What do you think heats up faster on the stove: a liter of water in a saucepan or the saucepan itself weighing 1 kilogram? The mass of the bodies is the same, it can be assumed that heating will occur at the same rate.
But that was not the case! You can do an experiment - put an empty saucepan on the fire for a few seconds, just don’t burn it, and remember to what temperature it heated up. And then pour into the pan exactly the same weight of water as the weight of the pan. In theory, the water should heat up to the same temperature as an empty pan in twice as long, since in this case they both heat up - both the water and the pan.
However, even if you wait three times longer, you will be convinced that the water will still heat up less. It will take water almost ten times longer to reach the same temperature as a pan of the same weight. Why is this happening? What prevents water from heating up? Why should we waste extra gas heating water when cooking? Because there is a physical quantity called the specific heat capacity of a substance.
Specific heat capacity of a substance
This value shows how much heat must be transferred to a body weighing one kilogram in order for its temperature to increase by one degree Celsius. Measured in J/(kg * ˚С). This value exists not because of its own whim, but because of the difference in the properties of various substances.
The specific heat of water is about ten times higher than the specific heat of iron, so the pan will heat up ten times faster than the water in it. It is curious that the specific heat capacity of ice is half that of water. Therefore, the ice will heat up twice as fast as water. Melting ice is easier than heating water. As strange as it may sound, it is a fact.
Calculation of heat amount
Specific heat capacity is designated by the letter c And used in the formula for calculating the amount of heat:
Q = c*m*(t2 - t1),
where Q is the amount of heat,
c - specific heat capacity,
m - body weight,
t2 and t1 are the final and initial body temperatures, respectively.
Specific Heat Capacity Formula: c = Q / m*(t2 - t1)
You can also express from this formula:
- m = Q / c*(t2-t1) - body weight
- t1 = t2 - (Q / c*m) - initial body temperature
- t2 = t1 + (Q / c*m) - final body temperature
- Δt = t2 - t1 = (Q / c*m) - temperature difference (delta t)
What about the specific heat capacity of gases? Everything is more confusing here. With solids and liquids the situation is much simpler. Their specific heat capacity is a constant, known, and easily calculated value. As for the specific heat capacity of gases, this value is very different in different situations. Let's take air as an example. The specific heat capacity of air depends on its composition, humidity, and atmospheric pressure.
At the same time, as the temperature increases, the gas increases in volume, and we need to enter another value - constant or variable volume, which will also affect the heat capacity. Therefore, when calculating the amount of heat for air and other gases, special graphs of the specific heat capacity of gases are used depending on various factors and conditions.
The amount of energy that must be supplied to 1 g of a substance to increase its temperature by 1°C. By definition, in order to increase the temperature of 1 g of water by 1°C, 4.18 J is required. Ecological encyclopedic dictionary.... ... Ecological dictionary
specific heat- - [A.S. Goldberg. English-Russian energy dictionary. 2006] Energy topics in general EN specific heatSH ...
SPECIFIC HEAT- physical a quantity measured by the amount of heat required to heat 1 kg of a substance by 1 K (cm). SI unit of specific heat capacity (cm) per kilogram kelvin (J kg∙K)) ... Big Polytechnic Encyclopedia
specific heat- savitoji šiluminė talpa statusas T sritis fizika atitikmenys: engl. heat capacity per unit mass; mass heat capacity; specific heat capacity vok. Eigenwärme, f; spezifische Wärme, f; spezifische Wärmekapazität, f rus. mass heat capacity, f;… … Fizikos terminų žodynas
See Heat capacity... Great Soviet Encyclopedia
specific heat- specific heat... Dictionary of chemical synonyms I
specific heat capacity of gas- - Topics oil and gas industry EN gas specific heat ... Technical Translator's Guide
specific heat capacity of oil- - Topics oil and gas industry EN oil specific heat ... Technical Translator's Guide
specific heat capacity at constant pressure- - [A.S. Goldberg. English-Russian energy dictionary. 2006] Topics: energy in general EN specific heat at constant pressurecpconstant pressure specific heat ... Technical Translator's Guide
specific heat capacity at constant volume- - [A.S. Goldberg. English-Russian energy dictionary. 2006] Topics energy in general EN specific heat at constant volumeconstant volume specific heatCv ... Technical Translator's Guide
Books
- Physical and geological foundations of the study of water movement in deep horizons, V.V. Trushkin. In general, the book is devoted to the law of self-regulation of the temperature of water with a host body, discovered by the author in 1991. At the beginning of the book, a review of the state of knowledge of the problem of movement of deep...
