specific heat of copper experiment

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Specific Heat of Copper

Description & background information.

• Lab description (1995)
• platinum resistance thermometer
• diffusion pump
• cooling trap

Pre-Lab Questions

Results to be reported.

• ©2007 The University of Utah
• Your Department or Organization Address
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• The University of Utah

Specific Heat

• 1.1.1 Thermal Energy Equation
• 1.1.2 Law of Dulong and Petit
• 1.1.3 Einstein-Debye Model
• 1.2 Computational Model
• 2.2 Middling
• 2.3 Difficult
• 3 Connectedness
• 5.1 Further reading
• 5.2 External links
• 6 References

The Specific Heat Capacity of a substance, also known as the Specific Heat, is defined as the amount of energy required to raise the temperature of one gram of the substance by one degree Celsius. Specific Heat Capacity is important, as it can determine the thermal interaction a material has with other materials. We can test the validity of models with Specific Heat Capacity since it is experimentally measurable. Also, the Specific Heat Capacity of a substance depends on its phase (solid, liquid, gas, or plasma) and its molecular structure. At its core, Specific Heat Capacity is based on the idea that different materials will store Heat differently, due to varying masses, molecular structure, and number of particles per unit mass. Finally, Specific Heat Capacity is an intensive property, meaning that the amount of the substance does not affect this property, only the composition of the substance does. It is worth noting the Specific Heat Capacity of a substance usually changes slightly with Temperature, as can be seen in the table for air on the right. However, in our studies, we will consider it as a constant.

There are a few quantities that are closely related to the Specific Heat Capacity of a substance:

• The concept of Heat Capacity is an extensive property (dependent on how much of the substance is present) that is integral to understanding how the Temperature of a substance rises and falls. Heat Capacity is the ratio of energy added or removed from a substance to the Temperature change observed in that substance. Typically, heat capacities are expressed in terms of the amount of heat (kJ, J, or kCal) that needs to be added to raise the temperature of a substance by 1 degree (Celsius, Fahrenheit, Kelvin)
• Specific Heat Capacity is an intensive property as mentioned previously. Conversely, Heat Capacity is an extensive property, meaning it does depend on the amount of substance present. In other words, the Specific Heat Capacity for 1 kg of iron is the same as that of 100 kg of iron, but the Heat Capacity would be different for these two amounts, since it takes more Heat to raise 100 kg of iron by one degree than it does to raise 1 kg of iron by one degree. To determine the Heat Capacity of a quantity of substance, simply multiply the Specific Heat Capacity by the amount of substance present

• Typical units of Heat Capacities are J/g, kJ/kg, and BTU/lb-mass. The SI unit of Heat Capacity is J/g
• Molar Heat Capacity is similar to Specific Heat Capacity . It expresses the amount of Heat required to raise one gram-mole of a substance by one degree Celsius
• It is expressed in J/mol-°C. The Molar Heat Capacity of water is 75.37 J/mol-°C

Mathematical Model

There are a few ways to find the Specific Heat Capacity of a material or system, such as the Thermal Energy equation, the Law of Dulong and Petit, or the Einstein-Debye Model.

Thermal Energy Equation

The relationship between the Heat and Temperature change of a system is best defined by the Specific Heat constant [math]\displaystyle{ C }[/math] in the equation below:

For a review of the meaning of this equation, view Thermal Energy Equation .

It is important to note this equation does not apply if a phase change occurs (say from a liquid state to a gaseous state).

Rearranging this equation gives us a way to calculate the Specific Heat Capacity of the system:

We can in fact see a dependence on the Temperature of the system here. We can rewrite this as:

If we know the dependence of the Specific Heat Capacity on Temperature, we can solve for the change in Thermal Energy.

• Law of Dulong and Petit

The Law of Dulong and Petit is a Thermodynamic law discovered in 1819 by the French physicists Pierre Louis Dulong and Alexis Thérèse Petit. It yields the expression for the Molar Specific Heat Capacity of certain chemical elements. They found, through experiments, that the Mass Specific Heat Capacity for many elements was close to a constant value, after it had been adjusted to reflect the relative atomic weight of the element.

Basically, Dulong and Petit found that the Specific Heat Capacity of a mole of numerous solid elements is about 3R, where R is the universal gas constant. Dulong and Petit were unaware of the relationship to R, since it had not yet been defined. The value of 3R is about [math]\displaystyle{ 25 \ \frac{J}{mol \cdot K} }[/math] , and Dulong and Petit found that this was the approximate Molar Specific Heat Capacity of some solid elements per mole of atoms they contained.

Einstein-Debye Model

Einstein and Debye each developed a model for Specific Heat Capacity separately. Einstein's model stated that low energy excitation of a solid material was caused by the oscillation of a single atom, whereas Debye's model stated that phonons or collective modes iterating through a material caused excitations. However, these two models are able to be put together to find the Specific Heat Capacity given by the following formula:

For low temperatures, Einstein and Debye found that the Law of Dulong and Petit was not applicable. At lower temperatures, it was found that atomic interactions were deemed significant in calculating the Molar Specific Heat Capacity of an object:

According to the Einstein Debye Model for Copper and Aluminum, specific heat varies a lot at lower temperatures and goes much below the Dulong-Petit Model. This is due to increased effects on specific heat by interatomic forces. However, for very high temperatures, the Einstein-Debye Model cannot be used. In fact, at high temperatures, Einstein's expression of specific heat reduces to the Dulong-Petit mathematical expression.

The Einstein Debye Equation is below:

For high temperatures it may be reduced like this:

Computational Model

To the right is a table containing the Specific Heat Capacity for a variety of atoms that will be useful for the examples.

350 grams of an unknown substance is heated from 22ºC to 173ºC with 34,700 Joules of energy. There is no phase change.

1,200 grams of coffee is sitting on a table is at a Temperature of [math]\displaystyle{ T_{co_{0}} = 93ºC }[/math] . Assume the Specific Heat Capacity of coffee is [math]\displaystyle{ 4.12 \ \frac{J}{gºC} }[/math] . The coffee is mixed with 55.3 grams of cream at [math]\displaystyle{ T_{cr_{0}} = 5ºC }[/math] . The Specific Heat Capacity of creamer is [math]\displaystyle{ 3.8 \ \frac{J}{gºC} }[/math] .

