The specific heat capacity of the body is equal to. Specific heat capacity: calculation of the amount of heat
Specific heat capacity 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 his discovery, a new round of development of thermodynamics began, the science of energy transition processes that relate to heat and the functioning of the system.
Usually, specific heat capacity and the basics of thermodynamics are used in the manufacture radiators and systems designed for cooling vehicles, as well as in chemistry, nuclear engineering and aerodynamics. If you want to know how the specific heat capacity is calculated, then check out the proposed article.
Before proceeding with the direct calculation of the parameter, you should familiarize yourself with the formula and its components.
The formula for calculating the specific heat capacity is as follows:
- с = Q/(m*∆T)
Knowledge of the quantities and their symbolic designations used in the calculation 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 denoting a gradual change in the temperature of a substance. The symbol "Δ" is pronounced like a delta.
ΔT = t2–t1, where
- t1 is the primary temperature;
- t2 is the final temperature after the change.
m is the mass of the substance used for heating (g).
Q - the amount of heat (J / J)
Based on CR, other equations can be derived:
- Q \u003d m * cp * ΔT - the amount of heat;
- m = Q/cr * (t2 - t1) - the mass of the substance;
- t1 = t2–(Q/цp*m) – primary temperature;
- t2 = t1+(Q/цp*m) – final temperature.
Instructions for calculating the parameter
- Take the calculation formula: Heat capacity \u003d Q / (m * ∆T)
- Write out the original data.
- Plug them into the formula.
- Do the calculation and get the result.
As an example, let's calculate an unknown substance weighing 480 grams and having 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:
We write out 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:
с=Q/(m*∆T)=35 thousand J/(480 g*235º)=35 thousand J/(112800 g*º)=0.31 J/g*º.
Calculation
Let's perform 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:
- ΔTv = t2–t1 = 80–24 = 56ºC
- ΔТо = t2–t1 = 180–20 =160ºC
Then we find the specific heat capacity:
- c \u003d Q / (m * ΔTv) \u003d 28 thousand J / (500 g * 56ºC) \u003d 28 thousand J / (28 thousand g * ºC) \u003d 1 J / g * ºC.
- с=Q/(m*ΔТо)=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 the input heat of 28 thousand J, tin will heat up faster than water, since its heat capacity is less.
Heat capacity is possessed not only by gases, liquids and solids, but also by food.
How to calculate the heat capacity of food
When calculating the power capacity the equation will take the following form:
c=(4.180*w)+(1.711*p)+(1.928*f)+(1.547*c)+(0.908*a), where:
- w is the amount of water in the product;
- p is the amount of proteins in the product;
- f is the percentage of fat;
- c is the percentage of carbohydrates;
- a is the percentage of inorganic components.
Determine the heat capacity of processed cream cheese Viola. To do this, we write out the desired 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:
- c=(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 process of heating the metal is faster than that of water, since it has C P 2.5 times less;
- if possible, transform the results obtained to a higher order, if conditions allow;
- in order to check the results, you can use the Internet and look with for the calculated substance;
- under equal experimental conditions, more significant temperature changes will be observed in materials with low specific heat.
(or heat transfer).
Specific heat capacity of a substance.
Heat capacity is the amount of heat absorbed by the body when heated by 1 degree.
The heat capacity of the body is indicated by a capital Latin letter With.
What determines the heat capacity of a body? First of all, from its mass. It is clear that heating, for example, 1 kilogram of water will require more heat than heating 200 grams.
What about the kind of substance? Let's do an experiment. Let's take two identical vessels and, pouring water weighing 400 g into one of them, and vegetable oil weighing 400 g into the other, we will begin to heat them with the help of identical burners. By observing the readings of thermometers, we will see that the oil heats up quickly. To heat water and oil to the same temperature, the water must be heated longer. But the longer we heat the water, the more heat it receives from the burner.
Thus, to heat the same mass of different substances to the same temperature, different amounts of heat are required. The amount of heat required to heat a body and, consequently, its heat capacity depend on the kind of substance of which this body is composed.
So, for example, to increase the temperature of water with a mass of 1 kg by 1 ° C, an amount of heat equal to 4200 J is required, and to heat the same mass of sunflower oil by 1 ° C, an amount of heat equal to 1700 J is required.
The physical quantity showing how much heat is required to heat 1 kg of a substance by 1 ºС is called specific heat this substance.
Each substance has its own specific heat capacity, which is denoted by the Latin letter c and is measured in joules per kilogram-degree (J / (kg ° C)).