The table shows the thermophysical properties of helium He in the gaseous state depending on temperature and pressure. The thermophysical properties and density of helium in the table are given at temperatures from 0 to 1000 ° C and pressure from 1 to 100 atmospheres.
It should be noted that such properties of helium as thermal diffusivity and kinematic viscosity depend significantly on temperature, increasing their values by an order of magnitude when heated by 1000 degrees. With increasing pressure, these properties of helium decrease their values, while the density of helium increases significantly.
Under normal conditions, the density of helium is 0.173 kg/m3(at a temperature of 0°C and normal atmospheric pressure). With increasing helium pressure, its density increases proportionally, for example at 10 atm. The density of helium will already be 1.719 kg/m 3 (at the same temperature). With further compression of this gas to 100 atm. The density of helium will become equal to 16.45 kg/m 3. Thus, there is an almost hundredfold increase in the density of helium relative to the initial value (at atmospheric pressure).
As is known, the lowest density has a gas such as, and helium ranks second among gases in terms of density.
Helium is considered the most optimal gas for filling balloons used in aeronautics, since, unlike hydrogen, it does not create an explosive mixture with air.
Since the density of helium is much less than air, at the same temperatures, balloons and balloons filled with helium have good lifting force. The sufficiently low density of helium makes it possible to create unmanned high-altitude balloons for weather and scientific research.
How high can a helium balloon rise? as it gains altitude, it begins to decrease and at altitudes of about 33...36 km it will become equal to the density of helium in the balloon, and its rise will stop.
The table shows the following properties of helium:
- helium density γ , kg/m 3 ;
- specific heat S p , kJ/(kg deg);
- thermal conductivity coefficient λ , W/(m deg);
- dynamic viscosity μ , ;
- thermal diffusivity a , m 2 /s;
- kinematic viscosity ν , m 2 /s;
- Prandtl number Pr .
Note: Be careful! Thermal conductivity in the table is indicated to the power of 10 2. Don't forget to divide by 100.
Thermal conductivity of helium at normal atmospheric pressure.
The thermal conductivity values of helium at normal atmospheric pressure depending on temperature are given in the table.
Thermal conductivity (in the dimension W/(m deg)) is indicated for helium gas in the temperature range from -203 to 1727 °C.
Note: Be careful! The thermal conductivity of helium in the table is indicated to the power of 10 3. Don't forget to divide by 1000. According to the thermal conductivity table, it can be seen that its values increase with increasing helium temperature.
Thermal conductivity of helium at high temperatures.
The table shows the thermal conductivity of helium at normal atmospheric pressure and at high temperatures.
The thermal conductivity of helium in the gaseous state is given in the temperature range 2500...6000 K.
Note: Be careful! The thermal conductivity of helium in the table is indicated to the power of 10 3. Do not forget to divide by 1000. The value of the thermal conductivity coefficient of helium increases with increasing temperature and reaches a value of 1.2 W/(m deg) at 6000 K.
Thermal conductivity of liquid helium at low temperatures.
The thermal conductivity values of liquid helium at normal atmospheric pressure and extremely low temperatures are given.
The thermal conductivity of helium in the liquid state is given in the table for a temperature of 2.3...4.2 K (-270.7...-268.8°C).
Note: Be careful! The thermal conductivity of helium in the table is indicated to the power of 10 3. Remember to divide by 1000. The thermal conductivity of helium increases with increasing temperature and in the liquid state at low temperatures.
Thermal conductivity of helium depending on pressure and temperature.
The table shows the thermal conductivity of helium depending on pressure and temperature.
Thermal conductivity (dimension W/(m deg)) is indicated for helium gas in the temperature range from 0 to 1227 °C and pressure from 1 to 300 atm.
Note: Be careful! The thermal conductivity of helium in the table is indicated to the power of 10 3. Don't forget to divide by 1000. The thermal conductivity of helium has a slight tendency to increase with increasing gas pressure.
