what is the cooling curve experiment

Investigating Cooling Curves for Stearic Acid ( Oxford AQA IGCSE Physics )

Revision note.

Physics Content Creator

Required Practical: Investigating Cooling Curves for Stearic Acid

Aim of the experiment.

The aim of this experiment is to obtain a cooling curve for stearic acid as it is cooled and use the curve to find the melting point of stearic acid

Independent variable = Time, t

Dependent variable = Temperature, T

Control variables :

Volume of acid,  V

Initial mass of acid, m

Equipment List

Resolution of measuring equipment:

Thermometer = 1 °C

Stopwatch = 0.01 s

Equipment for obtaining the cooling curve of stearic acid

Secure the boiling tube of boiling stearic acid in place with the clamp stand inside the beaker of boiling water. Position the thermometer and start your stopwatch

Record the temperature in degrees Celsius every minute until it reaches about 40 °C

For each reading state in the results table whether the stearic acid looks more like a solid or a liquid

Example Results Table

Analysis of results

Plot a graph of temperature against time

Draw a line for the cooling curve passing through all the indicated points

If the cooling curve shows a horizontal straight line then this indicates the time when the stearic acid is changing state

An example of a cooling curve for stearic acid

Evaluating the experiment

Systematic Errors:

Make sure the measurements on the thermometer and stopwatch are taken at eye level to avoid parallax error

Random Errors:

Make sure the thermometer is fully in contact with the stearic acid and not in contact with the glass boiling tube

Repeat the experiment several times

Make sure the thermometer and the boiling tube are in the same depth every time

Safety considerations

Wear goggles during this experiment in case the stearic acid splatters

Wear gloves during this experiment in case the stearic acid spills

Stand up while carrying out the experiment making sure not to knock the clamp stand and boiling tube or beaker

Place a mat or a soft material below the beaker to absorb any acid that might spill

Use a G clamp to secure the clamp stand to the desk so that the boiling tube does not fall over

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Author: Ann Howell

Ann obtained her Maths and Physics degree from the University of Bath before completing her PGCE in Science and Maths teaching. She spent ten years teaching Maths and Physics to wonderful students from all around the world whilst living in China, Ethiopia and Nepal. Now based in beautiful Devon she is thrilled to be creating awesome Physics resources to make Physics more accessible and understandable for all students, no matter their schooling or background.

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Melting and freezing stearic acid

In association with Nuffield Foundation

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In this experiment, a solid turns into a liquid and then the liquid turns into a solid. The energy changes are examined

The students will take the temperature of stearic acid at regular intervals as they heat and cool it. They can observe the melting and freezing points of the acid and can plot a   graph. This experiment could also be done using data-logging equipment

This practical takes quite a long time to carry out. Students can begin by simply recording their data but, once they get the hang of what they are doing, most should be able to plot the graph at the same time as taking readings. If data-loggers are being used then students will need another activity to be doing alongside the experiment.

• Eye protection
• Beaker (250 cm 3 )
• Boiling tube (note 1)
• Thermometer (0–100˚C)
• Clamp, stand and boss
• Bunsen burner
• Heat resistant mat

Apparatus notes

• If, after the practical, the boiling tubes are left containing both the stearic acid and the thermometer, immerse all the boiling tubes in hot water to remove the thermometers. The stearic acid can then be stored in the boiling tubes and recycled several times.
• Stearic acid (octadecanoic acid)

Health, safety and technical notes

• Read our standard health and safety guidance
• Wear eye protection.
• Stearic acid (octadecanoic acid), CH 3 (CH 2 ) 16 COOH(s) – see CLEAPSS Hazcard HC038b . The stearic acid in this practical can be used again and again. Have enough to quarter fill a boiling tube for each student
• Put about 150 cm 3  water into the beaker.
• Heat it on a tripod and gauze until the water just starts to boil.
• Set up the apparatus as shown in the diagram and start the timer. Keep the water boiling, but not boiling vigorously.
• Using a suitable results table, record the temperature of the stearic acid every minute until it reaches about 70˚C. Note on your results table the point at which you see the solid start to melt.
• Use the clamp stand to lift the tube from the hot water. Record the temperature every minute as the stearic acid cools down until it reaches about 50˚C. Note on your results table the temperature at which you see the stearic acid begin to solidify.

