow process that is full of opportunity for mistakes, trial and error, and improvement. Scientific theories take years to develop and even longer for the scientific community to accept them. Science is an art and should be played with. Most earth-shattering discoveries were made by accident or with a different end goal in mind. For experiments such as the one described above, it is good practice to aim for an 80% accuracy for an at home value. When experimenting, it is key to have an open mind, roll with punches and most of all be safe.

5.0 References

‘Theory of Heat’ – Maxwell, James Clerk – page 57-67 – Westport, Conn., Greenwood Press – 1970: https://archive.org/details/theoryheat04maxwgoog/page/n77 -Talks about conservation of heat, the form, function and a bit of history of calorimeters, and the Method of Mixture. It is also a good book to understand heat in general, and it’s free.

Author: Cole Boucher, Junior Technical Writer at Thermtest

Can’t find the right product for your testing?

Measure the Specific Heat of Water and Other Fluids

Introduction: Measure the Specific Heat of Water and Other Fluids

Step 1: Equipment

• Digital postal scale
• Plastic cup that will hold at least 250ml
• Variable power supply]
• Digital thermometer with probe
• 7.5 ohm, 5W resistor] (or something close)
• Short length of wire
• Clock showing time in seconds (not shown)
• 250 ml of cold water (tap water will do, distilled is better) (not shown)

Step 2: Setup

1. Prepare some cold water (preferably distilled) by putting it in a container in a refrigerator for an hour or so. You want to start with cold water so that your experimental data will include temperatures on either side of the ambient temperature. 2. Put the cup on the digital scale and zero the scale. 3. Pour cold water into the cup until the scale reads at least 250 g. Record the mass "M" of water that you actually added. 3. Strip a short length of the insulation from the wire and connect the resistor in series with the variable power supply by twisting the bare wires around the resistor leads. 4. Put the resistor into the cup of water so that the resistor is submerged 5. Record the ambient temperature "Ta" and then put the temperature probe into the water as well.

Step 3: Procedure

1. Turn on the power supply and adjust the voltage to around 7.5 volts (or the value of the resistor that you used). Record the voltage "V" and the current "I". Note that with a DC power supply there is no danger of electric shock. You can handle the bare wires by hand. 2. Record the water temperature every 2 minutes until the temperature is about 5 degrees C above the ambient room temperature. Aside: Ohm's law states that Voltage (V), Current (I), and Resistance (R) are related by the formula V = I*R (where V is in volts, I is in Amps, and R is in Ohms). You measured V and I. Therefore you can calculate R = V/I and compare this with the known value of your resistor. In my case, V=7.5V, and I=1.00A. Therefore R=V/I = 7.5 Ohms which is the value of the resistor that I used. Hurray!

Step 4: Results

The ambient temperature was 20.1 degrees C. One of the datapoints was 20.2 degrees C which is very close to ambient. To minimize error due to heat transfer to or from the surroundings, lets look at the data from 10 minutes before till 10 minutes after this datapoint.

Step 5: Conclusion

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Specific heat.

Students use a temperature sensor to experimentally determine the identity of a metal based on its specific heat capacity.

Supports NGSS Performance Expectation HS-PS3-1: Create a computational model to calculate the change in the energy of one component in a system when the change in energy of the other component(s) and energy flows in and out of the system are known.

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