The focus of our article is the amount of heat. We will look at the concept internal energy, which transforms when this value changes. We will also show some examples of the application of calculations in human activity.
Heat
With any word native language Each person has their own associations. They are determined personal experience and irrational feelings. What do you usually think of when you hear the word “warmth”? Soft blanket, working central heating radiator in winter, first sunlight spring, cat Or a mother’s look, a friend’s comforting word, timely attention.
Physicists mean a very specific term by this. And very important, especially in some sections of this complex but fascinating science.
Thermodynamics
It is not worth considering the amount of heat in isolation from the simplest processes on which the law of conservation of energy is based - nothing will be clear. Therefore, first let us remind our readers of them.
Thermodynamics considers any thing or object as a combination of a very large number of elementary parts - atoms, ions, molecules. Its equations describe any change in the collective state of the system as a whole and as part of the whole when macroparameters change. The latter refers to temperature (denoted as T), pressure (P), concentration of components (usually C).
Internal energy
Internal energy is a rather complex term, the meaning of which is worth understanding before talking about the amount of heat. It denotes the energy that changes when the value of the macroparameters of an object increases or decreases and does not depend on the reference system. It is part of the total energy. It coincides with it in conditions when the center of mass of the thing under study is at rest (that is, there is no kinetic component).
When a person feels that an object (say a bicycle) has become hot or cold, it shows that all the molecules and atoms that make up this system, experienced a change in internal energy. However, the constant temperature does not mean the preservation of this indicator.
Work and heat
The internal energy of any thermodynamic system can be transformed in two ways:
- by doing work on it;
- during heat exchange with the environment.
The formula for this process looks like this:
dU=Q-A, where U is internal energy, Q is heat, A is work.
Let the reader not be deceived by the simplicity of the expression. The rearrangement shows that Q=dU+A, however, the introduction of entropy (S) brings the formula to the form dQ=dSxT.
Since in this case the equation takes the form of a differential one, the first expression requires the same. Next, depending on the forces acting in the object under study and the parameter that is being calculated, the required ratio is derived.
Let's take a metal ball as an example of a thermodynamic system. If you press on it, throw it up, drop it into a deep well, then this means doing work on it. Outwardly, all these harmless actions will not cause any harm to the ball, but its internal energy will change, albeit very slightly.
The second method is heat exchange. Now we come to main goal of this article: a description of what the amount of heat is. This is a change in the internal energy of a thermodynamic system that occurs during heat exchange (see formula above). It is measured in joules or calories. Obviously, if you hold the ball over a lighter, in the sun, or simply in a warm hand, it will heat up. And then you can use the change in temperature to find the amount of heat that was communicated to him.
Why gas is the best example of a change in internal energy, and why schoolchildren don’t like physics because of this
Above we described changes in the thermodynamic parameters of a metal ball. They are not very noticeable without special devices, and the reader can only take the word about the processes occurring with the object. It's another matter if the system is gas. Press on it - it will be visible, heat it - the pressure will rise, lower it underground - and it can be easily recorded. Therefore, in textbooks, gas is most often used as a visual thermodynamic system.
But, alas, in modern education Not much attention is paid to real experiments. Scientist who writes Toolkit, understands perfectly what he's talking about we're talking about. It seems to him that, using the example of gas molecules, all thermodynamic parameters will be in the right way demonstrated. But a student who is just discovering this world is bored hearing about an ideal flask with a theoretical piston. If the school had real research laboratories and allocated hours to work in them, things would be different. So far, unfortunately, the experiments are only on paper. And, most likely, this is precisely the reason why people consider this branch of physics to be something purely theoretical, far from life and unnecessary.
Therefore, we decided to use the bicycle already mentioned above as an example. A person presses on the pedals and does work on them. In addition to imparting torque to the entire mechanism (thanks to which the bicycle moves in space), the internal energy of the materials from which the levers are made changes. The cyclist presses the handles to turn, and again does the work.