At low temperatures, the Specific Heat Capacities of solids are typically proportional to [math]\displaystyle{ T^3 }[/math] . The first understanding of this behavior was due to the Dutch physicist Peter Debye, who in 1912, treated atomic oscillations with the quantum theory that Max Planck had recently used for radiation. For instance, a good approximation for the Specific Heat Capacity of salt, NaCl, is [math]\displaystyle{ C = 3.33 \times 10^4 \ \frac{J}{kg \cdot K}\left(\frac{T}{321K}\right)^3 }[/math] . The constant [math]\displaystyle{ 321 K }[/math] is called the Debye temperature of NaCl, [math]\displaystyle{ \theta_D }[/math] , and the formula works well when [math]\displaystyle{ T \lt 0.04\theta_D }[/math] .

Connectedness

The Specific Heat Capacity most commonly known is the Specific Heat Capacity of water, which is about 4.12 J/g°C or 1 calorie/g°C. The specific heat of water is higher than any other common substance. Water has a very large specific heat on a per-gram basis, meaning that it takes a lot more added heat to cause a change in its temperature. Since the specific heat of water is so high, water can be used for temperature regulation. Due to the difference in atomic structures, the specific heat per gram of water is much higher than that of a metal substance. It is possible to predict the specific heat of any material, as long as you know about its atomic structure, as a rise in temperature is the increase in energy at the atomic level of substances. Generally, it is more more useful to compare molar specific heats of substances.

It is easy to notice that water's specific heat capacity is much larger than anything else, but why? The answer is due to water's intermolecular forces. Since a water molecule is made up of one oxygen atom(negative charge) and two hydrogen atoms(slight positive charges), water has hydrogen bonds which result in the "sticking" of water molecules. Because of these hydrogen bonds, it requires a lot of energy to heat up water molecules, because not only do you have to use energy to increase the movement of the particle, but also to break the hydrogen bonds. As a result water has a high specific heat capacity because it takes a lot of energy to break the hydrogen bonds.

This is not an exclusive trait to water, however. The stronger the intermolecular forces of an object, generally the higher the specific heat capacity. Traditionally, gases and liquids have a higher specific heat capacity than solids. In addition, specific heat capacity is also related to the amount of kinetic energy possible in a molecule. Therefore, molecules with more available movement(liquids and gases), there is more room for the heat to "go". Because it is related to kinetic energy, as the external temperature approaches absolute zero, so does specific heat capacity.

But why is this important?

A large body of water can absorb and store a huge amount of heat from the sun in the daytime, such as during summer, while only warming up a few degrees. During night and Winter, the gradually cooling water can warm the air. This is the reason coastal areas generally have milder climates than inland rtegions. The high specific heat of water also tends to stabilize ocean temperatures, creating a favorable environment for marine life. Thus because of its high specific heat, the water that covers most of Earth keeps temperature fluctuations on land and in water within limits that permit life. Also, because organisms are primarily made of water, they are more able to resist changes in their own temperature than if they were made of a liquid with a lower specific heat.

Specific heat and Thermodynamics are used often in chemistry, nuclear engineering, aerodynamics, and mechanical engineering. It is also used in everyday life in the radiator and cooling system of a car.

Specific heat can have a lot to do with prosthetic manufacturing, which is a focus in Biomedical Engineering. Prosthetics materials must be durable and easy to manipulate in a normal range of temperatures. In order to created medical devices, specific heats must be known, especially for welding or molding things, which require a specific temperature to be effective. At higher temperatures, the Dulong-Petit law must be used to calculate the specific heat of an object. Especially for solid metal objects, which would be used in prosthetics, Dulong-Petit is very useful.

Cooking materials such as pots and pans are made to have a low specific heat so that they need less heat to raise their temperature. This allows for faster cooking processes. The handles of these cooking utensils are made of substances with high specific heats so that their temperature won’t rise too much if a large amount of heat is absorbed.

Have you ever noticed that sand on the beach can burn your feet but the ocean water is cool and refreshing? Sand has a lower specific heat than ocean water. So when the sun is beating down, the temperature of the land increases faster than that of the sea.

Insulation is made of materials with high specific heat so that they won't change temperature easily. For example, wood has a high specific heat. A wooden house helps keep the inside cooler during summer because it requires lots of heat to change its temperature. Builders can choose certain materials which allows us to build houses for specific locations and altitudes.

Dr. Joseph Black, of the University of Glasgow, was first credited with developing the concept of latent heat and Specific Heat Capacity in the mid 18th century. This allowed the study of Thermodynamics to be further looked in to. Before Specific Heat Capacity was known, scientists referred to Heat as some sort of invisible liquid. Black, while studying super-cooled water, noticed that when shaken, it instantly turned into ice. This lead him to the concept of "stored Heat," in that shaking it released some form of Heat. This was further developed into the idea that different substances responded to Heat changes differently. He performed an experiment by placing ice and super-cooled water in a room, and the water rapidly rose in temperature while the ice did not. This implied that more Heat was required to raise the Temperature of water than of ice. Black claimed, "If the complete change of ice and snow into water required only the further addition of a very small quantity of heat, the mass, though of considerable size, ought all to be melted in a few minutes or seconds more. Were this really the case, the consequences would be dreadful." After the establishment of the idea of Specific Heat Capacity and latent heat, scientists began to think of Heat as a system's change in internal energy. This is very important as the concept of Specific Heat Capacity has helped lead to the vast development of the field of Thermodynamics.

Further reading

• Kinds of Matter
• Boiling Point
• Melting Point
• Thermal Energy
• First Law of Thermodynamics
• Second Law of Thermodynamics and Entropy
• Internal Energy
• Temperature
• Change of State
• Entropy and Enthalpy Calculations
• Elementary Principles of Chemical Processes (3rd Edition) By: Richard M. Felder & Ronald M. Rousseau
• Encyclopædia Britannica, 2015, "Heat capacity"
• Biology, 7th Edition by Neil A. Campbell and Jane B. Reece

External links

• WiseGeek.org
• KhanAcademy.org
• Wikipedia: Heat Capacity
• HyperPhysics Specific Heat
• Wikipedia: Heat Equation
• Specific Heat Simulation
• OpenStax Vol 2 Heat Transfer, Specific Heat, and Calorimetry
• HyperPhysics Concepts
• Eric Weisstein's World of Physics: Specific Heat
• WIkiHow: How to Calculate Specific Heat
• Applications of Specific Heat
• Uses of Specific Heat
• Heat Capacity, Specific Heat, and Enthalpy
• Table of Specific Heats
• Heat Capacity and Water
• Joseph Black and Latent Heat
• Introduction to Chemical Engineering Thermodynamics Seventh Edition. J. M. Smith, H. C. Van Ness, Michael M. Abbott
• Matter & Interactions Vol I. Chabay Sherwood

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11.2 Heat, Specific Heat, and Heat Transfer

Section learning objectives.

