Calculation of the amount of heat required to heat a body or released by it during cooling. Heat formula

In practice, thermal calculations are often used. For example, when constructing buildings, it is necessary to take into account how much heat the entire heating system should give to the building. You should also know how much heat will escape into the surrounding space through windows, walls, and doors.

We will show with examples how to carry out simple calculations.

So, you need to find out how much heat the copper part received when heated. Its mass was 2 kg, and the temperature increased from 20 to 280 °C. First, using Table 1, we determine the specific heat capacity of copper with m = 400 J / kg °C). This means that heating a copper part weighing 1 kg by 1 °C will require 400 J. To heat a copper part weighing 2 kg by 1 °C, the amount of heat required is 2 times greater - 800 J. The temperature of the copper part must be increased by more than 1 °C, and at 260 °C, this means that 260 times more heat will be required, i.e. 800 J 260 = 208,000 J.

If we denote the mass as m, the difference between the final (t 2) and initial (t 1) temperatures - t 2 - t 1, we obtain a formula for calculating the amount of heat:

Q = cm(t 2 - t 1).

Example 1. An iron cauldron weighing 5 kg is filled with water weighing 10 kg. How much heat must be transferred to the boiler with water to change its temperature from 10 to 100 °C?

When solving the problem, you need to take into account that both bodies - the boiler and the water - will heat up together. Heat exchange occurs between them. Their temperatures can be considered the same, i.e. the temperature of the boiler and water changes by 100 °C - 10 °C = 90 °C. But the amounts of heat received by the boiler and water will not be the same. After all, their masses and specific heat capacities are different.

Heating water in a pot

Example 2. We mixed water weighing 0.8 kg at a temperature of 25 °C and water at a temperature of 100 °C weighing 0.2 kg. The temperature of the resulting mixture was measured, and it turned out to be 40 °C. Calculate how much heat the hot water gave up when cooling and the cold water received when heated. Compare these amounts of heat.

Let's write down the conditions of the problem and solve it.

We see that the amount of heat given off by hot water and the amount of heat received by cold water are equal. This is not a random result. Experience shows that if heat exchange occurs between bodies, then the internal energy of all heating bodies increases by as much as the internal energy of cooling bodies decreases.

When conducting experiments, it usually turns out that the energy given off by hot water is greater than the energy received by cold water. This is explained by the fact that part of the energy is transferred to the surrounding air, and part of the energy is transferred to the vessel in which the water was mixed. The equality of energy given and received will be more accurate, the less energy loss is allowed in the experiment. If you calculate and take into account these losses, the equality will be exact.

Questions

1. What do you need to know to calculate the amount of heat received by a body when heated?
2. Explain with an example how the amount of heat imparted to a body when it is heated or released when it is cooled is calculated.
3. Write a formula to calculate the amount of heat.
4. What conclusion can be drawn from the experiment of mixing cold and hot water? Why are these energies not equal in practice?

Exercise 8

1. How much heat is required to heat 0.1 kg of water by 1 °C?
2. Calculate the amount of heat required to heat: a) a cast iron iron weighing 1.5 kg to change its temperature by 200 °C; b) an aluminum spoon weighing 50 g from 20 to 90 °C; c) a brick fireplace weighing 2 tons from 10 to 40 °C.
3. How much heat was released when water with a volume of 20 liters cooled, if the temperature changed from 100 to 50 °C?

In this lesson we will learn how to calculate the amount of heat required to heat a body or released by it when cooling. To do this, we will summarize the knowledge that was acquired in previous lessons.

In addition, we will learn, using the formula for the amount of heat, to express the remaining quantities from this formula and calculate them, knowing other quantities. An example of a problem with a solution for calculating the amount of heat will also be considered.

This lesson is devoted to calculating the amount of heat when a body is heated or released when cooled.

The ability to calculate the required amount of heat is very important. This may be needed, for example, when calculating the amount of heat that needs to be imparted to water to heat a room.