Source: Royal Society of Chemistry

Apparatus set-up for the melting and freezing stearic acid experiment

Teaching notes

Remind students not to attempt to move the thermometer in the solid stearic acid, as it will break.

Energy must be supplied to melt a solid; this same energy is released when the liquid re-solidifies.

This presents a good opportunity to demonstrate how to maintain a steady temperature using a Bunsen burner. This can be achieved by sliding the Bunsen burner aside as the boiling becomes too vigorous; slide it back as the water stops boiling. It is not essential that the water bath is boiling. Students could be provided with another thermometer, and asked to maintain a lower temperature, say 80 °C.

A temperature sensor attached to a computer can be used in place of a thermometer. It can plot the temperature change on a graph and show this as it occurs. A slight modification of the experiment can yield an intriguing result: When the test tube is cooling place it in an insulated cup containing a few cm 3  of water. Use a second temperature sensor to monitor the temperature of the water. The water temperature should rise as the stearic acid cools and it should continue to rise even as it changes state.

A slight alternative to this experiment is to plot only the cooling curve. Place all the boiling tubes with stearic acid into a large beaker. Place some hot water in the beaker and continue to heat with a Bunsen burner. Remove from the heat when all the stearic acid has melted. Students can place a thermometer into the stearic acid and place the boiling tube into a test tube rack or beaker. They take the temperature every 30 seconds or every minute and plot a graph. Many students will anticipate that the stearic acid will continue to cool to zero – it is useful to discuss why the stearic acid stops cooling when it reaches room temperature.

In either version of the experiment it is good practice for students to draw a graph of their results. There should be a clear horizontal line in the graph which corresponds to the change of state, however many school samples of stearic acid are not very pure and hence the line is often not perfectly horizontal. The exact melting and freezing points of the stearic acid may not be exactly the same and will depend on the purity of the product and where it was purchased from, but are usually around 55–70 ˚C.

Additional information

This is a resource from the  Practical Chemistry project , developed by the Nuffield Foundation and the Royal Society of Chemistry.

Practical Chemistry activities accompany  Practical Physics  and  Practical Biology . 

© Nuffield Foundation and the Royal Society of Chemistry

• 11-14 years
• 14-16 years
• Practical experiments
• Properties of matter

Specification

• Melting and freezing take place at the melting point, boiling and condensing take place at the boiling point
• Describe how heating a system will change the energy stored within the system and raise its temperature or produce changes of state.
• Describe how, when substances melt, freeze, evaporate, condense or sublimate, mass is conserved but that these physical changes differ from chemical changes because the material recovers its original properties if the change is reversed.
• 2.3 Explain the changes in arrangement, movement and energy of particles during these interconversions
• 6. Investigate the properties of different materials including solubilities, conductivity, melting points and boiling points.

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Cooling Curves

A cooling curve is a line graph that represents the change of phase of matter, typically from a gas to a solid or a liquid to a solid. It shows how the temperature changes as a substance is cooled down.

" Cooling Curves " appears in:

Study guides ( 1 ).

• AP Chemistry - 6.5 Phase Changes and Energy

Related terms

Phase Change : This is when substances transition between solid, liquid, and gas states. It's like moving from one floor to another in our staircase analogy.

Thermal Equilibrium : This refers to when all parts of a system are at the same temperature. It's like everyone on one floor of our staircase being at the same level.

Endothermic Process : This is when heat energy is absorbed from surroundings during a chemical reaction. Imagine having to exert more effort (absorb more energy) to climb up the stairs.

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12.5: Interpretation of Cooling Curves

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• Dissemination of IT for the Promotion of Materials Science (DoITPoMS)
• University of Cambridge

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The melting temperature of any pure material (a one-component system) at constant pressure is a single unique temperature. The liquid and solid phases exist together in equilibrium only at this temperature. When cooled, the temperature of the molten material will steadily decrease until the melting point is reached.

At this point the material will start to crystallise, leading to the evolution of latent heat at the solid liquid interface, maintaining a constant temperature across the material. Once solidification is complete, steady cooling resumes. The arrest in cooling during solidification allows the melting point of the material to be identified on a time-temperature curve.

Most systems consisting of two or more components exhibit a temperature range over which the solid and liquid phases are in equilibrium. Instead of a single melting temperature, the system now has two different temperatures, the liquidus temperature and the solidus temperature which are needed to describe the change from liquid to solid.