By the end of this section, you will be able to do the following:

• Explain heat, heat capacity, and specific heat
• Distinguish between conduction, convection, and radiation
• Solve problems involving specific heat and heat transfer

Teacher Support

The learning objectives in this section will help your students master the following standards:

• (F) contrast and give examples of different processes of thermal energy transfer, including conduction, convection, and radiation.

Section Key Terms

[BL] [OL] [AL] Review concepts of heat, temperature, and mass.

[AL] Check prior knowledge of conduction and convection.

Heat Transfer, Specific Heat, and Heat Capacity

We learned in the previous section that temperature is proportional to the average kinetic energy of atoms and molecules in a substance, and that the average internal kinetic energy of a substance is higher when the substance’s temperature is higher.

If two objects at different temperatures are brought in contact with each other, energy is transferred from the hotter object (that is, the object with the greater temperature) to the colder (lower temperature) object, until both objects are at the same temperature. There is no net heat transfer once the temperatures are equal because the amount of heat transferred from one object to the other is the same as the amount of heat returned. One of the major effects of heat transfer is temperature change: Heating increases the temperature while cooling decreases it. Experiments show that the heat transferred to or from a substance depends on three factors—the change in the substance’s temperature, the mass of the substance, and certain physical properties related to the phase of the substance.

The equation for heat transfer Q is

where m is the mass of the substance and Δ T is the change in its temperature, in units of Celsius or Kelvin. The symbol c stands for specific heat , and depends on the material and phase. The specific heat is the amount of heat necessary to change the temperature of 1.00 kg of mass by 1.00 ºC. The specific heat c is a property of the substance; its SI unit is J/(kg ⋅ ⋅ K) or J/(kg ⋅ ⋅ °C °C ). The temperature change ( Δ T Δ T ) is the same in units of kelvins and degrees Celsius (but not degrees Fahrenheit). Specific heat is closely related to the concept of heat capacity . Heat capacity is the amount of heat necessary to change the temperature of a substance by 1.00 °C °C . In equation form, heat capacity C is C = m c C = m c , where m is mass and c is specific heat. Note that heat capacity is the same as specific heat, but without any dependence on mass. Consequently, two objects made up of the same material but with different masses will have different heat capacities. This is because the heat capacity is a property of an object, but specific heat is a property of any object made of the same material.

Values of specific heat must be looked up in tables, because there is no simple way to calculate them. Table 11.2 gives the values of specific heat for a few substances as a handy reference. We see from this table that the specific heat of water is five times that of glass, which means that it takes five times as much heat to raise the temperature of 1 kg of water than to raise the temperature of 1 kg of glass by the same number of degrees.

[BL] [OL] [AL] Explain that this formula only works when there is no change in phase of the substance. The transfer of thermal energy, heat, and phase change will be covered later in the chapter.

Misconception Alert

The units of specific heat are J/(kg ⋅ °C ⋅ °C ) and J/(kg ⋅ ⋅ K). However, degrees Celsius and Kelvins are not always interchangeable. The formula for specific heat uses a difference in temperature and not absolute temperature. This is the reason that degrees Celsius may be used in place of Kelvins.

Temperature Change of Land and Water

What heats faster, land or water? You will answer this question by taking measurements to study differences in specific heat capacity.

• Open flame—Tie back all loose hair and clothing before igniting an open flame. Follow all of your teacher's instructions on how to ignite the flame. Never leave an open flame unattended. Know the location of fire safety equipment in the laboratory.
• Sand or soil
• Oven or heat lamp
• Two small jars
• Two thermometers

Instructions

• Place equal masses of dry sand (or soil) and water at the same temperature into two small jars. (The average density of soil or sand is about 1.6 times that of water, so you can get equal masses by using 50 percent more water by volume.)
• Heat both substances (using an oven or a heat lamp) for the same amount of time.
• Record the final temperatures of the two masses.
• Now bring both jars to the same temperature by heating for a longer period of time.
• Remove the jars from the heat source and measure their temperature every 5 minutes for about 30 minutes.
• The pond will reach 0 °C first because of water’s greater specific heat.
• The field will reach 0 °C first because of soil’s lower specific heat.
• They will reach 0° C at the same time because they are exposed to the same weather.
• The water will take longer to heat as well as to cool. This tells us that the specific heat of water is greater than that of land.

Conduction, Convection, and Radiation

Whenever there is a temperature difference, heat transfer occurs. Heat transfer may happen rapidly, such as through a cooking pan, or slowly, such as through the walls of an insulated cooler.

There are three different heat transfer methods: conduction , convection , and radiation . At times, all three may happen simultaneously. See Figure 11.3 .

Conduction is heat transfer through direct physical contact. Heat transferred between the electric burner of a stove and the bottom of a pan is transferred by conduction. Sometimes, we try to control the conduction of heat to make ourselves more comfortable. Since the rate of heat transfer is different for different materials, we choose fabrics, such as a thick wool sweater, that slow down the transfer of heat away from our bodies in winter.

As you walk barefoot across the living room carpet, your feet feel relatively comfortable…until you step onto the kitchen’s tile floor. Since the carpet and tile floor are both at the same temperature, why does one feel colder than the other? This is explained by different rates of heat transfer: The tile material removes heat from your skin at a greater rate than the carpeting, which makes it feel colder.

[BL] [OL] [AL] Ask students what the current temperature in the classroom is. Ask them if all the objects in the room are at the same temperature. Once this is established, ask them to place their hand on their desk or on a metal object. Does it feel colder? Why? If their desk is Formica laminate, then it will feel cool to their hand because the laminate is a good conductor of heat and draws heat from their hand creating a sensation of “cold” due to heat leaving the body.