Rice. 1. The amount of heat that must be imparted to the water to heat the room

Or to calculate the amount of heat that is released when fuel is burned in various engines:

Rice. 2. The amount of heat that is released when fuel is burned in the engine

This knowledge is also needed, for example, to determine the amount of heat that is released by the Sun and falls on the Earth:

Rice. 3. The amount of heat released by the Sun and falling on the Earth

To calculate the amount of heat, you need to know three things (Fig. 4):

• body weight (which can usually be measured using a scale);
• the temperature difference by which a body must be heated or cooled (usually measured using a thermometer);
• specific heat capacity of the body (which can be determined from the table).

Rice. 4. What you need to know to determine

The formula by which the amount of heat is calculated looks like this:

The following quantities appear in this formula:

The amount of heat measured in joules (J);

The specific heat capacity of a substance is measured in ;

- temperature difference, measured in degrees Celsius ().

Let's consider the problem of calculating the amount of heat.

Task

A copper glass with a mass of grams contains water with a volume of liter at a temperature. How much heat must be transferred to a glass of water so that its temperature becomes equal to ?

Rice. 5. Illustration of the problem conditions

First we write down a short condition ( Given) and convert all quantities to the international system (SI).

Solution:

First, determine what other quantities we need to solve this problem. Using the table of specific heat capacity (Table 1) we find (specific heat capacity of copper, since by condition the glass is copper), (specific heat capacity of water, since by condition there is water in the glass). In addition, we know that to calculate the amount of heat we need a mass of water. According to the condition, we are given only the volume. Therefore, from the table we take the density of water: (Table 2).

Table 1. Specific heat capacity of some substances,

Table 2. Densities of some liquids

Now we have everything we need to solve this problem.

Note that the final amount of heat will consist of the sum of the amount of heat required to heat the copper glass and the amount of heat required to heat the water in it:

Let's first calculate the amount of heat required to heat a copper glass:

Before calculating the amount of heat required to heat water, let’s calculate the mass of water using a formula that is familiar to us from grade 7:

Now we can calculate:

Then we can calculate:

Let's remember what kilojoules mean. The prefix "kilo" means .

Answer:.

For the convenience of solving problems of finding the amount of heat (the so-called direct problems) and quantities associated with this concept, you can use the following table.

The focus of our article is the amount of heat. We will consider the concept of internal energy, which is transformed when this quantity changes. We will also show some examples of the use of calculations in human activity.

Heat

Every person has their own associations with any word in their native language. They are determined by personal experience and irrational feelings. What do you usually think of when you hear the word “warmth”? A soft blanket, a working central heating radiator in winter, the first sunlight in spring, a 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 a 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 warmed up or cooled down, this indicates that all the molecules and atoms that make up that system have 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 the 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 experiences. The scientist who writes the methodological manual understands perfectly what is at stake. It seems to him that using the example of gas molecules, all thermodynamic parameters will be properly 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 that 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 - increases energy exchange. However, calculating the amount of heat in all these cases will show a very small, imperceptible value. Therefore, it seems that there are no manifestations of thermodynamic physics in real life.

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. Every 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 in order not to search blindly, hoping to accidentally stumble upon the desired composition, you have to do a lot of calculations, including those that include changes in the amount of heat.

721. Why is water used to cool some mechanisms?
Water has a high specific heat capacity, which facilitates good heat removal from the mechanism.

722. In which case is it necessary to spend more energy: to heat one liter of water by 1 °C or to heat one hundred grams of water by 1 °C?
To heat a liter of water, the greater the mass, the more energy needs to be spent.

723. Cupronickel silver and silver forks of equal mass were lowered into hot water. Will they receive the same amount of heat from the water?
A cupronickel fork will receive more heat because the specific heat of cupronickel is greater than that of silver.

724. A piece of lead and a piece of cast iron of the same mass were hit three times with a sledgehammer. Which piece got hotter?
Lead will heat up more because its specific heat capacity is lower than cast iron and it takes less energy to heat the lead.

725. One flask contains water, the other contains kerosene of the same mass and temperature. An equally heated iron cube was dropped into each flask. What will heat up to a higher temperature - water or kerosene?
Kerosene.