The liquidus temperature is the temperature above which the system is entirely liquid, and the solidus is the temperature below which the system is completely solid. Between these two points the liquid and solid phases are in equilibrium. When the liquidus temperature is reached, solidification begins and there is a reduction in cooling rate caused by latent heat evolution and a consequent reduction in the gradient of the cooling curve.

Upon the completion of solidification the cooling rate alters again allowing the temperature of the solidus to be determined. As can be seen on the diagram below, these changes in gradient allow the liquidus temperature T L , and the solidus temperature T S to be identified.

When cooling a material of eutectic composition, solidification of the whole sample takes place at a single temperature. This results in a cooling curve similar in shape to that of a single-component system with the system solidifying at its eutectic temperature.

When solidifying hypoeutectic or hypereutectic alloys, the first solid to form is a single phase which has a composition different to that of the liquid. This causes the liquid composition to approach that of the eutectic as cooling occurs. Once the liquid reaches the eutectic temperature it will have the eutectic composition and will freeze at that temperature to form a solid eutectic mixture of two phases.

Formation of the eutectic causes the system to cease cooling until solidification is complete. The resulting cooling curve shows the two stages of solidification with a section of reduced gradient where a single phase is solidifying and a plateau where eutectic is solidifying.

By taking a series of cooling curves for the same system over a range of compositions the liquidus and solidus temperatures for each composition can be determined allowing the solidus and liquidus to be mapped to determine the phase diagram.

Below are cooling curves for the same system recorded for different compositions and then displaced along the time axis. The red regions indicate where the material is liquid, the blue regions indicate where the material is solid and the green regions indicate where the solid and liquid phases are in equilibrium.

By removing the time axis from the curves and replacing it with composition, the cooling curves indicate the temperatures of the solidus and liquidus for a given composition.

This allows the solidus and liquidus to be plotted to produce the phase diagram:

Cooling curve: practical

I can conduct an investigation to create and analyse cooling curves and identify where state changes are happening on temperature vs time graphs.

Lesson details

Key learning points.

• State changes are visible as plateaus on cooling curves.
• As it cools, a substance transfers energy into the surroundings by heating and it may condense or freeze.
• Energy is released to the surroundings as a substance cools and this is observed as a decrease in substance temperature.
• Graphs provide a visual representation of data for easier analysis and help identify trends/patterns.

Common misconception

All substances freeze when really cold, like water.

Show state changes using particle diagrams / kinetic energy model. Challenge pupils to identify state given real world temperature data.

Plateau - A plateau is a section of a graph that does not change value (stays at the same level for a period of time).

Melting point - The melting point of a substance is the temperature at which it changes from solid state to a liquid state.

Freezing - Freezing is the process of a substance changing from a liquid state to a solid state.

Temperature - Measured using a thermometer (commonly in °C). Temperature is an indirect measure of the energy of the particles in a substance.

Content guidance

• Equipment requiring safe usage.

Supervision

Adult supervision recommended.

This content is © Oak National Academy Limited ( 2024 ), licensed on Open Government Licence version 3.0 except where otherwise stated. See Oak's terms & conditions (Collection 2).

Starter quiz

6 questions.

substances in the solid state -  

particles have the strongest forces of attraction

substances in the liquid state -  

particles have some (weakened) forces of attraction

substances in the gas state -  

particles have fully overcome forces of attraction

Cooling Curves

Section 1: Understanding Cooling Curves

• A cooling curve graphically represents the change in state of matter as a substance cools and transitions from gas to solid.
• The curve reflects changes in temperature over time as the substance releases heat.

Section 2: Cooling Curve Components

• Initial drop : The region of the curve starting from the top that indicates the cooling of the substance in its gaseous state.
• Plateau : The flat segment of the curve where the substance is transitioning from one state to another, during which the temperature does not change.
• Subsequent drop : The region of the curve following each plateau also represents cooling, but in the next state of matter.

Section 3: Interpreting Cooling Curves

• First plateau: Gas to liquid (Condensation)
• Second plateau: Liquid to solid (Freezing)
• The length of each plateau is proportional to the amount of energy released during that transition.

Section 4: Practical Application of Cooling Curves

• Cooling curves can help to determine the melting and boiling points of substances.
• The process of creating a cooling curve involves careful monitoring of temperature changes in controlled conditions.