The internal energy of the outer coating (plastic or metal) increases. A person rides out into a clearing under the bright sun - the bicycle heats up, its amount of heat changes. Stops to rest in the shade of an old oak tree and the system cools, losing calories or joules. Increases speed - energy exchange increases. However, calculating the amount of heat in all these cases will show a very small, imperceptible value. Therefore, it seems that the manifestations of thermodynamic physics in real life No.
Application of calculations for changes in the amount of heat
The reader will probably say that all this is very educational, but why are we so tormented at school with these formulas? And now we will give examples in which areas of human activity they are directly needed and how this concerns anyone in their everyday life.
First, look around you and count: how many metal objects surround you? Probably more than ten. But before becoming a paper clip, a carriage, a ring or a flash drive, any metal undergoes smelting. Each plant that processes, say, iron ore, must understand how much fuel is required in order to optimize costs. And when calculating this, it is necessary to know the heat capacity of the metal-containing raw material and the amount of heat that needs to be imparted to it in order for all technological processes to occur. Since the energy released by a unit of fuel is calculated in joules or calories, the formulas are needed directly.
Or another example: most supermarkets have a department with frozen goods - fish, meat, fruit. Where raw materials from animal meat or seafood are transformed into semi-finished products, they must know how much electricity refrigeration and freezing units will consume per ton or unit of finished product. To do this, you need to calculate how much heat a kilogram of strawberries or squid loses when cooled by one degree Celsius. And in the end, this will show how much electricity a freezer of a certain power will consume.
Planes, ships, trains
Above we showed examples of relatively motionless, static objects to which a certain amount of heat is imparted or from which, on the contrary, a certain amount of heat is taken away. For objects that move in conditions of constantly changing temperature during operation, calculations of the amount of heat are important for another reason.
There is such a thing as “metal fatigue”. It also includes maximum permissible loads at a certain rate of temperature change. Imagine an airplane taking off from the humid tropics into the frozen upper atmosphere. Engineers have to work hard to ensure that it does not fall apart due to cracks in the metal that appear when the temperature changes. They are looking for an alloy composition that can withstand real loads and have a large margin of safety. And so as not to search blindly, hoping to accidentally stumble upon the desired composition, you have to do a lot of calculations, including those involving changes in the amount of heat.
Learning Objective: Introduce the concepts of heat quantity and specific heat capacity.
Developmental goal: To cultivate attentiveness; teach to think, draw conclusions.
1. Updating the topic
2. Explanation of new material. 50 min.
You already know that the internal energy of a body can change both by doing work and by heat transfer (without doing work).
The energy that a body gains or loses during heat transfer is called the amount of heat. (write in notebook)
This means that the units for measuring the amount of heat are also Joules ( J).
We conduct an experiment: two glasses in one with 300 g of water, and in the other with 150 g, and an iron cylinder weighing 150 g. Both glasses are placed on the same tile. After some time, thermometers will show that the water in the vessel in which the body is located heats up faster.
This means that heating 150 g of iron requires less heat than heating 150 g of water.
The amount of heat transferred to a body depends on the type of substance from which the body is made. (write in notebook)
We propose the question: is the same amount of heat required to heat bodies to the same temperature? equal mass, but consisting of different substances?
We conduct an experiment with Tyndall's device to determine specific heat capacity.
We conclude: bodies of different substances, but of the same mass, give up when cooled and require the same number of degrees when heated different quantities warmth.
We draw conclusions:
1. To heat bodies of equal mass, consisting of different substances, to the same temperature, different amounts of heat are required.
2. Bodies of equal mass, consisting of different substances and heated to the same temperature. When cooled by the same number of degrees, different amounts of heat are released.
We conclude that the amount of heat required to heat a unit mass of different substances by one degree will vary.
We give the definition of specific heat capacity.
A physical quantity numerically equal to the amount of heat that must be transferred to a body weighing 1 kg in order for its temperature to change by 1 degree is called the specific heat capacity of a substance.
