Define the specific heat capacity of a substance. Specific heat capacity of gases and vapors

/(kg K), etc.

Specific heat capacity is usually denoted by the letters c or WITH, often with indexes.

The specific heat capacity is affected by the temperature of the substance and other thermodynamic parameters. For example, measuring the specific heat capacity of water will give different results at 20 °C and 60 °C. In addition, specific heat capacity depends on how the thermodynamic parameters of the substance (pressure, volume, etc.) are allowed to change; for example, specific heat capacity at constant pressure ( C P) and at constant volume ( C V), generally speaking, are different.

Formula for calculating specific heat capacity:

$c=\backslash frac(Q)(\; m\backslash Delta\; T),$ Where c- specific heat capacity, Q- the amount of heat received by a substance when heated (or released when cooled), m- mass of the heated (cooled) substance, Δ T- the difference between the final and initial temperatures of the substance.

Specific heat capacity can depend (and in principle, strictly speaking, always, more or less strongly, depends) on temperature, therefore the following formula with small (formally infinitesimal) values ​​is more correct: $\backslash delta\; T$ And $\backslash delta\; Q$:

$c(T)\; =\; \backslash frac\; 1\; (m)\; \backslash left(\backslash frac(\backslash delta\; Q)(\backslash delta\; T)\backslash right).$

Specific heat values ​​for some substances

(For gases, the specific heat capacity in an isobaric process (C p) is given)

Table I: Standard Specific Heat Capacity Values

Substance	State of aggregation	Specific heat capacity, kJ/(kg K)
air (dry)	gas	1,005
air (100% humidity)	gas	1,0301
aluminum	solid	0,903
beryllium	solid	1,8245
brass	solid	0,37
tin	solid	0,218
copper	solid	0,385
molybdenum	solid	0,250
steel	solid	0,462
diamond	solid	0,502
ethanol	liquid	2,460
gold	solid	0,129
graphite	solid	0,720
helium	gas	5,190
hydrogen	gas	14,300
iron	solid	0,444
lead	solid	0,130
cast iron	solid	0,540
tungsten	solid	0,134
lithium	solid	3,582
	liquid	0,139
nitrogen	gas	1,042
petroleum oils	liquid	1,67 - 2,01
oxygen	gas	0,920
quartz glass	solid	0,703
water 373 K (100 °C)	gas	2,020
water	liquid	4,187
ice	solid	2,060
beer wort	liquid	3,927
Values ​​are based on standard conditions unless otherwise noted.

Table II: Specific Heat Capacity Values ​​for Some Building Materials

Substance	Specific heat capacity kJ/(kg K)
asphalt	0,92
solid brick	0,84
sand-lime brick	1,00
concrete	0,88
crown glass (glass)	0,67
flint (glass)	0,503
window glass	0,84
granite	0,790
soapstone	0,98
gypsum	1,09
marble, mica	0,880
sand	0,835
steel	0,47
the soil	0,80
wood	1,7

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Notes

Literature

• Tables of physical quantities. Handbook, ed. I. K. Kikoina, M., 1976.
• Sivukhin D.V. General course in physics. - T. II. Thermodynamics and molecular physics.
• E. M. Lifshits // under. ed. A. M. Prokhorova Physical Encyclopedia. - M.: “Soviet Encyclopedia”, 1998. - T. 2.<

Excerpt characterizing Specific Heat Capacity

- Does it work? – Natasha repeated.
– I’ll tell you about myself. I had one cousin...
- I know - Kirilla Matveich, but he’s an old man?
– It wasn’t always an old man. But here’s what, Natasha, I’ll talk to Borya. He doesn't need to travel so often...
- Why shouldn’t he, if he wants to?
- Because I know that this will not end in anything.
- Why do you know? No, mom, you don't tell him. What nonsense! - Natasha said in the tone of a person from whom they want to take away his property.
“Well, I won’t get married, so let him go, if he’s having fun and I’m having fun.” – Natasha smiled and looked at her mother.
“Not married, just like that,” she repeated.
- How is this, my friend?
- Yes, yes. Well, it’s very necessary that I don’t get married, but... so.
“Yes, yes,” the countess repeated and, shaking her whole body, laughed with a kind, unexpected old woman’s laugh.
“Stop laughing, stop,” Natasha shouted, “you’re shaking the whole bed.” You look terribly like me, the same laugher... Wait... - She grabbed both hands of the countess, kissed the little finger bone on one - June, and continued to kiss July, August on the other hand. - Mom, is he very much in love? How about your eyes? Were you so in love? And very sweet, very, very sweet! But it’s not quite to my taste - it’s narrow, like a table clock... Don’t you understand?... Narrow, you know, gray, light...
- Why are you lying! - said the countess.
Natasha continued:
- Do you really not understand? Nikolenka would understand... The earless one is blue, dark blue with red, and he is quadrangular.
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She couldn't sleep for a long time. She kept thinking that no one could understand everything that she understood and that was in her.
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The next day, the countess, inviting Boris to her place, talked with him, and from that day he stopped visiting the Rostovs.