Heat capacity of liquid helium depending on temperature.
The table shows the specific (mass) values heat capacity of liquid helium in a state of saturation depending on temperature.
As is known, helium can only exist in a liquid state at very low temperatures, approaching absolute zero.
The heat capacity of liquid helium (dimension kJ/(kg deg)) is given in the temperature range from 1.8 to 5.05 K.
Sources:
1.
2. .
3. Physical quantities. Directory. A.P. Babichev, N.A. Babushkina, A.M. Bratkovsky and others; Ed. I.S. Grigorieva, E.Z. Meilikhova. - M.: Energoatomizdat, 1991. - 1232 p.
Physics and thermal phenomena is a fairly extensive section that is thoroughly studied in the school course. Not the last place in this theory is given to specific quantities. The first of these is specific heat capacity.
However, insufficient attention is usually paid to the interpretation of the word “specific”. Students simply remember it as a given. What does it mean?
If you look into Ozhegov’s dictionary, you can read that such a quantity is defined as a ratio. Moreover, it can be performed in relation to mass, volume or energy. All these quantities must be taken equal to one. Specific heat capacity is related to what?
To the product of mass and temperature. Moreover, their values must be equal to one. That is, the divisor will contain the number 1, but its dimension will combine kilogram and degree Celsius. This must be taken into account when formulating the definition of specific heat capacity, which is given a little below. There is also a formula from which it is clear that these two quantities are in the denominator.
What is it?
The specific heat capacity of a substance is introduced at the moment when the situation with its heating is considered. Without it, it is impossible to know how much heat (or energy) will be required for this process. And also calculate its value when the body cools. By the way, these two amounts of heat are equal to each other in modulus. But they have different signs. So, in the first case it is positive, because energy needs to be expended and it is transferred to the body. The second cooling situation gives a negative number because heat is released and the internal energy of the body decreases.
This physical quantity is denoted by the Latin letter c. It is defined as a certain amount of heat required to heat one kilogram of a substance by one degree. In a school physics course, this degree is the one taken on the Celsius scale.
How to count it?
If you want to know what the specific heat capacity is, the formula looks like this:
c = Q / (m * (t 2 - t 1)), where Q is the amount of heat, m is the mass of the substance, t 2 is the temperature that the body acquired as a result of heat exchange, t 1 is the initial temperature of the substance. This is formula number 1.
Based on this formula, the unit of measurement of this quantity in the international system of units (SI) turns out to be J/(kg*ºС).
How to find other quantities from this equality?
Firstly, the amount of heat. The formula will look like this: Q = c * m * (t 2 - t 1). Only it is necessary to substitute values in SI units. That is, mass in kilograms, temperature in degrees Celsius. This is formula number 2.
Secondly, the mass of a substance that cools or heats up. The formula for it will be: m = Q / (c * (t 2 - t 1)). This is formula number 3.
Thirdly, temperature change Δt = t 2 - t 1 = (Q / c * m). The sign “Δ” is read as “delta” and denotes a change in a quantity, in this case temperature. Formula No. 4.
Fourthly, the initial and final temperatures of the substance. The formulas valid for heating a substance look like this: t 1 = t 2 - (Q / c * m), t 2 = t 1 + (Q / c * m). These formulas are No. 5 and 6. If the problem is about cooling a substance, then the formulas are: t 1 = t 2 + (Q / c * m), t 2 = t 1 - (Q / c * m). These formulas are No. 7 and 8.
What meanings can it have?
It has been established experimentally what values it has for each specific substance. Therefore, a special specific heat capacity table has been created. Most often it contains data that is valid under normal conditions.
What is the laboratory work involved in measuring specific heat capacity?
In the school physics course it is defined for a solid body. Moreover, its heat capacity is calculated by comparison with that which is known. The easiest way to do this is with water.
During the work, it is necessary to measure the initial temperatures of water and a heated solid. Then lower it into the liquid and wait for thermal equilibrium. The entire experiment is carried out in a calorimeter, so energy losses can be neglected.
Then you need to write down the formula for the amount of heat that water receives when heated from a solid. The second expression describes the energy that a body gives off when cooling. These two values are equal. Through mathematical calculations, it remains to determine the specific heat capacity of the substance that makes up the solid.
