About thermal energy in simple language! Quantity of heat. Units of heat quantity
By definition, a calorie is the amount of heat required to heat one cubic centimeter water by 1 degree Celsius. A gigacalorie, used to measure thermal energy in thermal power engineering and utilities, is a billion calories. There are 100 centimeters in 1 meter, therefore, in one cubic meter there are 100 x 100 x 100 = 1000000 centimeters. Thus, to heat a cube of water by
1 degree, a million calories or 0.001 Gcal will be required.
In my city, the price of heating is 1132.22 rubles / Gcal, and the price of hot water is 71.65 rubles / cubic meter, price cold water 16.77 rub/cub.m.
How many Gcal are spent to heat 1 cubic meter of water?
I think so
s x 1132.22 = 71.65 - 16.77 and thus solve the equations to find out what s (Gcal) is equal to, that is, equal to 0.0484711452 Gcal
I have my doubts, I think I'm making the wrong decision
ANSWER:
I don't find any errors in your calculation.
Naturally, the above tariffs should not include the cost of wastewater (sewage).
An approximate calculation for the city of Izhevsk according to the old standards looks like this:
0.19 Gcal per person per month (this norm has now been abolished, but there is no other one, it’s suitable for example) / 3.6 cubic meters. per person per month (hot water consumption rate) = 0.05278 Gcal per 1 cubic meter. (this is how much heat is needed to heat 1 cubic meter of cold water to the standard temperature of hot water, which, let me remind you, is 60 degrees C).
For a more accurate calculation of the amount of thermal energy for heating water using the direct method based on physical quantities(and not the other way based on the approved tariffs for hot water supply) - I recommend using template for calculating the tariff for hot water (REK UR). The calculation formula, among other things, uses the temperature of cold water in the summer and winter (heating) periods and the duration of these periods.
Tags: gigacalorie, hot water
- We pay for hot water services, the temperature is significantly lower than the standard. What to do?
- The duration of the DHW disconnection period established by the Rules is not illegal - decision of the Supreme Court of the Russian Federation (2017)
- Initiative to establish more fair tariffs and methods for metering hot water consumption
- On the procedure for recalculating the amount of payment for heating and hot water during outages - clarification of Rospotrebnadzor for SD
- About accounting for coolant in closed system heating supply - letter of the Ministry of Construction of the Russian Federation dated March 31, 2015 No. 9116-OD/04
- UR - On reducing fees for heating and hot water - letter from the Ministry of Energy of the UR dated 08/17/2015 No. 11-10/5661
- What is the standard period for verification of a common house heating and hot water meter?
- Dirty hot water from the tap. Where to contact?
- Can the water meter in an apartment be increased for the entire entrance? How to pay? Monthly readings - 42 cubic meters
- The procedure for maintaining separate cost accounting in the field of water supply and sanitation - order of the Ministry of Construction of the Russian Federation dated January 25, 2014 No. 22/pr
- payment for water and electricity in an apartment without accommodation
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- Huge payments for a dorm room (17.3 sq.m.)
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“...- How many parrots can fit in you, such is your height.
- I really need it! I won’t swallow so many parrots!...”
From the film “38 Parrots”
According to the international rules of SI (International System of Units), the amount of thermal energy or quantity of heat is measured in Joules [J], and there are also multiple units kiloJoule [kJ] = 1000 J., MegaJoule [MJ] = 1,000,000 J, GigaJoule [ GJ] = 1,000,000,000 J. etc. This unit of measurement of thermal energy is the main international unit and is most often used in scientific and scientific-technical calculations.
However, all of us know or have heard at least once another unit of measurement of the amount of heat (or simply heat) is the calorie, as well as the kilocalorie, Megacalorie and Gigacalorie, which is what the prefixes kilo, Giga and Mega mean, see the example with Joules above. In our country, historically, when calculating tariffs for heating, be it heating with electricity, gas or pellet boilers, it is customary to consider the cost of exactly one Gigacalorie of thermal energy.
So what is Gigacalorie, kiloWatt, kiloWatt*hour or kiloWatt/hour and Joules and how are they related to each other?, you will learn in this article.
So, the basic unit of thermal energy is, as already mentioned, the Joule. But before talking about units of measurement, it is necessary, in principle, to explain at the everyday level what thermal energy is and how and why to measure it.