Section 5: Safety Considerations with Cooling Curves

• When conducting experiments to generate cooling curves, ensure all safety protocols are considered.
• Care must be taken when heating substances initially, and substances should be handled appropriately as they cool.

Section 6: Precision in Creating and Reading Cooling Curves

• Accurate data collection is paramount in the experiment to ensure the curve reflects the substance’s true characteristics.
• The cooling curve must be read accurately to provide valid conclusions regarding the substance’s boiling and melting points.

Dissemination of IT for the Promotion of Materials Science (DoITPoMS)

Interpretation of cooling curves

The melting temperature of any pure material (a one-component system) at constant pressure is a single unique temperature. The liquid and solid phases exist together in equilibrium only at this temperature. When cooled, the temperature of the molten material will steadily decrease until the melting point is reached.

At this point the material will start to crystallise, leading to the evolution of latent heat at the solid liquid interface, maintaining a constant temperature across the material. Once solidification is complete, steady cooling resumes. The arrest in cooling during solidification allows the melting point of the material to be identified on a time-temperature curve.

Most systems consisting of two or more components exhibit a temperature range over which the solid and liquid phases are in equilibrium. Instead of a single melting temperature, the system now has two different temperatures, the liquidus temperature and the solidus temperature which are needed to describe the change from liquid to solid.

The liquidus temperature is the temperature above which the system is entirely liquid, and the solidus is the temperature below which the system is completely solid. Between these two points the liquid and solid phases are in equilibrium. When the liquidus temperature is reached, solidification begins and there is a reduction in cooling rate caused by latent heat evolution and a consequent reduction in the gradient of the cooling curve.

Upon the completion of solidification the cooling rate alters again allowing the temperature of the solidus to be determined. As can be seen on the diagram below, these changes in gradient allow the liquidus temperature T L , and the solidus temperature T S to be identified.

When cooling a material of eutectic composition, solidification of the whole sample takes place at a single temperature. This results in a cooling curve similar in shape to that of a single-component system with the system solidifying at its eutectic temperature.

When solidifying hypoeutectic or hypereutectic alloys, the first solid to form is a single phase which has a composition different to that of the liquid. This causes the liquid composition to approach that of the eutectic as cooling occurs. Once the liquid reaches the eutectic temperature it will have the eutectic composition and will freeze at that temperature to form a solid eutectic mixture of two phases.

Formation of the eutectic causes the system to cease cooling until solidification is complete. The resulting cooling curve shows the two stages of solidification with a section of reduced gradient where a single phase is solidifying and a plateau where eutectic is solidifying.

By taking a series of cooling curves for the same system over a range of compositions the liquidus and solidus temperatures for each composition can be determined allowing the solidus and liquidus to be mapped to determine the phase diagram.

Below are cooling curves for the same system recorded for different compositions and then displaced along the time axis. The red regions indicate where the material is liquid, the blue regions indicate where the material is solid and the green regions indicate where the solid and liquid phases are in equilibrium.

By removing the time axis from the curves and replacing it with composition, the cooling curves indicate the temperatures of the solidus and liquidus for a given composition.

This allows the solidus and liquidus to be plotted to produce the phase diagram:

Practical Science

Table of Contents

Cooling Curve

A cooling curve is a graph that shows the change in temperature of a substance over time as it cools down from a melted state to a solid state. This type of experiment can be useful for studying the cooling behaviour of substances and can also provide information about the purity and crystalline structure of the substance. In this experiment, we will produce a cooling curve of stearic acid using a water bath set to 80 degrees Celsius and a 250 ml beaker.

• Stearic acid (25 g)
• 100 ml beaker
• Thermometer
• Stirring rod
• Stopwatch or timer

Safety Precautions:

• Wear appropriate protective goggles.
• Be cautious when handling hot materials and equipment.
• Make sure to dispose of all chemicals properly.

Experimental Procedure:

• Set up the water bath by filling a large beaker with water and placing it on a hotplate. Heat the water to 80 degrees Celsius.
• Add 25 grams of stearic acid to a 100 ml beaker.
• Place the beaker in the water bath and melt the stearic acid by stirring with a stirring rod.
• Once the stearic acid has melted completely, remove the beaker from the water bath and start the cooling process.
• Record the temperature of the stearic acid every 10 seconds using a thermometer until it solidifies completely.
• Plot the temperature versus time data on a graph to produce the cooling curve.