Most often it is proposed to compare it with table values in order to try to guess what substance the body under study is made of.
Task No. 1
Condition. The temperature of the metal varies from 20 to 24 degrees Celsius. At the same time, its internal energy increased by 152 J. What is the specific heat of the metal if its mass is 100 grams?
Solution. To find the answer, you will need to use the formula written under number 1. All the quantities necessary for the calculations are there. Just first you need to convert the mass into kilograms, otherwise the answer will be wrong. Because all quantities must be those accepted in SI.
There are 1000 grams in one kilogram. This means that 100 grams must be divided by 1000, you get 0.1 kilograms.
Substitution of all quantities gives the following expression: c = 152 / (0.1 * (24 - 20)). The calculations are not particularly difficult. The result of all actions is the number 380.
Answer: s = 380 J/(kg * ºС).
Problem No. 2
Condition. Determine the final temperature to which water with a volume of 5 liters will cool if it was taken at 100 ºС and released 1680 kJ of heat into the environment.
Solution. It’s worth starting with the fact that energy is given in a non-systemic unit. Kilojoules need to be converted to joules: 1680 kJ = 1680000 J.
To find the answer, you need to use formula number 8. However, mass appears in it, and in the problem it is unknown. But the volume of liquid is given. This means that we can use the formula known as m = ρ * V. The density of water is 1000 kg/m3. But here the volume will need to be substituted in cubic meters. To convert them from liters, you need to divide by 1000. Thus, the volume of water is 0.005 m 3.
Substituting the values into the mass formula gives the following expression: 1000 * 0.005 = 5 kg. You will need to look up the specific heat capacity in the table. Now you can move on to formula 8: t 2 = 100 + (1680000 / 4200 * 5).
The first action is to multiply: 4200 * 5. The result is 21000. The second is division. 1680000: 21000 = 80. The last one is subtraction: 100 - 80 = 20.
Answer. t 2 = 20 ºС.
Problem No. 3
Condition. There is a beaker weighing 100 g. 50 g of water is poured into it. The initial temperature of the water with the glass is 0 degrees Celsius. How much heat is required to bring water to a boil?
Solution. A good place to start is by introducing a suitable designation. Let the data related to the glass have an index of 1, and for water - an index of 2. In the table, you need to find the specific heat capacities. The beaker is made of laboratory glass, so its value c 1 = 840 J/ (kg * ºC). The data for water is: c 2 = 4200 J/ (kg * ºС).
Their masses are given in grams. You need to convert them to kilograms. The masses of these substances will be designated as follows: m 1 = 0.1 kg, m 2 = 0.05 kg.
The initial temperature is given: t 1 = 0 ºС. It is known about the final value that it corresponds to the point at which water boils. This is t 2 = 100 ºС.
Since the glass heats up along with the water, the required amount of heat will be the sum of two. The first, which is required to heat the glass (Q 1), and the second, which is used to heat the water (Q 2). To express them you will need a second formula. It must be written down twice with different indices, and then sum them up.
It turns out that Q = c 1 * m 1 * (t 2 - t 1) + c 2 * m 2 * (t 2 - t 1). The common factor (t 2 - t 1) can be taken out of the bracket to make it easier to calculate. Then the formula that will be required to calculate the amount of heat will take the following form: Q = (c 1 * m 1 + c 2 * m 2) * (t 2 - t 1). Now you can substitute the quantities known in the problem and calculate the result.
Q = (840 * 0.1 + 4200 * 0.05) * (100 - 0) = (84 + 210) * 100 = 294 * 100 = 29400 (J).
Answer. Q = 29400 J = 29.4 kJ.
When introducing the concept of heat capacity, we did not pay attention to one essential circumstance: heat capacities depend not only on the properties of the substance, but also on the process by which heat transfer occurs.
If you heat a body at constant pressure, it will expand and do work. To heat a body by 1 K at constant pressure, it needs to transfer more heat than with the same heating at constant volume.
Liquid and solid bodies expand slightly when heated, and their heat capacities at constant volume and constant pressure differ little. But for gases this difference is significant. Using the first law of thermodynamics, one can find the relationship between the heat capacities of a gas at constant volume and constant pressure.