We all know from childhood that in order to keep warm (get thermal energy) you need to set something on fire, so we all burned fires, the traditional fuel for a fire is wood. Thus, obviously, when burning fuel (any: firewood, coal, pellets, natural gas, diesel fuel) thermal energy (heat) is released. But to heat, for example, different volumes of water are required different quantities firewood (or other fuel). It is clear that to heat two liters of water, a few fires are enough, and to prepare half a bucket of soup for the entire camp, you need to stock up on several bundles of firewood. In order not to measure such strict technical quantities as the amount of heat and the heat of combustion of fuel with bundles of firewood and buckets of soup, heating engineers decided to bring clarity and order and agreed to invent a unit for the amount of heat. In order for this unit to be the same everywhere, it was defined as follows: to heat one kilogram of water by one degree under normal conditions (atmospheric pressure) requires 4,190 calories, or 4.19 kilocalories, therefore, to heat one gram of water a thousand times less heat will be enough – 4.19 calories.
The calorie is related to the international unit of thermal energy, the Joule, by the following relationship:
1 calorie = 4.19 Joules.
Thus, to heat 1 gram of water by one degree, 4.19 Joules of thermal energy will be required, and to heat one kilogram of water, 4,190 Joules of heat will be required.
In technology, along with the unit of measurement of thermal (and any other) energy, there is a unit of power and, in accordance with the international system (SI), this is the Watt. The concept of power also applies to heating devices. If a heating device is capable of delivering 1 Joule of thermal energy in 1 second, then its power is 1 Watt. Power is the ability of a device to produce (create) a certain amount of energy (in our case, thermal energy) per unit of time. Let's return to our example with water, to heat one kilogram (or one liter, in the case of water, a kilogram is equal to a liter) of water by one degree Celsius (or Kelvin, it makes no difference), we need a power of 1 kilocalorie or 4,190 J of thermal energy. To heat one kilogram of water in 1 second of time by 1 degree, we need a device with the following power:
4190 J./1 s. = 4,190 W. or 4.19 kW.
If we want to heat our kilogram of water by 25 degrees in the same second, then we will need twenty-five times more power, i.e.
4.19*25 =104.75 kW.
Thus, we can conclude that the pellet boiler has a capacity of 104.75 kW. heats 1 liter of water by 25 degrees in one second.
Since we got to watts and kilowatts, we should say a word about them. As already mentioned, Watt is a unit of power, including the thermal power of the boiler, but in addition to pellet boilers and gas boilers, humanity is also familiar with electric boilers, the power of which is measured, of course, in the same kilowatts and they consume neither pellets nor gas, and electricity, the amount of which is measured in kilowatt hours. Correct writing energy units kilowatt*hour (namely, kilowatt multiplied by hour, not divided), writing kW/hour is an error!
In electric boilers Electric Energy is converted into heat (the so-called Joule heat), and if the boiler consumed 1 kW*hour of electricity, then how much heat did it produce? To answer this simple question, you need to do a simple calculation.
Let's convert kiloWatts into kiloJoules/seconds (kiloJoule per second), and hours into seconds: there are 3,600 seconds in one hour, we get:
1 kW*hour = [1 kJ/s]*3600 s.=1,000 J *3600 s = 3,600,000 Joules or 3.6 MJ.
So,
1 kW*hour = 3.6 MJ.
In turn, 3.6 MJ/4.19 = 0.859 Mcal = 859 kcal = 859,000 cal. Energy (thermal).
Now let's move on to Gigacalories, the price of which is various types Heating engineers like to count fuels.
1 Gcal = 1,000,000,000 cal.
1,000,000,000 cal. = 4.19*1,000,000,000 = 4,190,000,000 J. = 4,190 MJ. = 4.19 GJ.
Or, knowing that 1 kW*hour = 3.6 MJ, let’s recalculate 1 Gigacalorie per kilowatt*hour:
1 Gcal = 4190 MJ/3.6 MJ = 1,163 kW*hours!
If, after reading this article, you decide to consult with a specialist from our company on any issue related to heat supply, then you Here!
Source: teplo-en.ru
Humanity knows few types of energy - mechanical energy (kinetic and potential), internal energy (thermal), field energy (gravitational, electromagnetic and nuclear), chemical. It is worth highlighting the energy of the explosion...
Vacuum energy and dark energy, which still exists only in theory. In this article, the first in the “Heat Engineering” section, I will try in simple and accessible language, using a practical example, to talk about the most important type of energy in people’s lives - about thermal energy and about giving birth to her in time thermal power.
A few words to understand the place of thermal engineering as a branch of the science of obtaining, transferring and using thermal energy. Modern thermal engineering has emerged from general thermodynamics, which in turn is one of the branches of physics. Thermodynamics is literally “warm” plus “power”. Thus, thermodynamics is the science of the “temperature change” of a system.
An external influence on a system, which changes its internal energy, can be the result of heat exchange. Thermal energy, which is acquired or lost by the system as a result of such interaction with the environment, is called amount of heat and is measured in SI units in Joules.