The specific heat capacity of the same substance in different aggregate states (solid, liquid and gaseous) is different. For example, the specific heat capacity of water is 4200 J/(kg ºС), and the specific heat capacity of ice is 2100 J/(kg ºС); aluminum in the solid state has a specific heat capacity of 920 J/(kg - °C), and in the liquid state it is 1080 J/(kg - °C).
Note that water has a very high specific heat capacity. Therefore, the water in the seas and oceans, heating up in summer, absorbs a large amount of heat from the air. Due to this, in those places that are located near large bodies of water, summer is not as hot as in places far from water.
Calculation of the amount of heat required to heat the body or released by it during cooling.
From the foregoing, it is clear that the amount of heat necessary to heat the body depends on the type of substance of which the body consists (i.e., its specific heat capacity) and on the mass of the body. It is also clear that the amount of heat depends on how many degrees we are going to increase the temperature of the body.
So, to determine the amount of heat required to heat the body or released by it during cooling, you need to multiply the specific heat of the body by its mass and by the difference between its final and initial temperatures:
Q = cm (t 2 - t 1 ) ,
where Q- quantity of heat, c is the specific heat capacity, m- body mass , t 1 - initial temperature, t 2 is the final temperature.
When the body is heated t 2 > t 1 and hence Q > 0 . When the body is cooled t 2and< t 1 and hence Q< 0 .
If the heat capacity of the whole body is known With, Q is determined by the formula:
Q \u003d C (t 2 - t 1 ) .
Physics and thermal phenomena is a rather extensive section, which is thoroughly studied in the school course. Not the last place in this theory is given to specific quantities. The first of these is the specific heat capacity.
However, the interpretation of the word "specific" is usually given insufficient attention. Students simply memorize it as a given. And what does it mean?
If you look into Ozhegov's dictionary, you can read that such a value is defined as a ratio. Moreover, it can be performed for mass, volume or energy. All these quantities must necessarily be taken equal to unity. The relation to what is given in the specific heat capacity?
To the product of mass and temperature. Moreover, their values must necessarily 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 lower. There is also a formula from which it can be seen that these two quantities are in the denominator.
What it is?
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 need to be spent on this process. And also calculate its value when the body is cooled. By the way, these two quantities of heat are equal to each other in modulus. But they have different signs. So, in the first case, it is positive, because energy must be spent and it is transferred to the body. The second situation with cooling 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 the course of school physics, this degree is the one that is 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 \u003d 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 transfer, t 1 is the initial temperature of the substance. This is formula #1.
Based on this formula, the unit of measurement of this quantity in the international system of units (SI) is J / (kg * ºС).
How to find other quantities from this equation?
First, the amount of heat. The formula will look like this: Q \u003d c * m * (t 2 - t 1). Only in it it is necessary to substitute values in units included in the SI. That is, mass is in kilograms, temperature is in degrees Celsius. This is formula #2.
Secondly, the mass of a substance that cools or heats up. The formula for it will be: m \u003d Q / (c * (t 2 - t 1)). This is formula number 3.
Thirdly, the change in temperature Δt \u003d t 2 - t 1 \u003d (Q / c * m). The sign "Δ" is read as "delta" and denotes a change in magnitude, in this case temperature. Formula number 4.
Fourth, the initial and final temperatures of the substance. Formulas that are valid for heating a substance look like this: t 1 \u003d t 2 - (Q / c * m), t 2 \u003d t 1 + (Q / c * m). These formulas have numbers 5 and 6. If the problem is about cooling a substance, then the formulas are: t 1 \u003d t 2 + (Q / c * m), t 2 \u003d t 1 - (Q / c * m). These formulas have numbers 7 and 8.
What meanings can it have?
It has been experimentally established what values it has for each specific substance. Therefore, a special table of specific heat capacity has been created. Most often, it gives data that are valid under normal conditions.
What is the laboratory work on measuring specific heat?
In a school physics course, it is determined for a solid body. Moreover, its heat capacity is calculated by comparing with the one that is known. The easiest way to do this is with water.
In the process of performing the work, it is required to measure the initial temperatures of the water and the 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 body. The second expression describes the energy that the body gives off when it cools. These two values are equal. By mathematical calculations, it remains to determine the specific heat capacity of the substance that makes up the solid body.
Most often, it is proposed to compare it with tabular values in order to try to guess what substance the body under study is made of.
Task #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 capacity of the metal if its mass is 100 grams?
Decision. To find the answer, you will need to use the formula written under number 1. There are all the quantities necessary for the calculations. Only first you need to convert the mass to kilograms, otherwise the answer will be wrong. Because all quantities must be those that are accepted in SI.
There are 1000 grams in one kilogram. So, 100 grams must be divided by 1000, you get 0.1 kilograms.