Data Analysis:

• Observe the cooling curve and identify any plateau regions. These plateau regions indicate that the stearic acid is undergoing a phase change and solidifying.
• Determine the melting point of the stearic acid from the cooling curve. The melting point is the temperature at which the stearic acid changes from a solid to a liquid.
• Calculate the cooling rate of the stearic acid by dividing the temperature change by the time interval for each section of the cooling curve.
• Analyse the graph to determine the purity and crystalline structure of the stearic acid. A pure substance will exhibit a sharp melting point and a crystalline structure will have a characteristic pattern in the cooling curve.

Data Processing:

Calculate how much heat energy was released by stearic acid during freezing using this formula:

Where Q is heat energy in joules (J), m is mass in grams (g), and Lf is latent heat of fusion in J/g. The value of Lf for stearic acid is 247 J/g.

Conclusion:

In this experiment, we have produced a cooling curve of stearic acid using a water bath set to 80 degrees Celsius and a 250 ml beaker. The cooling curve provides information about the cooling behaviour, melting point, purity, and crystalline structure of the stearic acid. The experimental procedure outlined above can be used by IB DP chemistry students to gain practical experience in the laboratory and to better understand the behaviour of materials during phase changes.

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11.7: Heating Curve for Water

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• Page ID 46976

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Freezing, condensation, and deposition, which are the reverse of fusion, sublimation, and vaporization—are exothermic. Thus heat pumps that use refrigerants are essentially air-conditioners running in reverse. Heat from the environment is used to vaporize the refrigerant, which is then condensed to a liquid in coils within a house to provide heat. The energy changes that occur during phase changes can be quantified by using a heating or cooling curve.

Heating Curves

Figure \(\PageIndex{3}\) shows a heating curve, a plot of temperature versus heating time, for a 75 g sample of water. The sample is initially ice at 1 atm and −23°C; as heat is added, the temperature of the ice increases linearly with time. The slope of the line depends on both the mass of the ice and the specific heat ( C s ) of ice, which is the number of joules required to raise the temperature of 1 g of ice by 1°C. As the temperature of the ice increases, the water molecules in the ice crystal absorb more and more energy and vibrate more vigorously. At the melting point, they have enough kinetic energy to overcome attractive forces and move with respect to one another. As more heat is added, the temperature of the system does not increase further but remains constant at 0°C until all the ice has melted. Once all the ice has been converted to liquid water, the temperature of the water again begins to increase. Now, however, the temperature increases more slowly than before because the specific heat capacity of water is greater than that of ice. When the temperature of the water reaches 100°C, the water begins to boil. Here, too, the temperature remains constant at 100°C until all the water has been converted to steam. At this point, the temperature again begins to rise, but at a faster rate than seen in the other phases because the heat capacity of steam is less than that of ice or water.

Thus the temperature of a system does not change during a phase change . In this example, as long as even a tiny amount of ice is present, the temperature of the system remains at 0°C during the melting process, and as long as even a small amount of liquid water is present, the temperature of the system remains at 100°C during the boiling process. The rate at which heat is added does not affect the temperature of the ice/water or water/steam mixture because the added heat is being used exclusively to overcome the attractive forces that hold the more condensed phase together. Many cooks think that food will cook faster if the heat is turned up higher so that the water boils more rapidly. Instead, the pot of water will boil to dryness sooner, but the temperature of the water does not depend on how vigorously it boils.

The temperature of a sample does not change during a phase change.

If heat is added at a constant rate, as in Figure \(\PageIndex{3}\), then the length of the horizontal lines, which represents the time during which the temperature does not change, is directly proportional to the magnitude of the enthalpies associated with the phase changes. In Figure \(\PageIndex{3}\), the horizontal line at 100°C is much longer than the line at 0°C because the enthalpy of vaporization of water is several times greater than the enthalpy of fusion.

A superheated liquid is a sample of a liquid at the temperature and pressure at which it should be a gas. Superheated liquids are not stable; the liquid will eventually boil, sometimes violently. The phenomenon of superheating causes “bumping” when a liquid is heated in the laboratory. When a test tube containing water is heated over a Bunsen burner, for example, one portion of the liquid can easily become too hot. When the superheated liquid converts to a gas, it can push or “bump” the rest of the liquid out of the test tube. Placing a stirring rod or a small piece of ceramic (a “boiling chip”) in the test tube allows bubbles of vapor to form on the surface of the object so the liquid boils instead of becoming superheated. Superheating is the reason a liquid heated in a smooth cup in a microwave oven may not boil until the cup is moved, when the motion of the cup allows bubbles to form.