If you are not a heating engineer and do not deal with thermal engineering issues on a daily basis, then when you encounter them, sometimes without experience it can be very difficult to quickly understand them. Without experience, it is difficult to even imagine the dimensions of the required values of the amount of heat and thermal power. How many Joules of energy are needed to heat 1000 cubic meters of air from a temperature of -37˚С to +18˚С?.. What power of the heat source is needed to do this in 1 hour?.. Today we can answer these not the most difficult questions “immediately” “Not everyone is an engineer. Sometimes specialists even remember the formulas, but only a few can apply them in practice!
After reading this article to the end, you will be able to easily solve real industrial and everyday problems related to heating and cooling of various materials. Understanding the physical essence of heat transfer processes and knowledge of simple basic formulas are the main blocks in the foundation of knowledge in heat engineering!
The amount of heat during various physical processes.
Majority known substances may at different temperatures and pressure to be in solid, liquid, gaseous or plasma states. Transition from one state of aggregation to another occurs at constant temperature(provided that pressure and other environmental parameters do not change) and is accompanied by the absorption or release of thermal energy. Despite the fact that 99% of matter in the Universe is in the plasma state, we will not consider this state of aggregation in this article.
Let's look at the graph shown in the figure. It shows the temperature dependence of a substance T on the amount of heat Q, brought to a certain closed system containing a certain mass of a specific substance.
1. A solid with a temperature T1, heat to temperature Tmel, spending on this process an amount of heat equal to Q1 .
2. Next, the melting process begins, which occurs at a constant temperature Tpl(melting point). To melt the entire mass of a solid, it is necessary to expend thermal energy in the amount Q2 - Q1 .
3. Next, the liquid resulting from the melting of the solid is heated to the boiling point (gas formation) Tkp, spending on this amount of heat equal Q3-Q2 .
4. Now at a constant boiling point Tkp the liquid boils and evaporates, turning into a gas. To transform the entire mass of liquid into gas, it is necessary to expend thermal energy in the amount Q4-Q3.
5. At the last stage, the gas is heated from temperature Tkp up to a certain temperature T2. In this case, the amount of heat consumed will be Q5-Q4. (If we heat the gas to the ionization temperature, the gas will turn into plasma.)
Thus, heating the original solid on temperature T1 up to temperature T2 we spent thermal energy in the amount Q5, transferring a substance through three states of aggregation.
Moving in the opposite direction, we will remove the same amount of heat from the substance Q5, having gone through the stages of condensation, crystallization and cooling from temperature T2 up to temperature T1. Of course we are considering closed system without energy loss to the external environment.
Note that a transition from the solid state to the gaseous state is possible, bypassing the liquid phase. This process is called sublimation, and the reverse process is called desublimation.
So, we realized that the processes of transitions between aggregate states of matter are characterized by energy consumption at a constant temperature. When heating a substance that is in one unchanged state of aggregation, the temperature rises and thermal energy is also consumed.
Main heat transfer formulas.
The formulas are very simple.
Quantity of heat Q in J is calculated using the formulas:
1. From the heat consumption side, that is, from the load side:
1.1. When heating (cooling):
Q = m * c *(T2 -T1)
m – mass of substance in kg
With - specific heat capacity of a substance in J/(kg*K)
1.2. When melting (freezing):
Q = m * λ
λ – specific heat of melting and crystallization of a substance in J/kg
1.3. During boiling, evaporation (condensation):
Q = m * r
r – specific heat of gas formation and condensation of a substance in J/kg
2. From the heat production side, that is, from the source side:
2.1. When fuel burns:
Q = m * q
q – specific heat of combustion of fuel in J/kg
2.2. When converting electricity into thermal energy (Joule-Lenz law):
Q =t *I *U =t *R *I ^2=(t /R)*U^2
t – time in s
I – effective current value in A
U – effective voltage value in V
R – Load resistance in ohms
We conclude that the amount of heat is directly proportional to the mass of the substance during all phase transformations and, during heating, additionally directly proportional to the temperature difference. Proportionality coefficients ( c , λ , r , q ) for each substance they have their own meanings and are determined empirically (taken from reference books).
Thermal power N in W is the amount of heat transferred to the system in a certain time:
N=Q/t
The faster we want to heat the body to a certain temperature, the greater the power the source of thermal energy should be - everything is logical.
Calculation of an applied problem in Excel.