Enter the unit of measurement for specific heat capacity: 1J/kg*degree.
Physical meaning of the term : Specific heat capacity shows by what amount the internal energy of 1g (kg) of a substance changes when it is heated or cooled by 1 degree.
Let's look at the table of specific heat capacities of some substances.
We solve the problem analytically
How much heat is required to heat a glass of water (200 g) from 20 0 to 70 0 C.
To heat 1 g per 1 g, 4.2 J is required.
And to heat 200 g by 1 g, it will take 200 more - 200 * 4.2 J.
And to heat 200 g by (70 0 -20 0) it will take another (70-20) more - 200 * (70-20) * 4.2 J
Substituting the data, we get Q = 200 * 50 * 4.2 J = 42000 J.
Let us write the resulting formula in terms of the corresponding quantities
4. What determines the amount of heat received by a body when heated?
Please note that the amount of heat required to heat any body is proportional to the mass of the body and the change in its temperature.,
There are two cylinders of equal mass: iron and brass. Is the same amount of heat required to heat them the same number of degrees? Why?
What amount of heat is needed to heat 250 g of water from 20 o to 60 0 C.
What is the relationship between calorie and joule?
A calorie is the amount of heat required to heat 1 g of water by 1 degree.
1 cal = 4.19 = 4.2 J
1kcal=1000cal
1kcal=4190J=4200J
3. Problem solving. 28 min.
If cylinders of lead, tin and steel weighing 1 kg heated in boiling water are placed on ice, they will cool and part of the ice under them will melt. How will the internal energy of the cylinders change? Under which cylinder will it melt? more ice, under which – less?
A heated stone weighing 5 kg. Cooling in water by 1 degree, it transfers 2.1 kJ of energy to it. What is the specific heat capacity of the stone?
When hardening a chisel, it was first heated to 650 0, then lowered into oil, where it cooled to 50 0 C. What amount of heat was released if its mass was 500 grams.
How much heat was used to heat a steel blank for the compressor crankshaft weighing 35 kg from 20 0 to 1220 0 C.
Independent work
What type of heat transfer?
Students fill out the table.
- The air in the room is heated through the walls.
- Through open window, into which warm air enters.
- Through glass that lets in the sun's rays.
- The earth is heated by the sun's rays.
- The liquid is heated on the stove.
- The steel spoon is heated by the tea.
- The air is heated by the candle.
- The gas moves near the fuel-generating parts of the machine.
- Heating a machine gun barrel.
- Boiling milk.
5. Homework: Peryshkin A.V. “Physics 8” § §7, 8; collection of problems 7-8 Lukashik V.I. No. 778-780, 792,793 2 min.
As is known, during various mechanical processes a change in mechanical energy occurs W meh. A measure of the change in mechanical energy is the work of forces applied to the system:
\(~\Delta W_(meh) = A.\)
During heat exchange, a change in the internal energy of the body occurs. A measure of the change in internal energy during heat transfer is the amount of heat.
Quantity of heat is a measure of the change in internal energy that a body receives (or gives up) during the process of heat exchange.
Thus, both work and the amount of heat characterize the change in energy, but are not identical to energy. They do not characterize the state of the system itself, but determine the process of energy transition from one type to another (from one body to another) when the state changes and significantly depend on the nature of the process.
The main difference between work and the amount of heat is that work characterizes the process of changing the internal energy of a system, accompanied by the transformation of energy from one type to another (from mechanical to internal). The amount of heat characterizes the process of transfer of internal energy from one body to another (from more heated to less heated), not accompanied by energy transformations.
Experience shows that the amount of heat required to heat a body mass m on temperature T 1 to temperature T 2, calculated by the formula
\(~Q = cm (T_2 - T_1) = cm \Delta T, \qquad (1)\)
Where c- specific heat capacity of the substance;
\(~c = \frac(Q)(m (T_2 - T_1)).\)
The SI unit of specific heat capacity is joule per kilogram Kelvin (J/(kg K)).