Some materials simply conduct thermal energy faster than others. In general, metals (like copper, aluminum, gold, and silver) are good heat conductors, whereas materials like wood, plastic, and rubber are poor heat conductors.

Figure 11.4 shows particles (either atoms or molecules) in two bodies at different temperatures. The (average) kinetic energy of a particle in the hot body is higher than in the colder body. If two particles collide, energy transfers from the particle with greater kinetic energy to the particle with less kinetic energy. When two bodies are in contact, many particle collisions occur, resulting in a net flux of heat from the higher-temperature body to the lower-temperature body. The heat flux depends on the temperature difference Δ T = T hot − T cold Δ T = T hot − T cold . Therefore, you will get a more severe burn from boiling water than from hot tap water.

Convection is heat transfer by the movement of a fluid. This type of heat transfer happens, for example, in a pot boiling on the stove, or in thunderstorms, where hot air rises up to the base of the clouds.

Tips For Success

In everyday language, the term fluid is usually taken to mean liquid. For example, when you are sick and the doctor tells you to “push fluids,” that only means to drink more beverages—not to breath more air. However, in physics, fluid means a liquid or a gas . Fluids move differently than solid material, and even have their own branch of physics, known as fluid dynamics , that studies how they move.

As the temperature of fluids increase, they expand and become less dense. For example, Figure 11.4 could represent the wall of a balloon with different temperature gases inside the balloon than outside in the environment. The hotter and thus faster moving gas particles inside the balloon strike the surface with more force than the cooler air outside, causing the balloon to expand. This decrease in density relative to its environment creates buoyancy (the tendency to rise). Convection is driven by buoyancy—hot air rises because it is less dense than the surrounding air.

Sometimes, we control the temperature of our homes or ourselves by controlling air movement. Sealing leaks around doors with weather stripping keeps out the cold wind in winter. The house in Figure 11.5 and the pot of water on the stove in Figure 11.6 are both examples of convection and buoyancy by human design. Ocean currents and large-scale atmospheric circulation transfer energy from one part of the globe to another, and are examples of natural convection.

Radiation is a form of heat transfer that occurs when electromagnetic radiation is emitted or absorbed. Electromagnetic radiation includes radio waves, microwaves, infrared radiation, visible light, ultraviolet radiation, X-rays, and gamma rays, all of which have different wavelengths and amounts of energy (shorter wavelengths have higher frequency and more energy).

[BL] [OL] Electromagnetic waves are also often referred to as EM waves. We perceive EM waves of different frequencies differently. Just as we are able to see certain frequencies as visible light, we perceive certain others as heat.

You can feel the heat transfer from a fire and from the sun. Similarly, you can sometimes tell that the oven is hot without touching its door or looking inside—it may just warm you as you walk by. Another example is thermal radiation from the human body; people are constantly emitting infrared radiation, which is not visible to the human eye, but is felt as heat.

Radiation is the only method of heat transfer where no medium is required, meaning that the heat doesn’t need to come into direct contact with or be transported by any matter. The space between Earth and the sun is largely empty, without any possibility of heat transfer by convection or conduction. Instead, heat is transferred by radiation, and Earth is warmed as it absorbs electromagnetic radiation emitted by the sun.

All objects absorb and emit electromagnetic radiation (see Figure 11.7 ). The rate of heat transfer by radiation depends mainly on the color of the object. Black is the most effective absorber and radiator, and white is the least effective. People living in hot climates generally avoid wearing black clothing, for instance. Similarly, black asphalt in a parking lot will be hotter than adjacent patches of grass on a summer day, because black absorbs better than green. The reverse is also true—black radiates better than green. On a clear summer night, the black asphalt will be colder than the green patch of grass, because black radiates energy faster than green. In contrast, white is a poor absorber and also a poor radiator. A white object reflects nearly all radiation, like a mirror.

Ask students to give examples of conduction, convection, and radiation.

Virtual Physics

Energy forms and changes.

In this animation, you will explore heat transfer with different materials. Experiment with heating and cooling the iron, brick, and water. This is done by dragging and dropping the object onto the pedestal and then holding the lever either to Heat or Cool. Drag a thermometer beside each object to measure its temperature—you can watch how quickly it heats or cools in real time.

Now let’s try transferring heat between objects. Heat the brick and then place it in the cool water. Now heat the brick again, but then place it on top of the iron. What do you notice?

Selecting the fast forward option lets you speed up the heat transfers, to save time.

• Water will take the longest, and iron will take the shortest time to heat, as well as to cool. Objects with greater specific heat would be desirable for insulation. For instance, woolen clothes with large specific heat would prevent heat loss from the body.
• Water will take the shortest, and iron will take the longest time to heat, as well as to cool. Objects with greater specific heat would be desirable for insulation. For instance, woolen clothes with large specific heat would prevent heat loss from the body.
• Brick will take shortest and iron will take longest time to heat up as well as to cool down. Objects with greater specific heat would be desirable for insulation. For instance, woolen clothes with large specific heat would prevent heat loss from the body.
• Water will take shortest and brick will take longest time to heat up as well as to cool down. Objects with greater specific heat would be desirable for insulation. For instance, woolen clothes with large specific heat would prevent heat loss from the body.

Have students consider the differences in the interactive exercise results if different materials were used. For example, ask them whether the temperature change would be greater or smaller if the brick were replaced with a block of iron with the same mass as the brick. Ask students to consider identical masses of the metals aluminum, gold, and copper. After they have stated whether the temperature change is greater or less for each metal, have them refer to Table 11.2 and check whether their predictions were correct.

Solving Heat Transfer Problems

Worked example, calculating the required heat: heating water in an aluminum pan.

A 0.500 kg aluminum pan on a stove is used to heat 0.250 L of water from 20.0 °C °C to 80.0 °C °C . (a) How much heat is required? What percentage of the heat is used to raise the temperature of (b) the pan and (c) the water?

The pan and the water are always at the same temperature. When you put the pan on the stove, the temperature of the water and the pan is increased by the same amount. We use the equation for heat transfer for the given temperature change and masses of water and aluminum. The specific heat values for water and aluminum are given in the previous table.

Because the water is in thermal contact with the aluminum, the pan and the water are at the same temperature.

