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 one is the specific heat capacity.

However, the interpretation of the word “specific” is usually not given enough attention. Students simply remember it as a given. 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 to the mass, volume or energy. All these values ​​must be taken to be equal to one. Attitude to what is given in the specific heat?

To the product of mass and temperature. Moreover, their values ​​must necessarily be equal to one. That is, the number 1 will be in the divider, but its dimension will combine the kilogram and degree Celsius. This is necessarily taken into account when formulating the definition of specific heat, 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 it is?

The specific heat 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 to spend on this process. And also to calculate its value when cooling the body. By the way, these two quantities of heat are equal to each other in absolute value. But they have different signs. So, in the first case, it is positive, because energy must 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 for heating one kilogram of a substance per degree. In the course of school physics, this degree is taken by the one that is taken on the Celsius scale.

## How to count it?

If you want to find out what the specific heat capacity is equal to, the formula looks like this:

C = Q / (m * (t2   - t1   )), where Q is the amount of heat, m is the mass of a substance, t2   - the temperature that the body has acquired as a result of heat exchange, t1   - the initial temperature of the substance. This is the formula number 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 values ​​from this equation?

First, the amount of heat. The formula will look like this: Q = c * m * (t2   - t1   ). Only in it it is necessary to substitute the values ​​in the units included in the SI. That is, the mass in kilograms, the temperature - in degrees Celsius. This is the formula number 2.

Second, the mass of a substance that cools or heats up. The formula to be: m = Q / (c * (t2   - t1   )). This is formula number 3.

Third, the temperature change Δt = t2   - t1   = (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 temperature of the substance. The formulas valid for heating a substance look like this: t1 = t2 – (Q / c * m), t2 = t1   + (Q / c * m). These formulas have numbers 5 and 6. If the problem deals with the cooling of a substance, then the formulas are: t1 = t2   + (Q / c * m), t2 = t1   - (Q / c * m). These formulas are No. 7 and 8.

## What values ​​can it have?

It has been experimentally established what values ​​it has for each specific substance. Therefore, a special table of specific heat was created. Most often, it gives data that are valid under normal conditions.

## What is the laboratory work on the measurement of specific heat?

In a school physics course, it is determined for a solid. Moreover, its heat capacity is calculated due to the comparison with that which is known. This is most easily done with water.

In the process of doing the work, it is required to measure the initial temperatures of the water and the heated solid. Then lower it into a liquid and wait for thermal equilibrium. The whole experiment is carried out in a calorimeter, so the energy loss can be neglected.

Then you want to write a formula for the amount of heat that gets water when heated from a solid body. The second expression describes the energy that gives the body during cooling. These two values are equal. By mathematical calculations it remains to determine the specific heat of a substance, which is solid.

Most often, it is suggested to compare it with tabular values ​​in order to try to guess which substance the body under study is made of.

## Problem number 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 a metal, if its mass is 100 grams?

Decision.   To find the answer, you will need to use the formula recorded at number 1. All values ​​necessary for calculations are. Only first you need to convert the mass to kilograms, otherwise the answer will be wrong. Because all quantities must be such that are accepted in the SI.

In one kilogram of 1000 grams. So, 100 grams need to be divided into 1000, you get 0.1 kilograms.

The substitution of all values ​​gives the following expression: c = 152 / (0.1 * (24 - 20)). Calculations are not particularly difficult. The result of all actions is the number 380.

Answer:   c = 380 J / (kg * ºС).

## Problem number 2

Condition.   Determine the final temperature to which the water of 5 liters will cool down if it was taken at 100 ºС and released 1680 kJ of heat into the environment.

Decision.   We should start with the fact that energy is given in a non-system unit. Kilojoules need to be translated into joules: 1680 kJ = 1680000 J.

To find the answer, you must use the formula number 8. However, the mass appears in it, but it is unknown in the problem. But given the amount of fluid. So, you can use the formula known as m = ρ * V. The density of water is 1000 kg / m 3. But here the volume will need to be substituted in cubic meters. To transfer them from liters, it is necessary to divide by 1000. Thus, the volume of water is 0.005 m 3.

Substitution of these values into the mass formula gives the expression: 1000 * 0,005 = 5 kg. the Specific heat will need to look in the table. You can now go to the formula 8: t2 = 100 + (1680000 / 4200 * 5).

The first action is supposed to perform multiplication: 4200 * 5. The result is 21000. The second is the division. 1680000. 21000 = 80. The last is the subtraction: 100 - 80 = 20.

## Problem # 3

Condition.   There is a beaker weighing 100 g. 50 g of water is poured into it. The initial temperature of the water with a glass is 0 degrees Celsius. How much heat is needed to bring the water to a boil?

Decision. You should start with in order to introduce a suitable notation. Let data relating to the glass will have an index of 1, and to water — index of 2. It is necessary to find specific heat capacity. Beaker made of laboratory glass, so its value with1 = 840 J / (kg * ºС). Data for water are: with2   = 4200 J / (kg * ºС).

Their masses are given in grams. You want to translate them into pounds. Masses of these substances will be designated as m1   = 0.1 kg, m2   = 0.05 kg.

Initial temperature given: t1   = 0 ºС. On the final it is known that it corresponds to that in which water boils. It's t2   = 100 ºС.

Since the glass is heated with water, the required amount of heat will consist of two. The first that is required to heat the glass (Q1   ), and the second, going to the heating water (Q2   ). Their expression will require a second formula. It must be recorded twice with different indices, and then make up their sum.

It turns out that Q = s1   * m1   * (t2   - t1   ) + c2   * m2   * (t2   - t1   ). Common factor (t2   - t1   ) can be taken out of the bracket so that it is more convenient to count. Then the formula that will be needed to calculate the amount of heat will take the following form: Q = (with1   * m1   + c2   * m2   ) * (t2   - t1   ). Now you can substitute the values ​​known in the problem and count 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.