Specific heat c is numerically equal to the amount of heat that must be imparted to a body weighing 1 kg in order to heat it by 1 K.
Heat capacity body C T is numerically equal to the amount of heat required to change body temperature by 1 K:
\(~C_T = \frac(Q)(T_2 - T_1) = cm.\)
The SI unit of heat capacity of a body is joule per Kelvin (J/K).
To transform a liquid into steam at a constant temperature, it is necessary to expend an amount of heat
\(~Q = Lm, \qquad (2)\)
Where L - specific heat vaporization. When steam condenses, the same amount of heat is released.
In order to melt a crystalline body weighing m at the melting point, the body needs to communicate the amount of heat
\(~Q = \lambda m, \qquad (3)\)
Where λ - specific heat of fusion. When a body crystallizes, the same amount of heat is released.
The amount of heat released during complete combustion of a mass of fuel m,
\(~Q = qm, \qquad (4)\)
Where q- specific heat of combustion.
The SI unit of specific heats of vaporization, melting and combustion is joule per kilogram (J/kg).
Literature
Aksenovich L. A. Physics in high school: Theory. Tasks. Tests: Textbook. benefits for institutions providing general education. environment, education / L. A. Aksenovich, N. N. Rakina, K. S. Farino; Ed. K. S. Farino. - Mn.: Adukatsiya i vyhavanne, 2004. - P. 154-155.
Exercise 81.
Calculate the amount of heat that will be released during the reduction of Fe 2 O 3 metallic aluminum if 335.1 g of iron was obtained. Answer: 2543.1 kJ.
Solution:
Reaction equation:
= (Al 2 O 3) - (Fe 2 O 3) = -1669.8 -(-822.1) = -847.7 kJ
Calculation of the amount of heat that is released when receiving 335.1 g of iron is made from the proportion:
(2 . 55,85) : -847,7 = 335,1 : X; x = (0847.7 . 335,1)/ (2 . 55.85) = 2543.1 kJ,
where 55.85 atomic mass of iron.
Answer: 2543.1 kJ.
Thermal effect of reaction
Task 82.
Gaseous ethyl alcohol C2H5OH can be obtained by the interaction of ethylene C 2 H 4 (g) and water vapor. Write the thermochemical equation for this reaction, having first calculated its thermal effect. Answer: -45.76 kJ.
Solution:
The reaction equation is:
C 2 H 4 (g) + H 2 O (g) = C2H 5 OH (g); = ?
The values of standard heats of formation of substances are given in special tables. Considering that the heats of formation of simple substances are conventionally assumed to be zero. Let us calculate the thermal effect of the reaction using a consequence of Hess’s law, we obtain:
= (C 2 H 5 OH) – [ (C 2 H 4) + (H 2 O)] =
= -235.1 -[(52.28) + (-241.83)] = - 45.76 kJ
Reaction equations in which their state of aggregation or crystal modification, as well as the numerical value of thermal effects are indicated next to the symbols of chemical compounds, are called thermochemical. In thermochemical equations, unless specifically stated, the values of thermal effects at constant pressure Q p are indicated equal to the change in enthalpy of the system. The value is usually given on the right side of the equation, separated by a comma or semicolon. The following abbreviated designations for the state of aggregation of a substance are accepted: G- gaseous, and- liquid, To
If heat is released as a result of a reaction, then< О. Учитывая сказанное, составляем термохимическое уравнение данной в примере реакции:
C 2 H 4 (g) + H 2 O (g) = C 2 H 5 OH (g); = - 45.76 kJ.
Answer:- 45.76 kJ.
Task 83.
Calculate the thermal effect of the reduction reaction of iron (II) oxide with hydrogen based on the following thermochemical equations:
a) EO (k) + CO (g) = Fe (k) + CO 2 (g); = -13.18 kJ;
b) CO (g) + 1/2O 2 (g) = CO 2 (g); = -283.0 kJ;
c) H 2 (g) + 1/2O 2 (g) = H 2 O (g); = -241.83 kJ.