726. Why are temperature fluctuations in winter and summer less sharp in cities on the seashore than in cities located inland?
Water heats up and cools down more slowly than air. In winter, it cools and moves warm air masses onto land, making the climate on the coast warmer.

727. The specific heat capacity of aluminum is 920 J/kg °C. What does this mean?
This means that to heat 1 kg of aluminum by 1 °C it is necessary to spend 920 J.

728. Aluminum and copper bars of the same mass 1 kg are cooled by 1 °C. How much will the internal energy of each block change? For which bar will it change more and by how much?

729. What amount of heat is needed to heat a kilogram of iron billet by 45 °C?

730. What amount of heat is required to heat 0.25 kg of water from 30 °C to 50 °C?

731. How will the internal energy of two liters of water change when heated by 5 °C?

732. What amount of heat is needed to heat 5 g of water from 20 °C to 30 °C?

733. What amount of heat is needed to heat an aluminum ball weighing 0.03 kg by 72 °C?

734. Calculate the amount of heat required to heat 15 kg of copper by 80 °C.

735. Calculate the amount of heat required to heat 5 kg of copper from 10 °C to 200 °C.

736. What amount of heat is required to heat 0.2 kg of water from 15 °C to 20 °C?

737. Water weighing 0.3 kg has cooled by 20 °C. How much has the internal energy of water decreased?

738. What amount of heat is needed to heat 0.4 kg of water at a temperature of 20 °C to a temperature of 30 °C?

739. What amount of heat is expended to heat 2.5 kg of water by 20 °C?

740. What amount of heat was released when 250 g of water cooled from 90 °C to 40 °C?

741. What amount of heat is required to heat 0.015 liters of water by 1 °C?

742. Calculate the amount of heat required to heat a pond with a volume of 300 m3 by 10 °C?

743. What amount of heat must be added to 1 kg of water to increase its temperature from 30 °C to 40 °C?

744. Water with a volume of 10 liters has cooled from a temperature of 100 °C to a temperature of 40 °C. How much heat was released during this?

745. Calculate the amount of heat required to heat 1 m3 of sand by 60 °C.

746. Air volume 60 m3, specific heat capacity 1000 J/kg °C, air density 1.29 kg/m3. How much heat is needed to raise it to 22°C?

747. Water was heated by 10 °C, expending 4.20 103 J of heat. Determine the amount of water.

748. 20.95 kJ of heat was imparted to water weighing 0.5 kg. What did the water temperature become if the initial water temperature was 20 °C?

749. A copper pan weighing 2.5 kg is filled with 8 kg of water at 10 °C. How much heat is needed to heat the water in the pan to a boil?

750. A liter of water at a temperature of 15 °C is poured into a copper ladle weighing 300 g. What amount of heat is needed to heat the water in the ladle to 85 °C?

751. A piece of heated granite weighing 3 kg is placed in water. Granite transfers 12.6 kJ of heat to water, cooling by 10 °C. What is the specific heat capacity of the stone?

752. Hot water at 50 °C was added to 5 kg of water at 12 °C, obtaining a mixture with a temperature of 30 °C. How much water did you add?

753. Water at 20 °C was added to 3 liters of water at 60 °C, obtaining water at 40 °C. How much water did you add?

754. What will be the temperature of the mixture if you mix 600 g of water at 80 °C with 200 g of water at 20 °C?

755. A liter of water at 90 °C was poured into water at 10 °C, and the water temperature became 60 °C. How much cold water was there?

756. Determine how much hot water heated to 60 °C should be poured into a vessel if the vessel already contains 20 liters of cold water at a temperature of 15 °C; the temperature of the mixture should be 40 °C.

757. Determine how much heat is required to heat 425 g of water by 20 °C.

758. How many degrees will 5 kg of water heat up if the water receives 167.2 kJ?

759. How much heat is required to heat m grams of water at temperature t1 to temperature t2?

760. 2 kg of water is poured into a calorimeter at a temperature of 15 °C. To what temperature will the calorimeter water heat up if a 500 g brass weight heated to 100 °C is lowered into it? The specific heat capacity of brass is 0.37 kJ/(kg °C).