Substitution of all values gives the following expression: c \u003d 152 / (0.1 * (24 - 20)). The calculations are not particularly difficult. The result of all actions is the number 380.
Answer: c \u003d 380 J / (kg * ºС).
Task #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.
Decision. It is worth starting with the fact that energy is given in a non-systemic unit. Kilojoules must be converted to joules: 1680 kJ = 1680000 J.
To find the answer, you need to use the formula number 8. However, the mass appears in it, and it is unknown in the problem. But given the volume of liquid. So, you can use the formula known as m \u003d ρ * V. The density of water is 1000 kg / m 3. But here the volume will need to be substituted in cubic meters. To convert them from liters, it is necessary 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 at the specific heat capacity in the table. Now you can move on to formula 8: t 2 \u003d 100 + (1680000 / 4200 * 5).
The first action is supposed to perform multiplication: 4200 * 5. The result is 21000. The second is division. 1680000: 21000 = 80. Last subtraction: 100 - 80 = 20.
Answer. t 2 \u003d 20 ºС.
Task #3
Condition. There is a chemical beaker with a mass of 100 g. 50 g of water is poured into it. The initial temperature of water with a glass is 0 degrees Celsius. How much heat is required to bring water to a boil?
Decision. You should start by introducing a suitable notation. Let the data related to the glass have index 1, and for water - index 2. In the table, you need to find the specific heat capacities. The chemical beaker is made of laboratory glass, so its value c 1 = 840 J / (kg * ºС). The data for water are as follows: s 2 \u003d 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 \u003d 0.1 kg, m 2 \u003d 0.05 kg.
The initial temperature is given: t 1 \u003d 0 ºС. It is known about the final that it corresponds to the one at which the water boils. This is t 2 \u003d 100 ºС.
Since the glass is heated together with water, the desired amount of heat will be the sum of the two. The first, which is required to heat the glass (Q 1), and the second, which goes to heat the water (Q 2). To express them, a second formula is required. It must be written twice with different indices, and then their sum must be added.
It turns out that Q \u003d 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 more convenient to count. Then the formula that is required to calculate the amount of heat will take the following form: Q \u003d (c 1 * m 1 + c 2 * m 2) * (t 2 - t 1). Now you can substitute the known values in the problem and calculate the result.
Q \u003d (840 * 0.1 + 4200 * 0.05) * (100 - 0) \u003d (84 + 210) * 100 \u003d 294 * 100 \u003d 29400 (J).
Answer. Q = 29400 J = 29.4 kJ.
Specific heat
Heat capacity is the amount of heat absorbed by a body when heated by 1 degree.The heat capacity of the body is indicated by a capital Latin letter WITH.
What determines the heat capacity of a body? First of all, from its mass. It is clear that heating, for example, 1 kilogram of water will require more heat than heating 200 grams.
What about the kind of substance? Let's do an experiment. Let's take two identical vessels and, pouring water weighing 400 g into one of them, and vegetable oil weighing 400 g into the other, we will begin to heat them with the help of identical burners. By observing the readings of thermometers, we will see that the oil heats up faster. To heat water and oil to the same temperature, the water must be heated longer. But the longer we heat the water, the more heat it receives from the burner.
Thus, to heat the same mass of different substances to the same temperature, different amounts of heat are required. The amount of heat required to heat a body and, consequently, its heat capacity depend on the kind of substance of which this body is composed.
So, for example, to increase the temperature of water with a mass of 1 kg by 1 °C, an amount of heat equal to 4200 J is required, and to heat the same mass of sunflower oil by 1 °C, an amount of heat equal to 1700 J is required.
The physical quantity showing how much heat is required to heat 1 kg of a substance by 1 ° C is called the specific heat of this substance.
Each substance has its own specific heat capacity, which is denoted by the Latin letter c and is measured in joules per kilogram-degree (J / (kg K)).
The specific heat capacity of the same substance in different aggregate states (solid, liquid and gaseous) is different. For example, the specific heat capacity of water is 4200 J/(kg K) , and the specific heat capacity of ice J/(kg K) ; aluminum in the solid state has a specific heat capacity of 920 J / (kg K), and in liquid - J / (kg K).
Note that water has a very high specific heat capacity. Therefore, the water in the seas and oceans, heating up in summer, absorbs a large amount of heat from the air. Due to this, in those places that are located near large bodies of water, summer is not as hot as in places far from water.