On December 31, on New Year's Eve 1810, le reveillon [night supper], there was a ball at Catherine's nobleman's house. The diplomatic corps and the sovereign were supposed to be at the ball.
On the Promenade des Anglais, the famous house of a nobleman glowed with countless lights. At the illuminated entrance with a red cloth stood the police, and not only gendarmes, but the police chief at the entrance and dozens of police officers. The carriages drove off, and new ones drove up with red footmen and footmen with feathered hats. Men in uniforms, stars and ribbons came out of the carriages; ladies in satin and ermine carefully stepped down the noisily laid down steps, and hurriedly and silently walked along the cloth of the entrance.
Almost every time a new carriage arrived, there was a murmur in the crowd and hats were taken off.
“Sovereign?... No, minister... prince... envoy... Don’t you see the feathers?...” said from the crowd. One of the crowd, better dressed than the others, seemed to know everyone, and called by name the most noble nobles of that time.
Already one third of the guests had arrived at this ball, and the Rostovs, who were supposed to be at this ball, were still hastily preparing to dress.
There was a lot of talk and preparation for this ball in the Rostov family, a lot of fears that the invitation would not be received, the dress would not be ready, and everything would not work out as needed.
Along with the Rostovs, Marya Ignatievna Peronskaya, a friend and relative of the countess, a thin and yellow maid of honor of the old court, leading the provincial Rostovs in the highest St. Petersburg society, went to the ball.
At 10 o'clock in the evening the Rostovs were supposed to pick up the maid of honor at the Tauride Garden; and yet it was already five minutes to ten, and the young ladies were not yet dressed.
Natasha was going to the first big ball in her life. That day she got up at 8 o'clock in the morning and was in feverish anxiety and activity all day. All her strength, from the very morning, was aimed at ensuring that they all: she, mother, Sonya were dressed in the best possible way. Sonya and the Countess trusted her completely. The countess was supposed to be wearing a masaka velvet dress, the two of them were wearing white smoky dresses on pink, silk covers with roses in the bodice. The hair had to be combed a la grecque [in Greek].
Everything essential had already been done: the legs, arms, neck, ears were already especially carefully, like a ballroom, washed, perfumed and powdered; they were already wearing silk, fishnet stockings and white satin shoes with bows; the hairstyles were almost finished. Sonya finished dressing, and so did the Countess; but Natasha, who was working for everyone, fell behind. She was still sitting in front of the mirror with a peignoir draped over her slender shoulders. Sonya, already dressed, stood in the middle of the room and, pressing painfully with her small finger, pinned the last ribbon that squealed under the pin.

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.

Specific heat is the energy required to increase the temperature of 1 gram of a pure substance by 1°. The parameter depends on its chemical composition and state of aggregation: gaseous, liquid or solid. After its discovery, a new round of development began in thermodynamics, the science of energy transients that relate to heat and the functioning of the system.

Usually, specific heat capacity and basic thermodynamics are used in the manufacture radiators and systems designed for cooling automobiles, as well as in chemistry, nuclear engineering and aerodynamics. If you want to know how specific heat capacity is calculated, then read the proposed article.

Before you begin directly calculating the parameter, you should familiarize yourself with the formula and its components.

The formula for calculating specific heat capacity is as follows:

• c = Q/(m*∆T)

Knowledge of quantities and their symbolic designations used in calculations is extremely important. However, it is necessary not only to know their visual appearance, but also to clearly understand the meaning of each of them. The calculation of the specific heat capacity of a substance is represented by the following components:

ΔT is a symbol indicating a gradual change in the temperature of a substance. The symbol "Δ" is pronounced delta.

ΔT = t2–t1, where

• t1 – primary temperature;
• t2 – final temperature after change.

m – mass of the substance used during heating (g).

Q – amount of heat (J/J)

Based on CR, other equations can be derived:

• Q = m*кp*ΔT – amount of heat;
• m = Q/cr*(t2 - t1) – mass of substance;
• t1 = t2–(Q/tp*m) – primary temperature;
• t2 = t1+(Q/tp*m) – final temperature.

Instructions for calculating the parameter

1. Take the calculation formula: Heat capacity = Q/(m*∆T)
2. Write down the original data.
3. Substitute them into the formula.
4. Carry out the calculation and get the result.