Heat capacity of gas at constant volume Let's find the molar heat capacity of gas at constant volume. According to the definition of heat capacity
where Δ T - temperature change. If the process occurs at a constant volume, then this heat capacity will be denoted by C v . Then
(5.6.1)
At constant volume, no work is done. Therefore, the first law of thermodynamics will be written as follows:
(5.6.2)
The change in energy of one mole of a sufficiently rarefied (ideal) monatomic gas is equal to: 
(see § 4.8).
Therefore, the molar heat capacity at a constant volume of a monatomic gas is equal to:
(5.6.3)
Heat capacity of gas at constant pressure
According to the definition of heat capacity at constant pressure WITH r
(5.6.4)
The work done by 1 mole of an ideal gas expanding at constant pressure is:
(5.6.5)
* From formula (5.6.5) it is clear that the universal gas constant is numerically equal to the work done by 1 mole of an ideal gas at constant pressure if its temperature increases by 1 K.
This follows from the expression for gas work at constant pressure A" =pΔ V and equations of state (for one mole) of an ideal gas pV = RT.
The internal energy of an ideal gas does not depend on volume. Therefore, even at constant pressure, the change in internal energy Δ U = C V Δ T, as with constant volume. Applying the first law of thermodynamics, we get:
(5.6.6)
Consequently, the molar heat capacities of an ideal gas are related by the relation
(5.6.7)
This formula was first obtained by R. Mayer and bears his name.
In the case of an ideal monatomic gas
(5.6.8)
Heat capacity of an ideal gas in an isothermal process
The concept of heat capacity can also be formally introduced for an isothermal process. Since during this process the internal energy of an ideal gas does not change, no matter how much heat is transferred to it, the heat capacity is infinite.
The molar heat capacity of an ideal gas at constant pressure is greater than the heat capacity at constant volume by the value of the universal gas constantR.
§ 5.7. Adiabatic process
We looked at isothermal, isobaric and isochoric processes. After becoming familiar with the first law of thermodynamics, it becomes possible to study another process,- This is a process that occurs in a system in the absence of heat exchange with surrounding bodies. (But the system can perform work on surrounding bodies.)
The process in a thermally insulated system is called adiabatic.
In an adiabatic process Q = 0 and according to law (5.5.3), the change in internal energy occurs only due to the work done:
(5.7.1)
Of course, it is impossible to surround the system with a shell that completely excludes heat exchange. But in a number of cases, real processes are very close to adiabatic. There are shells that have low thermal conductivity, for example double walls with a vacuum between them. This is how thermoses are made.
According to expression (5.7.1), when positive work is performed on a system, for example, when compressing a gas, its internal energy increases; the gas heats up. On the contrary, when expanding, the gas itself does positive work (A" > 0), but A< 0 and its internal energy decreases; the gas cools.
The dependence of gas pressure on its volume during an adiabatic process is depicted by a curve called the adiabatic (Fig. 5.9). The adiabat is necessarily steeper than the isotherm. Indeed, during an adiabatic process, the gas pressure decreases not only due to an increase in volume, as in an isothermal process, but also due to a decrease in its temperature.
Adiabatic processes are widely used in technology. They play a significant role in nature.
Heating air during rapid compression has found application in Diesel engines. These engines do not have systems for preparing and igniting the combustible mixture, which are necessary for conventional gasoline internal combustion engines. It is not the combustible mixture that is sucked into the cylinder, but atmospheric air. Towards the end of the compression stroke, liquid fuel is injected into the cylinder using a special nozzle (Fig. 5.10). At this point, the temperature of the compressed air is so high that the fuel ignites.
Since in a Diesel engine it is not the combustible mixture that is compressed, but air, the compression ratio of this engine is higher, which means the efficiency of Diesel engines is higher than that of conventional internal combustion engines. In addition, they can run on cheaper, low-grade fuel. However, the Diesel engine also has disadvantages: the need for high compression ratios and high operating pressure make these engines massive and, as a result, more inertial - they gain power more slowly. Diesel engines are more complex to manufacture and operate, however, they are gradually replacing conventional gasoline engines used in cars.
Gas cooling during adiabatic expansion occurs on a grand scale in the Earth's atmosphere. Heated air rises and expands as atmospheric pressure decreases with altitude. This expansion is accompanied by significant cooling. As a result, water vapor condenses and clouds form.