Specific heat capacity of solids
The table shows the average values of the specific heat capacity of substances in the temperature range from 0 to 10 ° C (if no other temperature is indicated)
| Substance | Specific heat capacity, kJ/(kg K) |
|---|---|
| Solid nitrogen (at t=-250°С) | 0,46
|
| Concrete (at t=20 °С) | 0,88
|
| Paper (at t=20 °С) | 1,50
|
| Solid air (at t=-193 °C) | 2,0
|
| Graphite |
0,75
|
| Oak tree |
2,40
|
| Tree pine, spruce |
2,70
|
| Rock salt |
0,92
|
| A rock |
0,84
|
| Brick (at t=0 °С) | 0,88
|
Specific heat capacity of liquids
| Substance | Temperature, °C | |
|---|---|---|
| Gasoline (B-70) |
20
|
2,05
|
| Water |
1-100
|
4,19
|
| Glycerol |
0-100
|
2,43
|
| Kerosene | 0-100
|
2,09
|
| Machine oil |
0-100
|
1,67
|
| Sunflower oil |
20
|
1,76
|
| Honey |
20
|
2,43
|
| Milk |
20
|
3,94
|
| Oil | 0-100
|
1,67-2,09
|
| Mercury |
0-300
|
0,138
|
| Alcohol |
20
|
2,47
|
| Ether |
18
|
3,34
|
Specific heat capacity of metals and alloys
| Substance | Temperature, °C | Specific heat capacity, k J/(kg K) |
|---|---|---|
| Aluminum |
0-200
|
0,92
|
| Tungsten |
0-1600
|
0,15
|
| Iron |
0-100
|
0,46
|
| Iron |
0-500
|
0,54
|
| Gold |
0-500
|
0,13
|
| Iridium |
0-1000
|
0,15
|
| Magnesium |
0-500
|
1,10
|
| Copper |
0-500
|
0,40
|
| Nickel |
0-300
|
0,50
|
| Tin |
0-200
|
0,23
|
| Platinum |
0-500
|
0,14
|
| Lead |
0-300
|
0,14
|
| Silver |
0-500
|
0,25
|
| Steel |
50-300
|
0,50
|
| Zinc |
0-300
|
0,40
|
| Cast iron |
0-200
|
0,54
|
Specific heat capacity of molten metals and liquefied alloys
| Substance | Temperature, °C | Specific heat capacity, k J/(kg K) |
|---|---|---|
| Nitrogen |
-200,4
|
2,01
|
| Aluminum |
660-1000
|
1,09
|
| Hydrogen |
-257,4
|
7,41
|
| Air |
-193,0
|
1,97
|
| Helium |
-269,0
|
4,19
|
| Gold |
1065-1300
|
0,14
|
| Oxygen |
-200,3
|
1,63
|
| Sodium |
100
|
1,34
|
| Tin |
250
|
0,25
|
| Lead |
327
|
0,16
|
| Silver |
960-1300
|
0,29
|
Specific heat capacity of gases and vapors
at normal atmospheric pressure
| Substance | Temperature, °C | Specific heat capacity, k J/(kg K) |
|---|---|---|
| Nitrogen |
0-200
|
1,0
|
| Hydrogen |
0-200
|
14,2
|
| water vapor |
100-500
|
2,0
|
| Air |
0-400
|
1,0
|
| Helium |
0-600
|
5,2
|
| Oxygen |
20-440
|
0,92
|
| Carbon monoxide(II) |
26-200
|
1,0
|
| Carbon monoxide(IV) | 0-600
|
1,0
|
| Alcohol vapor |
40-100
|
1,2
|
| Chlorine |
13-200
|
0,50
|
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 it wasn't there! You can do an experiment - put an empty saucepan on the fire for a few seconds, just do not burn it, and remember to what temperature it has heated up. And then pour water into the pan of exactly the same weight as the weight of the pan. In theory, the water should heat up to the same temperature as an empty pan in twice the time, since in this case both of them are heated - both the water and the pan.
However, even if you wait three times as long, make sure that the water is still less heated. It takes almost ten times longer for water to heat up to the same temperature as a pot of the same weight. Why is this happening? What stops water from heating up? Why should we waste extra gas to heat 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 with a mass of one kilogram in order for its temperature to increase by one degree Celsius. It is measured in J / (kg * ˚С). This value exists not on a whim, but because of the difference in the properties of various substances.
The specific heat of water is about ten times the specific heat of iron, so the pot will heat up ten times faster than the water in it. Curiously, the specific heat capacity of ice is half that of water. Therefore, ice will heat up twice as fast as water. Melting ice is easier than heating water. As strange as it sounds, it is a fact.
Calculation of the amount of heat
The specific heat capacity is denoted 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, respectively, the final and initial temperatures of the body.
Specific heat 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, 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 the composition, humidity, and atmospheric pressure.
At the same time, with an increase in temperature, the gas increases in volume, and we need to introduce one more value - a 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 values of the specific heat capacity of gases are used depending on various factors and conditions.
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