In life, it is often necessary to make a quick assessment calculation in order to understand whether it makes sense to continue studying a topic, doing a project and detailed, accurate, time-consuming calculations. Having made a calculation in a few minutes even with an accuracy of ±30%, you can accept the important management decision, which will be 100 times cheaper and 1000 times more efficient and ultimately 100,000 times more effective than performing an accurate calculation within a week, or even a month, by a group of expensive specialists...
Conditions of the problem:
We bring 3 tons of rolled metal from a warehouse on the street to the premises of the rolled metal preparation workshop with dimensions of 24m x 15m x 7m. There is ice on the metal rolling stock total mass 20kg. It's -37˚С outside. How much heat is needed to heat the metal to +18˚С; heat the ice, melt it and heat the water to +18˚С; heat the entire volume of air in the room, assuming that the heating was completely turned off before? What power should the heating system have if all of the above must be completed in 1 hour? (Very harsh and almost unrealistic conditions - especially regarding air!)
We will perform the calculation in the programMS Excel or in the programOOo Calc.
Check out the color formatting of cells and fonts on the “” page.
Initial data:
1. We write the names of the substances:
to cell D3: Steel
to cell E3: Ice
to cell F3: Ice/water
to cell G3: Water
to cell G3: Air
2. We enter the names of the processes:
to cells D4, E4, G4, G4: heat
to cell F4: melting
3. Specific heat capacity of substances c in J/(kg*K) we write for steel, ice, water and air, respectively
to cell D5: 460
to cell E5: 2110
to cell G5: 4190
to cell H5: 1005
4. Specific heat melting ice λ enter in J/kg
to cell F6: 330000
5. A lot of substances m We enter in kg respectively for steel and ice
to cell D7: 3000
to cell E7: 20
Since the mass does not change when ice turns into water, then
in cells F7 and G7: =E7 =20
We find the mass of air by multiplying the volume of the room by the specific gravity
in cell H7: =24*15*7*1.23 =3100
6. Process time t per minute we write only once for steel
to cell D8: 60
The time values for heating the ice, melting it and heating the resulting water are calculated from the condition that all these three processes must be completed in the same amount of time as allotted for heating the metal. Read accordingly
in cell E8: =E12/(($E$12+$F$12+$G$12)/D8) =9,7
in cell F8: =F12/(($E$12+$F$12+$G$12)/D8) =41,0
in cell G8: =G12/(($E$12+$F$12+$G$12)/D8) =9,4
The air should also warm up during the same allotted time, we read
in cell H8: =D8 =60,0
7. The initial temperature of all substances T1 We put it in ˚C
to cell D9: -37
to cell E9: -37
to cell F9: 0
to cell G9: 0
to cell H9: -37
8. The final temperature of all substances T2 We put it in ˚C
to cell D10: 18
to cell E10: 0
to cell F10: 0
to cell G10: 18
to cell H10: 18
I think there shouldn’t be any questions regarding items 7 and 8.
Calculation results:
9. Quantity of heat Q in KJ, required for each of the processes, we calculate
for heating steel in cell D12: =D7*D5*(D10-D9)/1000 =75900
for heating ice in cell E12: =E7*E5*(E10-E9)/1000 = 1561
for melting ice in cell F12: =F7*F6/1000 = 6600
for heating water in cell G12: =G7*G5*(G10-G9)/1000 = 1508
for heating air in cell H12: =H7*H5*(H10-H9)/1000 = 171330
We read out the total amount of thermal energy required for all processes
in merged cell D13E13F13G13H13: =SUM(D12:H12) = 256900
In cells D14, E14, F14, G14, H14, and the combined cell D15E15F15G15H15, the amount of heat is given in an arc unit of measurement - in Gcal (in gigacalories).
10. Thermal power N in kW required for each of the processes is calculated
for heating steel in cell D16: =D12/(D8*60) =21,083
for heating ice in cell E16: =E12/(E8*60) = 2,686
for melting ice in cell F16: =F12/(F8*60) = 2,686
for heating water in cell G16: =G12/(G8*60) = 2,686
for heating air in cell H16: =H12/(H8*60) = 47,592
The total thermal power required to complete all processes in time t calculated
in the merged cell D17E17F17G17H17: =D13/(D8*60) = 71,361
In cells D18, E18, F18, G18, H18, and the combined cell D19E19F19G19H19, the thermal power is given in an arc unit of measurement - in Gcal/hour.
This completes the calculation in Excel.
Conclusions:
Please note that heating air requires more than twice as much energy as heating the same mass of steel.
Heating water costs twice as much energy as heating ice. The melting process consumes many times more energy than the heating process (at a small temperature difference).
Heating water requires ten times more thermal energy than heating steel and four times more than heating air.