• Calculate the temperature difference. Δ T = T f − T i = 60.0 °C Δ T = T f − T i = 60.0 °C 11.8
• Calculate the mass of water using the relationship between density, mass, and volume. Density is mass per unit volume, or ρ = m V ρ = m V . Rearranging this equation, solve for the mass of water. m w = ρ ⋅ V = 1000  kg/m 3 × ( 0 .250 L× 0 .001 m 3 1 L ) =0 .250 kg m w = ρ ⋅ V = 1000  kg/m 3 × ( 0 .250 L× 0 .001 m 3 1 L ) =0 .250 kg 11.9
• Calculate the heat transferred to the water. Use the specific heat of water in the previous table. Q w = m w c w Δ T =   ( 0.250  kg ) ( 4186 J/kg °C ) ( 60 .0 °C )  = 62 .8 kJ Q w = m w c w Δ T =   ( 0.250  kg ) ( 4186 J/kg °C ) ( 60 .0 °C )  = 62 .8 kJ 11.10
• Calculate the heat transferred to the aluminum. Use the specific heat for aluminum in the previous table. Q A l = m A l c A l Δ T =   ( 0.500  kg ) ( 900 J/kg °C ) ( 60 .0 °C )  = 27 .0 ×10 3 J = 27 .0 kJ Q A l = m A l c A l Δ T =   ( 0.500  kg ) ( 900 J/kg °C ) ( 60 .0 °C )  = 27 .0 ×10 3 J = 27 .0 kJ 11.11
• Find the total transferred heat. Q T o t a l = Q w + Q A l = 62 .8 kJ + 27 .0 kJ = 89 .8 kJ Q T o t a l = Q w + Q A l = 62 .8 kJ + 27 .0 kJ = 89 .8 kJ 11.12

The percentage of heat going into heating the pan is

The percentage of heat going into heating the water is

In this example, most of the total heat transferred is used to heat the water, even though the pan has twice as much mass. This is because the specific heat of water is over four times greater than the specific heat of aluminum. Therefore, it takes a bit more than twice as much heat to achieve the given temperature change for the water than for the aluminum pan.

Water can absorb a tremendous amount of energy with very little resulting temperature change. This property of water allows for life on Earth because it stabilizes temperatures. Other planets are less habitable because wild temperature swings make for a harsh environment. You may have noticed that climates closer to large bodies of water, such as oceans, are milder than climates landlocked in the middle of a large continent. This is due to the climate-moderating effect of water’s large heat capacity—water stores large amounts of heat during hot weather and releases heat gradually when it’s cold outside.

Calculating Temperature Increase: Truck Brakes Overheat on Downhill Runs

When a truck headed downhill brakes, the brakes must do work to convert the gravitational potential energy of the truck to internal energy of the brakes. This conversion prevents the gravitational potential energy from being converted into kinetic energy of the truck, and keeps the truck from speeding up and losing control. The increased internal energy of the brakes raises their temperature. When the hill is especially steep, the temperature increase may happen too quickly and cause the brakes to overheat.

Calculate the temperature increase of 100 kg of brake material with an average specific heat of 800 J/kg ⋅ °C ⋅ °C from a 10,000 kg truck descending 75.0 m (in vertical displacement) at a constant speed.

We first calculate the gravitational potential energy ( Mgh ) of the truck, and then find the temperature increase produced in the brakes.

• Calculate the change in gravitational potential energy as the truck goes downhill. M g h = ( 10 , 000  kg ) (9 .80 m/s 2 ) ( 75 .0 m ) = 7.35 × 10 6 J M g h = ( 10 , 000  kg ) (9 .80 m/s 2 ) ( 75 .0 m ) = 7.35 × 10 6 J 11.15

where m is the mass of the brake material (not the entire truck). Insert the values Q = 7.35×10 6 J (since the heat transfer is equal to the change in gravitational potential energy), m = = 100 kg, and c = = 800 J/kg ⋅ ⋅ °C °C to find

This temperature is close to the boiling point of water. If the truck had been traveling for some time, then just before the descent, the brake temperature would likely be higher than the ambient temperature. The temperature increase in the descent would likely raise the temperature of the brake material above the boiling point of water, which would be hard on the brakes. This is why truck drivers sometimes use a different technique for called “engine braking” to avoid burning their brakes during steep descents. Engine braking is using the slowing forces of an engine in low gear rather than brakes to slow down.

Practice Problems

How much heat does it take to raise the temperature of 10.0 kg of water by 1.0 °C ?

Calculate the change in temperature of 1.0 kg of water that is initially at room temperature if 3.0 kJ of heat is added.

Check Your Understanding

Use these questions to assess student achievement of the section’s learning objectives. If students are struggling with a specific objective, these questions will help identify which and direct students to the relevant content.

• The mass difference between two objects causes heat transfer.
• The density difference between two objects causes heat transfer.
• The temperature difference between two systems causes heat transfer.
• The pressure difference between two objects causes heat transfer.
• The overall direction of heat transfer is from the higher-temperature object to the lower-temperature object.
• The overall direction of heat transfer is from the lower-temperature object to the higher-temperature object.
• The direction of heat transfer is first from the lower-temperature object to the higher-temperature object, then back again to the lower-temperature object, and so-forth, until the objects are in thermal equilibrium.
• The direction of heat transfer is first from the higher-temperature object to the lower-temperature object, then back again to the higher-temperature object, and so-forth, until the objects are in thermal equilibrium.
• conduction, radiation, and reflection
• conduction, reflection, and convection
• convection, radiation, and reflection
• conduction, radiation, and convection

True or false—Conduction and convection cannot happen simultaneously

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• Why Does Ice Float on Water
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Specific Heat and Heat Capacity

Specific heat is defined as the amount of heat required to raise the temperature of a unit mass of a substance by one degree Celsius. It plays a crucial role in understanding how different materials respond to heating and cooling and describes their ability to store and release thermal energy . For example, water has a higher specific heat than metals. It means that more heat is required to raise water’s temperature by one degree than metals’, as shown in the image below.

On the other hand, heat capacity is the amount of heat required to increase the temperature of the entire substance by one degree Celsius.