Answer: +27.99 kJ.
Solution:
The reaction equation for the reduction of iron (II) oxide with hydrogen has the form:
EeO (k) + H 2 (g) = Fe (k) + H 2 O (g); = ?
= (H2O) – [ (FeO)
The heat of formation of water is given by the equation
H 2 (g) + 1/2O 2 (g) = H 2 O (g); = -241.83 kJ,
and the heat of formation of iron (II) oxide can be calculated by subtracting equation (a) from equation (b).
=(c) - (b) - (a) = -241.83 – [-283.o – (-13.18)] = +27.99 kJ.
Answer:+27.99 kJ.
Task 84.
When gaseous hydrogen sulfide and carbon dioxide interact, water vapor and carbon disulfide CS 2 (g) are formed. Write the thermochemical equation for this reaction and first calculate its thermal effect. Answer: +65.43 kJ.
Solution:
G- gaseous, and- liquid, To-- crystalline. These symbols are omitted if the aggregative state of the substances is obvious, for example, O 2, H 2, etc.
The reaction equation is:
2H 2 S (g) + CO 2 (g) = 2H 2 O (g) + CS 2 (g); = ?
The values of standard heats of formation of substances are given in special tables. Considering that the heats of formation of simple substances are conventionally assumed to be zero. The thermal effect of a reaction can be calculated using a corollary of Hess's law:
= (H 2 O) + (СS 2) – [(H 2 S) + (СO 2)];
= 2(-241.83) + 115.28 – = +65.43 kJ.
2H 2 S (g) + CO 2 (g) = 2H 2 O (g) + CS 2 (g); = +65.43 kJ.
Answer:+65.43 kJ.
Thermochemical reaction equation
Task 85.
Write the thermochemical equation for the reaction between CO (g) and hydrogen, as a result of which CH 4 (g) and H 2 O (g) are formed. How much heat will be released during this reaction if 67.2 liters of methane were produced in terms of normal conditions? Answer: 618.48 kJ.
Solution:
Reaction equations in which their state of aggregation or crystal modification, as well as the numerical value of thermal effects are indicated next to the symbols of chemical compounds, are called thermochemical. In thermochemical equations, unless specifically stated, the values of thermal effects at constant pressure Q p equal to the change in enthalpy of the system are indicated. The value is usually given on the right side of the equation, separated by a comma or semicolon. The following abbreviated designations for the state of aggregation of a substance are accepted: G- gaseous, and- something, To- crystalline. These symbols are omitted if the aggregative state of the substances is obvious, for example, O 2, H 2, etc.
The reaction equation is:
CO (g) + 3H 2 (g) = CH 4 (g) + H 2 O (g); = ?
The values of standard heats of formation of substances are given in special tables. Considering that the heats of formation of simple substances are conventionally assumed to be zero. The thermal effect of a reaction can be calculated using a corollary of Hess's law:
= (H 2 O) + (CH 4) – (CO)];
= (-241.83) + (-74.84) – (-110.52) = -206.16 kJ.
The thermochemical equation will be:
22,4 : -206,16 = 67,2 : X; x = 67.2 (-206.16)/22?4 = -618.48 kJ; Q = 618.48 kJ.
Answer: 618.48 kJ.
Heat of formation
Task 86.
The thermal effect of which reaction is equal to the heat of formation. Calculate the heat of formation of NO based on the following thermochemical equations:
a) 4NH 3 (g) + 5O 2 (g) = 4NO (g) + 6H 2 O (l); = -1168.80 kJ;
b) 4NH 3 (g) + 3O 2 (g) = 2N 2 (g) + 6H 2 O (l); = -1530.28 kJ
Answer: 90.37 kJ.