761. There are pieces of copper, tin and aluminum of the same volume. Which of these pieces has the largest and which has the smallest heat capacity?

762. 450 g of water, the temperature of which was 20 °C, was poured into the calorimeter. When 200 g of iron filings heated to 100 °C were immersed in this water, the water temperature became 24 °C. Determine the specific heat capacity of sawdust.

763. A copper calorimeter weighing 100 g holds 738 g of water, the temperature of which is 15 °C. 200 g of copper were lowered into this calorimeter at a temperature of 100 °C, after which the temperature of the calorimeter rose to 17 °C. What is the specific heat capacity of copper?

764. A steel ball weighing 10 g is taken out of the oven and placed in water at a temperature of 10 °C. The water temperature rose to 25 °C. What was the temperature of the ball in the oven if the mass of water was 50 g? The specific heat capacity of steel is 0.5 kJ/(kg °C).

770. A steel cutter weighing 2 kg was heated to a temperature of 800 °C and then lowered into a vessel containing 15 liters of water at a temperature of 10 °C. To what temperature will the water in the vessel heat up?

(Indication: To solve this problem, it is necessary to create an equation in which the unknown temperature of the water in the vessel after lowering the cutter is taken as the unknown.)

771. What temperature will the water be obtained if you mix 0.02 kg of water at 15 °C, 0.03 kg of water at 25 °C and 0.01 kg of water at 60 °C?

772. For heating a well-ventilated class, the amount of heat required is 4.19 MJ per hour. Water enters the heating radiators at 80 °C and leaves them at 72 °C. How much water should be supplied to the radiators every hour?

773. Lead weighing 0.1 kg at a temperature of 100 °C was immersed in an aluminum calorimeter weighing 0.04 kg containing 0.24 kg of water at a temperature of 15 °C. After which the temperature in the calorimeter reached 16 °C. What is the specific heat of lead?

To learn how to calculate the amount of heat that is necessary to heat a body, let us first establish on what quantities it depends.

From the previous paragraph we already know that this amount of heat depends on the type of substance of which the body consists (i.e., its specific heat capacity):

Q depends on c.

But that is not all.

If we want to heat the water in the kettle so that it becomes only warm, then we will not heat it for long. And in order for the water to become hot, we will heat it longer. But the longer the kettle is in contact with the heater, the more heat it will receive from it. Consequently, the more the body temperature changes when heated, the greater the amount of heat that needs to be transferred to it.

Let the initial temperature of the body be tbegin, and the final temperature be tend. Then the change in body temperature will be expressed by the difference

Δt = t end – t start,

and the amount of heat will depend on this value:

Q depends on Δt.

Finally, everyone knows that heating, for example, 2 kg of water requires more time (and therefore more heat) than heating 1 kg of water. This means that the amount of heat required to heat a body depends on the mass of that body:

Q depends on m.

So, to calculate the amount of heat, you need to know the specific heat capacity of the substance from which the body is made, the mass of this body and the difference between its final and initial temperatures.

Let, for example, you need to determine how much heat is needed to heat an iron part weighing 5 kg, provided that its initial temperature is 20 °C, and the final temperature should be equal to 620 °C.

From Table 8 we find that the specific heat capacity of iron is c = 460 J/(kg*°C). This means that heating 1 kg of iron by 1 °C requires 460 J.

To heat 5 kg of iron by 1 °C, 5 times more heat will be required, i.e. 460 J * 5 = 2300 J.

To heat iron not by 1 °C, but by Δt = 600 °C, another 600 times more amount of heat will be required, i.e. 2300 J * 600 = 1,380,000 J. Exactly the same (modulo) amount of heat will be released and when this iron cools from 620 to 20 °C.

So, to find the amount of heat required to heat a body or released by it during cooling, you need to multiply the specific heat capacity of the body by its mass and by the difference between its final and initial temperatures:

When the body is heated, tcon > tstart and, therefore, Q > 0. When the body is cooled, tcon< t нач и, следовательно, Q < 0.

1. Give examples showing that the amount of heat received by a body when heated depends on its mass and temperature changes. 2. What formula is used to calculate the amount of heat required to heat a body or released by it when cooling?