Specific Heat and Heat Capacity Formula

The specific heat can be calculated from the amount of heat transferred into and out of a substance. The heat transfer equation provides a quantitative relationship between heat transfer, substance’s mass, specific heat, and temperature change. It is shown as follows:

• Q represents the heat transfer in Joules (J) or calories (cal).
• m denotes the mass of the substance in grams (g) or kilograms (kg) being heated or cooled.
• c is the specific heat
• ΔT represents the temperature change in degrees Celsius ( ∘ C) of the substance

This formula assumes no phase change occurs during heating or cooling processes and applies specifically to substances with constant specific heat within a given temperature range.

Rearranging the above equation, one can find the expression for specific heat.

The heat capacity (C) can be calculated by multiplying the specific heat with the mass. Therefore,

The unit of specific heat is Joules per gram per degree Celsius or J/g∙ ∘ C. Another unit of specific heat is calories per gram per degree Celsius or J/cal∙ ∘ C. The temperature change (∆T) in the Celsius (C) scale is the same as that in the Kelvin (K) scale, although the temperature values differ. Therefore, one can replace ∘ C with K. In that case, the SI unit of specific heat is Joules per kilogram per degree Kelvin or J/kg∙K.

The unit of heat capacity is J/ ∘ C or cal/ ∘ C.

Molar Specific Heat

The specific heat of a substance can also be described in terms of its molar amount. In that case, we use a term called molar specific heat. It is defined as the amount of heat required to change the temperature of one mole of a substance by one degree. It is represented in the unit of Joules per mole per degree Celsius or J/mol ∘ C.

Specific Heat Table

The specific heat values can be calculated for different substances from the above formula. Below is a table that shows the values for a few common substances:

Water has a particularly high specific heat compared to many other substances. Its specific heat capacity is 4.184 J/g°C, which means it takes 4.184 Joules of energy to raise the temperature of 1 gram of water by 1 degree Celsius. Let us discuss the significance of this remarkable property of water.

Specific Heat of Water

Water has an exceptionally high capacity to absorb and retain heat energy without undergoing large temperature changes. This property is significant in thermoregulation in nature and human-made systems.

Thermoregulation is the process by which living organisms maintain their internal body temperature within a narrow range despite fluctuations in the external environment. In nature, bodies of water such as oceans, lakes, and rivers act as thermal regulators by absorbing excess heat during the daytime and slowly releasing it at night. This helps maintain stable temperatures in surrounding areas and supports diverse ecosystems.

Understanding water’s specific heat is crucial for designing efficient heating and cooling artificial systems like car radiators and hot water pipes. Water’s ability to store large amounts of thermal energy makes it an ideal medium for transferring heat between different areas while minimizing temperature fluctuations.

Difference Between Specific Heat and Heat Capacity

Below is a table summarizing the difference between specific heat and heat capacity.

Example Problems with Solutions

Problem 1 :  A 200 g piece of aluminum initially at 80°C is dropped into 400 mL of water at 20°C. If the final temperature of the system is 40°C, calculate the heat transferred to the water. The specific heat of aluminum is 0.9 J/g∙°C, and the specific heat of water is 4.18 J/g∙°C.

The heat transfer equation is given by:

Heat lost by aluminum is:

\[ q_{\text{Al}} = m_{\text{Al}} \cdot c_{\text{Al}} \cdot \Delta T_{\text{Al}} \]

\[ q_{\text{Al}} = (200 \, \text{g}) \cdot (0.9 \, \text{J/g°C}) \cdot (40 \, \text{°C} – 80 \, \text{°C}) \]

\[ q_{\text{Al}} = -7200 \, \text{J} \]

The negative sign implies that aluminum loses heat to water. Therefore, the heat transferred to water is 7200 J .

Problem 2 : A 150 g piece of copper is heated from 20°C to 100°C. Calculate the heat energy absorbed by the copper. The specific heat of copper is 0.39 J/g∙°C.

\[ q = m \cdot c \cdot \Delta T \]

\[ q_{\text{copper}} = m_{\text{copper}} \cdot c_{\text{copper}} \cdot \Delta T_{\text{copper}} \]

\[ q_{\text{copper}} = (150 \, \text{g}) \cdot (0.39 \, \text{J/g°C}) \cdot (100 \, \text{°C} – 20 \, \text{°C}) \]

\[ q_{\text{copper}} = 4680 \, \text{J} \]

Therefore, the heat absorbed by copper is 4680 J .

• Specific Heat – Hyperphysics.phy-astr.gsu.edu
• Specific Heat Capacity – Study.com
• What is Specific Heat? – Chemistrytalk.org
• Specific Heat –  Phys.libretexts.org
• Specific Heats: the relation between temperature change and heat – Web.mit.edu
• Heat – Chemed.chem.purdue.edu
• Heat Capacity and Specific Heat – Chem.libretexts.org
• Specific Heat Definition – Thoughtco.com
• Specific Heat Capacity – Energyeducation.ca
• Specific heat and Heat Capacity – Cpanhd.sitehost.iu.edu

Article was last reviewed on Sunday, August 11, 2024

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Specific Heat Calculator

Table of contents

This specific heat calculator is a tool that determines the heat capacity of a heated or a cooled sample. Specific heat is the amount of thermal energy you need to supply to a sample weighing 1 kg to increase its temperature by 1 K . Read on to learn how to apply the heat capacity formula correctly to obtain a valid result.

💡 This calculator works in various ways, so you can also use it to, for example, calculate the heat needed to cause a temperature change (if you know the specific heat). To find specific heat from a complex experiment, calorimetry calculator might make the calculations much faster.

Prefer watching over reading? Learn all you need in 90 seconds with this video we made for you :

How to calculate specific heat

Determine whether you want to warm up the sample (give it some thermal energy) or cool it down (take some thermal energy away).

Insert the amount of energy supplied as a positive value. If you want to cool down the sample, insert the subtracted energy as a negative value. For example, say that we want to reduce the sample's thermal energy by 63,000 J. Then Q = − 63 ,  ⁣ 000  J Q = -63,\!000 \ \text{J} Q = − 63 , 000   J .

Decide the temperature difference between the initial and final state of the sample and type it into the heat capacity calculator. If the sample is cooled down, the difference will be negative, and if warmed up - positive. Let's say we want to cool the sample down by 3 degrees. Then Δ T = − 3  K \Delta T = -3 \ \text{K} Δ T = − 3   K . You can also select the Show initial and final temperatures checkbox to type the initial and final values of temperature manually.