Solution:
The standard heat of formation is equal to the heat of reaction of the formation of 1 mole of this substance from simple substances under standard conditions (T = 298 K; p = 1.0325.105 Pa). The formation of NO from simple substances can be represented as follows:
1/2N 2 + 1/2O 2 = NO
Given is reaction (a), which produces 4 mol of NO, and given reaction (b), which produces 2 mol of N2. Oxygen is involved in both reactions. Therefore, to determine the standard heat of formation of NO, we compose the following Hess cycle, i.e., we need to subtract equation (a) from equation (b):
Thus, 1/2N 2 + 1/2O 2 = NO; = +90.37 kJ.
Answer: 618.48 kJ.
Task 87.
Crystalline ammonium chloride is formed by the reaction of ammonia and hydrogen chloride gases. Write the thermochemical equation for this reaction, having first calculated its thermal effect. How much heat will be released if 10 liters of ammonia were consumed in the reaction, calculated under normal conditions? Answer: 78.97 kJ.
Solution:
Reaction equations in which their state of aggregation or crystal modification, as well as the numerical value of thermal effects are indicated next to the symbols of chemical compounds, are called thermochemical. In thermochemical equations, unless specifically stated, the values of thermal effects at constant pressure Q p equal to the change in enthalpy of the system are indicated. The value is usually given on the right side of the equation, separated by a comma or semicolon. The following have been accepted: To-- crystalline. These symbols are omitted if the aggregative state of the substances is obvious, for example, O 2, H 2, etc.
The reaction equation is:
NH 3 (g) + HCl (g) = NH 4 Cl (k). ;
The values of standard heats of formation of substances are given in special tables. Considering that the heats of formation of simple substances are conventionally assumed to be zero. The thermal effect of a reaction can be calculated using a corollary of Hess's law:
= ?
= (NH4Cl) – [(NH 3) + (HCl)];
The thermochemical equation will be:
= -315.39 – [-46.19 + (-92.31) = -176.85 kJ.
22,4 : -176,85 = 10 : The heat released during the reaction of 10 liters of ammonia in this reaction is determined from the proportion:
Answer: X; x = 10 (-176.85)/22.4 = -78.97 kJ; Q = 78.97 kJ.
78.97 kJ.
The concept of the amount of heat was formed in the early stages of the development of modern physics, when there were no clear ideas about the internal structure of matter, what energy is, what forms of energy exist in nature and about energy as a form of movement and transformation of matter. The amount of heat means physical quantity
equivalent to the energy transferred to a material body in the process of heat exchange.
Initially, it was assumed that the carrier of thermal energy was some completely weightless medium with the properties of a liquid. Numerous physical problems of heat transfer have been and are still being solved based on this premise. The existence of hypothetical caloric was the basis for many essentially correct constructions. It was believed that caloric is released and absorbed in the phenomena of heating and cooling, melting and crystallization. The correct equations for heat transfer processes were obtained based on incorrect physical concepts. There is a known law according to which the amount of heat is directly proportional to the mass of the body involved in heat exchange and the temperature gradient:
Where Q is the amount of heat, m is the body mass, and the coefficient With– a quantity called specific heat capacity. Specific heat capacity is a characteristic of a substance involved in a process.
Work in thermodynamics
As a result of thermal processes, purely mechanical work can be performed. For example, when a gas heats up, it increases its volume. Let's take a situation like the picture below:
In this case, the mechanical work will be equal to the force of gas pressure on the piston multiplied by the path traveled by the piston under pressure. Of course, this is the simplest case. But even in it one can notice one difficulty: the pressure force will depend on the volume of the gas, which means that we are not dealing with constants, but with variable quantities. Since all three variables: pressure, temperature and volume are related to each other, calculating work becomes significantly more complicated. There are some ideal, infinitely slow processes: isobaric, isothermal, adiabatic and isochoric - for which such calculations can be performed relatively simply. A graph of pressure versus volume is plotted and the work is calculated as an integral of the form.