As an example, let’s calculate an unknown substance weighing 480 grams with a temperature of 15ºC, which, as a result of heating (supplying 35 thousand J), increased to 250º.

According to the instructions given above, we perform the following actions:

Let's write down the initial data:

• Q = 35 thousand J;
• m = 480 g;
• ΔT = t2–t1 =250–15 = 235 ºC.

We take the formula, substitute the values ​​and solve:

c=Q/(m*∆T)=35 thousand J/(480 g*235º)=35 thousand J/(112800 g*º)=0.31 J/g*º.

Calculation

Let's do the calculation C P water and tin under the following conditions:

• m = 500 grams;
• t1 =24ºC and t2 = 80ºC – for water;
• t1 =20ºC and t2 =180ºC – for tin;
• Q = 28 thousand J.

First, we determine ΔT for water and tin, respectively:

• ΔТв = t2–t1 = 80–24 = 56ºC
• ΔTo = t2–t1 = 180–20 =160ºC

Then we find the specific heat capacity:

1. c=Q/(m*ΔTv)= 28 thousand J/(500 g *56ºC) = 28 thousand J/(28 thousand g*ºC) = 1 J/g*ºC.
2. c=Q/(m*ΔTo)=28 thousand J/(500 g*160ºC)=28 thousand J/(80 thousand g*ºC)=0.35 J/g*ºC.

Thus, the specific heat capacity of water was 1 J/g *ºC, and that of tin was 0.35 J/g*ºC. From this we can conclude that with an equal value of heat input of 28 thousand Joules, tin will heat up faster than water, since its heat capacity is lower.

Not only gases, liquids and solids, but also food products have heat capacity.

How to calculate the heat capacity of food

When calculating power capacity the equation will take the following form:

с=(4.180*w)+(1.711*p)+(1.928*f)+(1.547*c)+(0.908 *a), where:

• w – amount of water in the product;
• p – amount of proteins in the product;
• f – percentage of fat;
• c – percentage of carbohydrates;
• a is the percentage of inorganic components.

Let's determine the heat capacity of Viola cream cheese. To do this, write out the required values ​​​​from the composition of the product (weight 140 grams):

• water – 35 g;
• proteins – 12.9 g;
• fats – 25.8 g;
• carbohydrates – 6.96 g;
• inorganic components – 21 g.

Then we find with:

• с=(4.180*w)+(1.711*p)+(1.928*f)+(1.547*c)+(0.908*a)=(4.180*35)+(1.711*12.9)+(1.928*25 .8) + (1.547*6.96)+(0.908*21)=146.3+22.1+49.7+10.8+19.1=248 kJ/kg*ºC.

Always remember that:

• The heating process of metal is faster than that of water, since it has C P 2.5 times less;
• If possible, convert the results to a higher order if conditions permit;
• in order to check the results, you can use the Internet and look at the calculated substance;
• under equal experimental conditions, more significant temperature changes will be observed for materials with low specific heat capacity.

Devices and accessories used in work:

2. Weights.

3. Thermometer.

4. Calorimeter.

6. Calorimetric body.

7. Household tiles.

Goal of the work:

Learn to experimentally determine the specific heat capacity of a substance.

I. THEORETICAL INTRODUCTION.

Thermal conductivity- transfer of heat from more heated parts of the body to less heated ones as a result of collisions of fast molecules with slow ones, as a result of which fast molecules transfer part of their energy to slow ones.

The change in the internal energy of any body is directly proportional to its mass and the change in body temperature.

DU = cmDT (1)
Q = cmDT (2)

The quantity c characterizing the dependence of the change in the internal energy of a body during heating or cooling on the type of substance and external conditions is called specific heat capacity of the body.

(4)

The value C, which characterizes the dependence of a body to absorb heat when heated and is equal to the ratio of the amount of heat imparted to the body to the increase in its temperature, is called heat capacity of the body.

C = c × m. (5)
(6)
Q = CDT (7)

Molar heat capacity Cm, is the amount of heat required to heat one mole of a substance by 1 Kelvin

Cm = cM. (8)
C m = (9)

Specific heat capacity depends on the nature of the process in which it is heated.

Heat balance equation.

During heat exchange, the sum of the amounts of heat given off by all bodies whose internal energy decreases is equal to the sum of the amounts of heat received by all bodies whose internal energy increases.

SQ dept = SQ receive (10)

If the bodies form a closed system and only heat exchange occurs between them, then the algebraic sum of the received and given amounts of heat is equal to 0.

SQ dept + SQ receive = 0.

Example:

The heat exchange involves a body, a calorimeter, and a liquid. The body gives off heat, the calorimeter and liquid receive it.