Determine the mass of the sample. We will assume m = 5  kg m = 5 \ \text{kg} m = 5   kg .

Calculate specific heat as c = Q m Δ T c = \frac{Q}{m \Delta T} c = m Δ T Q ​ . In our example, it will be equal to:

c = − 63 , 000   J 5   k g ⋅   − 3   K = 4 ,  ⁣ 200   J k g ⋅ K c = \mathrm{\frac{-63,000 \ J}{5 \ kg \cdot \ -3 \ K}} = \mathrm{4,\!200 \ \frac{J}{kg \cdot K}} c = 5   kg ⋅   − 3   K − 63 , 000   J ​ = 4 , 200   kg ⋅ K J ​

This is the typical heat capacity of water.

Specific heat capacity formula

The formula for specific heat looks like this:

Q Q Q is the amount of supplied or subtracted heat (in joules), m m m is the mass of the sample, and Δ T \Delta T Δ T is the difference between the initial and final temperatures. Heat capacity is measured in J/(kg·K).

Typical values of specific heat

You don't need to use the heat capacity calculator for most common substances. The values of specific heat for some of the most popular ones are listed below.

• Ice: 2 ,  ⁣ 100   J / k g  ⁣ ⋅  ⁣ K \mathrm{2,\!100 \ {J}/{kg\! \cdot\! K}} 2 , 100   J / kg ⋅ K
• Water: 4 ,  ⁣ 200   J / k g  ⁣ ⋅  ⁣ K \mathrm{4,\!200 \ {J}/{kg\! \cdot\! K}} 4 , 200   J / kg ⋅ K
• Water vapor: 2 ,  ⁣ 000   J / k g  ⁣ ⋅  ⁣ K \mathrm{2,\!000 \ {J}/{kg\! \cdot\! K}} 2 , 000   J / kg ⋅ K
• Basalt: 840   J / k g  ⁣ ⋅  ⁣ K \mathrm{840 \ {J}/{kg\! \cdot\! K}} 840   J / kg ⋅ K
• Granite: 790   J / k g  ⁣ ⋅  ⁣ K \mathrm{790 \ {J}/{kg\! \cdot\! K}} 790   J / kg ⋅ K
• Aluminum: 890   J / k g  ⁣ ⋅  ⁣ K \mathrm{890 \ {J}/{kg\! \cdot\! K}} 890   J / kg ⋅ K
• Iron: 450   J / k g  ⁣ ⋅  ⁣ K \mathrm{450 \ {J}/{kg\! \cdot\! K}} 450   J / kg ⋅ K
• Copper: 380   J / k g  ⁣ ⋅  ⁣ K \mathrm{380 \ {J}/{kg\! \cdot\! K}} 380   J / kg ⋅ K
• Lead: 130   J / k g  ⁣ ⋅  ⁣ K \mathrm{130 \ {J}/{kg\! \cdot\! K}} 130   J / kg ⋅ K

Having this information, you can also calculate how much energy you need to supply to a sample to increase or decrease its temperature. For instance, you can check how much heat you need to bring a pot of water to a boil to cook some pasta. Or, you can use the water heating calculator for convenience, where all this information was already taken into account for you.

Wondering what the result actually means? Try our potential energy calculator to check how high you would raise the sample with this amount of energy. Or check how fast the sample could move with this kinetic energy calculator .

How to calculate specific heat capacity?

• Find the initial and final temperature as well as the mass of the sample and energy supplied.
• Subtract the final and initial temperature to get the change in temperature (ΔT).
• Multiply the change in temperature with the mass of the sample.
• Divide the heat supplied/energy with the product.
• The formula is C = Q / (ΔT × m) .

What is specific heat capacity at constant volume?

The specific heat capacity is the heat or energy required to change one unit mass of a substance of a constant volume by 1 °C . The formula is Cv = Q / (ΔT × m) .

What is the formula for specific heat?

The formula for specific heat capacity, C , of a substance with mass m , is C = Q /(m × ΔT) . Where Q is the energy added and ΔT is the change in temperature. The specific heat capacity during different processes, such as constant volume, Cv and constant pressure, Cp , are related to each other by the specific heat ratio, ɣ= Cp/Cv , or the gas constant R = Cp - Cv .

What are the units for specific heat capacity?

Specific heat capacity is measured in J/kg·K or J/kg·°C , as it is the heat or energy required during a constant volume process to change the temperature of a substance of unit mass by 1 °C or 1 K.

What is the specific heat capacity value of water?

The specific heat of water at 25 °C is 4,181.3 J/kg·K , the amount of heat required to raise the temperature of 1 kg of water by 1 Kelvin.

What are the imperial units for specific heat?

Specific heat is measured in BTU / lb °F in imperial units and in J/kg·K in SI units.

What is the specific heat capacity value of copper?

The specific heat of copper is 385 J/kg·K . You can use this value to estimate the energy required to heat a 100 g of copper by 5 °C, i.e., Q = m × Cp × ΔT = 0.1 × 385 × 5 = 192.5 J.

What is the specific heat capacity value of aluminum?

The specific heat of aluminum is 897 J/kg K . This value is almost 2.3 times of the specific heat of copper. You can use this value to estimate the energy required to heat a 500 g of aluminum by 5 °C, i.e., Q = m × Cp × ΔT = 0.5 × 897 × 5 = 2242.5 J.

Change of temperature (ΔT)

Substance (optional)

Specific heat capacity (c)

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.

In a laboratory experiment to measure specific heat capacity of copper, 0.05 kg of water at 60°C was poured into a copper calorimeter with a stirrer of mass 0.20 kg initially at 10°C. After stirring, the final temperature reached to 25°C. Specific heat of water is taken as 4200 J/kg°C. (a) What is the quantity of heat released per kg of water per 1°C fall in temperature? (b) Calculate the heat energy released by water in the experiment in cooling from 70°C to 45°C. (c) Assuming that the heat released by water is entirely used to raise the temperature of calorimeter from 10°C to 45°C, calculate the specific heat capacity of copper.

Electro-magnetism.

(a) Quantity of heat released per kg of water per 1°C fall in temperature is given by 4200 J/kg°C.