Cooling Curves

The cooling curve, a plot of temperature versus cooling time, in Figure \(\PageIndex{4}\) plots temperature versus time as a 75 g sample of steam, initially at 1 atm and 200°C, is cooled. Although we might expect the cooling curve to be the mirror image of the heating curve in Figure \(\PageIndex{3}\), the cooling curve is not an identical mirror image. As heat is removed from the steam, the temperature falls until it reaches 100°C. At this temperature, the steam begins to condense to liquid water. No further temperature change occurs until all the steam is converted to the liquid; then the temperature again decreases as the water is cooled. We might expect to reach another plateau at 0°C, where the water is converted to ice; in reality, however, this does not always occur. Instead, the temperature often drops below the freezing point for some time, as shown by the little dip in the cooling curve below 0°C. This region corresponds to an unstable form of the liquid, a supercooled liquid. If the liquid is allowed to stand, if cooling is continued, or if a small crystal of the solid phase is added (a seed crystal), the supercooled liquid will convert to a solid, sometimes quite suddenly. As the water freezes, the temperature increases slightly due to the heat evolved during the freezing process and then holds constant at the melting point as the rest of the water freezes. Subsequently, the temperature of the ice decreases again as more heat is removed from the system.

Supercooling effects have a huge impact on Earth’s climate. For example, supercooling of water droplets in clouds can prevent the clouds from releasing precipitation over regions that are persistently arid as a result. Clouds consist of tiny droplets of water, which in principle should be dense enough to fall as rain. In fact, however, the droplets must aggregate to reach a certain size before they can fall to the ground. Usually a small particle (a nucleus ) is required for the droplets to aggregate; the nucleus can be a dust particle, an ice crystal, or a particle of silver iodide dispersed in a cloud during seeding (a method of inducing rain). Unfortunately, the small droplets of water generally remain as a supercooled liquid down to about −10°C, rather than freezing into ice crystals that are more suitable nuclei for raindrop formation. One approach to producing rainfall from an existing cloud is to cool the water droplets so that they crystallize to provide nuclei around which raindrops can grow. This is best done by dispersing small granules of solid CO 2 (dry ice) into the cloud from an airplane. Solid CO 2 sublimes directly to the gas at pressures of 1 atm or lower, and the enthalpy of sublimation is substantial (25.3 kJ/mol). As the CO 2 sublimes, it absorbs heat from the cloud, often with the desired results.

Example \(\PageIndex{1}\): Cooling Hot Tea

If a 50.0 g ice cube at 0.0°C is added to 500 mL of tea at 20.0°C, what is the temperature of the tea when the ice cube has just melted? Assume that no heat is transferred to or from the surroundings. The density of water (and iced tea) is 1.00 g/mL over the range 0°C–20°C, the specific heats of liquid water and ice are 4.184 J/(g•°C) and 2.062 J/(g•°C), respectively, and the enthalpy of fusion of ice is 6.01 kJ/mol.

Given: mass, volume, initial temperature, density, specific heats, and \(ΔH_{fus}\)

Asked for: final temperature

Substitute the values given into the general equation relating heat gained to heat lost (Equation 5.39) to obtain the final temperature of the mixture.

When two substances or objects at different temperatures are brought into contact, heat will flow from the warmer one to the cooler. The amount of heat that flows is given by

\[q=mC_sΔT\]

where q is heat, m is mass, C s is the specific heat, and Δ T is the temperature change. Eventually, the temperatures of the two substances will become equal at a value somewhere between their initial temperatures. Calculating the temperature of iced tea after adding an ice cube is slightly more complicated. The general equation relating heat gained and heat lost is still valid, but in this case we also have to take into account the amount of heat required to melt the ice cube from ice at 0.0°C to liquid water at 0.0°C.