Q t = Q k + Q f

Q t = c t m t (T 2 – Q)

Q k = c k m k (Q – T 1)

Q f = c f m f (Q – T 1)

Where Q(tau) is the overall final temperature.

s t m t (T 2 -Q) = s to m to (Q- T 1) + s f m f (Q- T 1)

s t = ((Q - T 1)*(s to m to + s w m w)) / m t (T 2 - Q)

T = 273 0 + t 0 C

2. PROGRESS OF WORK.

ALL WEIGHINGS ARE CARRIED OUT WITH AN ACCURACY TO THE UP TO 0.1 g.

1. Determine by weighing the mass of the inner vessel, calorimeter m 1.

2. Pour water into the inner vessel of the calorimeter, weigh the inner glass together with the poured liquid m to.

3. Determine the mass of poured water m = m to - m 1

4. Place the inner vessel of the calorimeter in the outer one and measure the initial temperature of the water T 1.

5. Remove the test body from the boiling water, quickly transfer it to the calorimeter, determining T 2 - the initial temperature of the body, it is equal to the temperature of boiling water.

6. While stirring the liquid in the calorimeter, wait until the temperature stops increasing: measure the final (steady) temperature Q.

7. Remove the test body from the calorimeter, dry it with filter paper and determine its mass m 3 by weighing on a scale.

8. Enter the results of all measurements and calculations into the table. Perform calculations to the second decimal place.

9. Create a heat balance equation and find the specific heat capacity of the substance from it With.

10. Based on the results obtained in the application, determine the substance.

11. Calculate the absolute and relative error of the obtained result relative to the tabular result using the formulas:

;

12. Conclusion about the work done.

TABLE OF MEASUREMENT AND CALCULATION RESULTS

Specific heat capacity is a characteristic of a substance. That is, it is different for different substances. In addition, the same substance, but in different states of aggregation, has different specific heat capacity. Thus, it is correct to talk about the specific heat capacity of a substance (specific heat capacity of water, specific heat capacity of gold, specific heat capacity of wood, etc.).

The specific heat capacity of a particular substance shows how much heat (Q) must be transferred to it in order to heat 1 kilogram of this substance by 1 degree Celsius. Specific heat capacity is denoted by the Latin letter c. That is, c = Q/mt. Considering that t and m are equal to unity (1 kg and 1 °C), then the specific heat capacity is numerically equal to the amount of heat.

However, heat and specific heat capacity have different units of measurement. Heat (Q) in the Cu system is measured in Joules (J). And specific heat capacity is in Joules divided by kilogram multiplied by degrees Celsius: J/(kg °C).

If the specific heat capacity of a substance is, for example, 390 J/(kg °C), this means that if 1 kg of this substance is heated by 1 °C, it will absorb 390 J of heat. Or, in other words, to heat 1 kg of this substance by 1 °C, 390 J of heat must be transferred to it. Or, if 1 kg of this substance is cooled by 1 °C, then it will give off 390 J of heat.

If not 1, but 2 kg of a substance is heated by 1 °C, then twice as much heat must be transferred to it. So for the example above it will already be 780 J. The same will happen if 1 kg of substance is heated by 2 °C.

The specific heat capacity of a substance does not depend on its initial temperature. That is, if, for example, liquid water has a specific heat capacity of 4200 J/(kg °C), then heating by 1 °C even twenty-degree or ninety-degree water will equally require 4200 J of heat per 1 kg.

But ice has a specific heat capacity that is different from liquid water, almost two times less. However, in order to heat it by 1 °C, the same amount of heat per 1 kg will be required, regardless of its initial temperature.

Specific heat capacity also does not depend on the shape of the body that is made of a given substance. A steel bar and a steel sheet having the same mass will require the same amount of heat to heat them by the same number of degrees. Another thing is that the exchange of heat with the environment should be neglected. The sheet has a larger surface area than the bar, which means the sheet gives off more heat and will therefore cool faster. But under ideal conditions (when heat loss can be neglected), body shape does not matter. Therefore, they say that specific heat capacity is a characteristic of a substance, but not a body.

So, the specific heat capacity of different substances is different. This means that if different substances are given with the same mass and the same temperature, then in order to heat them to a different temperature, different amounts of heat must be transferred to them. For example, a kilogram of copper will require about 10 times less heat than water. That is, copper has a specific heat capacity that is approximately 10 times less than that of water. We can say that “less heat is placed in copper.”

The amount of heat that must be transferred to a body in order to heat it from one temperature to another is found using the following formula:

Q = cm(t k – t n)

Here tk and tn are the final and initial temperatures, m is the mass of the substance, c is its specific heat capacity. Specific heat capacity is usually taken from tables. From this formula the specific heat capacity can be expressed.