(b) Given, m w = 0.05 kg

m c = 0.20 kg

Specific heat capacity of water = 4200 J kg -1 K -1

final temperature of water = 25°C

Heat energy given out by water in lowering it's temperature from 70° C to 45° C = m x c x ΔT = 0.05 x 4200 x (70 - 45) = 0.05 x 4200 x 25 = 5250 J

(c) Heat energy taken by calorimeter when it raises it's temperature from 10°C to 45°C = m x c x change in temperature = 0.20 × c × (45 - 10) = 0.20 × c × 35 = 7c

If there is no loss of energy,

Heat energy gained = heat energy lost

Substituting the values we get,

5250 = 7 c ⇒ c = 5250 7 ⇒ c = 750  J Kg − 1 K − 1 5250 = 7c \\[0.5em] \Rightarrow c = \dfrac{5250}{7} \\[0.5em] \Rightarrow c = 750 \text{ J Kg}^{-1}\text{K}^{-1} 5250 = 7 c ⇒ c = 7 5250 ​ ⇒ c = 750  J Kg − 1 K − 1

Hence, specific heat capacity of copper = 750 J kg -1 K -1

Answered By

Related Questions

(a) a current of 1 a flows in a series circuit having an electric lamp and a conductor of 5 ω when connected to a 10 v battery. calculate the resistance of the electric lamp. (b) now, if a resistance of 10 ω is connected in parallel with this series combination, then what change (if any) in current flowing through 5 ω conductor and potential difference across the lamp will take place give reason., farmers fill their fields with water on a cold winter night. explain in brief along with a reason., (a) a cell is sending current in an external circuit. how does the terminal voltage compare with the emf of the cell (b) at what voltage is the alternating current supplied to our houses (c) how should the electric lamps in a building be connected, (a) how would you demonstrate magnetic effect of current by whom it was discovered (b) for the current carrying solenoid as shown below, draw magnetic field lines. explain that out of the three points a, b and c at which point, the field strength is maximum and at which point it is minimum..

• What is specific heat capacity?

Answer. It is defined as the capacity is defined as the quantity of heat needed for one gram of substance at 1℃.

• What are the parameters on which heat capacity depends.

Answer. Heat capacity is not constant and depends on various parameters like volume, temperature, and pressure.

• Define the heat capacity of copper?

Answer. The specific heat capacity of copper is 0.385 J/g-K and it is very low.

• What is a Hypsometer?

Answer: Hypsometer is an instrument designed to calculate the boiling point of water at a given altitude.

• What is the International System of Units (SI) of specific heat capacity?

Answer. Joule per kelvin per kilogram or J⋅kg−1⋅K−1

• What is calorimetry?

Answer. Calorimetry is the process of calculating the amount of heat released or absorbed during a chemical reaction.

• Is the heat loss seen in conduction, convection, and radiation?

Answer. Yes, heat loss is seen in conduction, convection, and radiation.

• What is the formula to calculate specific heat capacity?

Answer. Q = C m ∆t.

• What is the specific heat capacity of water?

Answer: The specific heat capacity of water is 4.2 J/g per Celsius degree or 1 calory per gram per Celsius degree.

• Define heat capacity.

Answer . Heat capacity is defined as the ratio of heat absorbed by a material to the temperature change.

• To determine the specific heat capacity of a given solid by the method of mixtures
• Chapter Wise Physics Class 11 MCQs
• Physics Class 11 Practicals list

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Optimizing Nanosatellite Power Generation Integrating Heat Pipes and Thermoelectric Generators for Low Earth Orbit Conditions

• Original Paper
• Published: 12 August 2024

Cite this article

• E. Viesser 1 ,
• C. E. B. Correa 1 ,
• L. K. Slongo 1 &
• K. V. Paiva   ORCID: orcid.org/0000-0003-4126-8778 1  

Electric power generation is crucial for efficiently operating spacecraft, including nanosatellites. Solar panels are widely used for this purpose; however, alternative technologies have been investigated to enhance nanosatellite power generation due to area limitations and low efficiency. This study conducted experimental research involving integrating two devices, heat pipes and thermoelectric generators (TEGs), to augment electric power generation in nanosatellites, particularly those of the 1U CubeSat standard. The study encompassed an assessment of the performance of these devices in various configurations and couplings within a nanosatellite. Furthermore, the feasibility of employing commercial copper–water heat pipes in nanosatellites was also explored. For this, a thermal vacuum chamber was developed that emulated space conditions and validated the experiments constructed. By utilizing solar irradiance data from the UFSC-developed CubeSat FloripaSat-I, it was possible to emulate the behavior of a solar panel, heat pipe, and TEG within the thermal vacuum chamber. The results presented in this study show that copper–water heat pipes coupled to solar panels increase electricity generation by 23%. When calculating the ratio between the energy generated by the device's mass, the value found was 10.05 mWh/g, demonstrating the viability of this approach. Experimental results demonstrated that copper–water heat pipes work in orbital conditions, even in temperatures below 0 °C. This adaptability is attributed to the transient regime in a low Earth orbit and the continuously imposed thermal loads, allowing its functionality in these conditions.

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Data availability

The datasets generated and/or analyzed during the current study are available by the corresponding author upon reasonable request.

Abbreviations

TEG total area, m 2

Alumina substrate area, m 2

Cross-sectional area of semiconductor bar, m 2

Alumina thermal conductivity, W/m K

Alumina substrate length, m

Length of semiconductor bar, m

TEG mass, g

Number of pairs

Maximum power, W

Thermal resistance, Ω

Average temperature, °C

Width of semiconductor bar, m

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We would like to thank FAPESC, CAPES, CNPq for supporting this research.

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E. Viesser, C. E. B. Correa, L. K. Slongo & K. V. Paiva

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Viesser, E., Correa, C.E.B., Slongo, L.K. et al. Optimizing Nanosatellite Power Generation Integrating Heat Pipes and Thermoelectric Generators for Low Earth Orbit Conditions. Int. J. Aeronaut. Space Sci. (2024). https://doi.org/10.1007/s42405-024-00798-2

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Received : 06 March 2024

Revised : 28 June 2024

Accepted : 18 July 2024

Published : 12 August 2024

DOI : https://doi.org/10.1007/s42405-024-00798-2

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