Exercise \(\PageIndex{1}\): Death by Freezing

Suppose you are overtaken by a blizzard while ski touring and you take refuge in a tent. You are thirsty, but you forgot to bring liquid water. You have a choice of eating a few handfuls of snow (say 400 g) at −5.0°C immediately to quench your thirst or setting up your propane stove, melting the snow, and heating the water to body temperature before drinking it. You recall that the survival guide you leafed through at the hotel said something about not eating snow, but you cannot remember why—after all, it’s just frozen water. To understand the guide’s recommendation, calculate the amount of heat that your body will have to supply to bring 400 g of snow at −5.0°C to your body’s internal temperature of 37°C. Use the data in Example \(\PageIndex{1}\)

200 kJ (4.1 kJ to bring the ice from −5.0°C to 0.0°C, 133.6 kJ to melt the ice at 0.0°C, and 61.9 kJ to bring the water from 0.0°C to 37°C), which is energy that would not have been expended had you first melted the snow.

• Physics Article
• Relationship Between Temperature Of Hot Body And Time By Plotting Cooling Curve

To Study the Relationship between the Temperature of a Hot Body and Time by Plotting a Cooling Curve

According to Newton’s law of cooling, the rate of cooling of a body is directly proportional to the difference in temperature of the body and the surrounding provided the difference in the temperature shouldn’t exceed 30°C. In this experiment, we will be validating the law by studying the relationship between the temperature of a hot body and time by plotting the cooling curve.

To study the relationship between the temperature of a hot body and time by plotting the cooling curve.

Apparatus and Materials Required

• Newton’s law of cooling apparatus (copper calorimeter with a wooden lid having two holes for inserting a thermometer and a stirrer and an open double – walled vessel)
• Two thermometers
• Clamp Stand
• Two rubber stoppers with holes
• Strong cotton threads

Newton’s law of cooling states that the rate of cooling of a body is directly proportional to the temperature difference between the body and the surrounding, provided the temperature difference is small.

Mathematically, it can be expressed as follows:

For a body of mass m , specific heat c , with a temperature T kept in the surrounding of temperature T 0 , the heat energy is given as follows:

Rate of cooling,

ms is a constant,

From the above relation, it is clear that as time increases, T decreases, ( T – T 0 ) decreases, as result the fall of temperature ( dT / dt ) must also decrease.

• Fill the space between the double wall of the enclosure with water and keep it on top of a table.
• Fill two-thirds of the calorimeter with water heated to about 80 °C.
• Suspend the calorimeter inside the enclosure with a thermometer in it. Cover it with a wooden lid with a hole in the middle.
• Suspend a thermometer from the clamp and stand into the enclosure water and the other thermometer in calorimeter water.
• Note the least count of thermometers.
• Set the stop clock to zero and note down its least count.
• Note the temperature T 0 of water in the enclosure.
• Start stirring the water in calorimeter so that it cools uniformly.
• When the calorimeter has convenient temperature reading, note it down and start and stop the clock watch
• Continue stirring and note the temperature after every few minutes. The temperature falls quickly in the first few minutes
• Note down the enclosure water temperature every five minutes.
• When the temperature fall becomes slow, note down the temperature at an interval of two minutes for ten minutes and then an interval of 5 minutes.
• Stop when the fall of temperature becomes very slow.
• Record your observation as given in the table below

Observation

The Least count of enclosure water thermometer = _____ °C

The Least count of calorimeter water thermometer = _____ °C

The Least count of stop clock watch = _____ s

Calculation

• The temperature of water in the enclosure will be found to be the same. If not, then take its mean.
• Find the temperature difference ( T – T 0 )
• Plot a graph between temperature T and time t as shown below. This graph is known as the cooling curve of the liquid.

The temperature is seen to fall quickly in the beginning and then the difference in temperature slowly decreases. This is in agreement with Newton’s law of cooling.

1. Why is water used as a cooling and heating agent?

Answer: Water has high specific heat capacity because of which it absorbs more quantity of heat than others.

2. State Newton’s law of cooling.

Answer: Newton’s law of cooling states that the rate of change of temperature of a body is directly proportional to the difference between its own temperature and the temperature of its Surrounding.

3. Why does a baby need more wrapping of woollen clothes than an adult?

Answer: Mass and the rate of fall of temperature are inversely proportional to each other. As the mass of a baby is lesser than an adult man, the baby loses a larger amount of heat than a grownup adult. Hence, a baby needs more wrapping.

4. Why is it easy to heat a hot pudding in a large plate than it is to heat it in a bowl?

Answer: The rate of cooling is directly proportional to the volume. Hence, a large plate cools faster than a bowl.

5. Define specific heat capacity.

Answer: The heat required to raise the temperature of unit mass by 1 o C is known as the specific heat capacity

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