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We recalled the concepts of “amount of heat” and “thermal power”, examined the fundamental formulas of heat transfer, and analyzed a practical example. I hope that my language was simple, clear and interesting.
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730. Why is water used to cool some mechanisms?
Water has a high specific heat capacity, which facilitates good heat removal from the mechanism.
731. 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 you need to expend.
732. 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.
733. 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.
734. 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 more high temperature– water or kerosene?
Kerosene.
735. 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.
736. 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.
737. 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?
738. What amount of heat is needed to heat a kilogram of iron billet by 45 °C?
739. What amount of heat is required to heat 0.25 kg of water from 30 °C to 50 °C?
740. How will the internal energy of two liters of water change when heated by 5 °C?
741. What amount of heat is needed to heat 5 g of water from 20 °C to 30 °C?
742. What amount of heat is needed to heat an aluminum ball weighing 0.03 kg by 72 °C?
743. Calculate the amount of heat required to heat 15 kg of copper by 80 °C.
744. Calculate the amount of heat required to heat 5 kg of copper from 10 °C to 200 °C.
745. What amount of heat is required to heat 0.2 kg of water from 15 °C to 20 °C?
746. Water weighing 0.3 kg has cooled by 20 °C. How much has the internal energy of water decreased?
747. 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?
748. What amount of heat is expended to heat 2.5 kg of water by 20 °C?
749. What amount of heat was released when 250 g of water cooled from 90 °C to 40 °C?
750. What amount of heat is required to heat 0.015 liters of water by 1 °C?
751. Calculate the amount of heat required to heat a pond with a volume of 300 m3 by 10 °C?
752. What amount of heat must be added to 1 kg of water to increase its temperature from 30 °C to 40 °C?
753. 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?
754. Calculate the amount of heat required to heat 1 m3 of sand by 60 °C.
755. 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?
756. Water was heated by 10 °C, expending 4.20 103 J of heat. Determine the amount of water.
757. 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?
758. 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?
759. 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?
760. 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?
761. 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?
762. 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?
763. 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?
764. 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?
765. 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.
766. Determine how much heat is required to heat 425 g of water by 20 °C.
767. How many degrees will 5 kg of water heat up if the water receives 167.2 kJ?
768. How much heat is required to heat m grams of water at temperature t1 to temperature t2?
769. 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).
770. 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?
771. 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.
772. 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?
773. 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).
776. Water weighing 0.95 g at a temperature of 80 °C was mixed with water weighing 0.15 g at a temperature of 15 °C. Determine the temperature of the mixture. 779. 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.)
780. What temperature will the water get 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?
781. 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?
782. 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?
(or heat transfer).
Specific heat capacity of a substance.
Heat capacity- this is the amount of heat absorbed by a body when heated by 1 degree.
The heat capacity of a body is indicated by a capital Latin letter WITH.
What does the heat capacity of a body depend on? 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 type of substance? Let's do an experiment. Let's take two identical vessels and, having poured water weighing 400 g into one of them, and vegetable oil weighing 400 g into the other, we will begin to heat them using identical burners. By observing the thermometer readings, 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, heating the same mass of different substances to the same temperature requires different amounts of heat. The amount of heat required to heat a body and, therefore, its heat capacity depend on the type of substance of which the body is composed.
So, for example, to increase the temperature of water weighing 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.
A physical quantity showing how much heat is required to heat 1 kg of a substance by 1 ºС is called specific heat capacity of this substance.
Each substance has its own specific heat capacity, which is denoted by the Latin letter c and measured in joules per kilogram degree (J/(kg °C)).
Specific heat capacity of the same substance in different states of aggregation(solid, liquid and gaseous) is different. For example, the specific heat capacity of water is 4200 J/(kg °C), and the specific heat capacity of ice is 2100 J/(kg °C); aluminum in the solid state has a specific heat capacity of 920 J/(kg - °C), and in the liquid state - 1080 J/(kg - °C).
Note that water has a very high specific heat capacity. Therefore, water in the seas and oceans, heating up in summer, absorbs a large amount of heat from the air. Thanks to this, in those places that are located near large bodies of water, summer is not as hot as in places far from the water.
Calculation of the amount of heat required to heat a body or released by it during cooling.
From the above it is clear that the amount of heat required to heat a 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 body temperature.
So, to determine 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:
Q = cm (t 2 - t 1 ) ,
Where Q- quantity of heat, c— specific heat capacity, m- body mass , t 1 — initial temperature, t 2 — final temperature.
When the body heats up t 2 > t 1 and therefore Q > 0 . When the body cools down t 2i< t 1 and therefore Q< 0 .
If the heat capacity of the entire body is known WITH, Q determined by the formula:
Q = C (t 2 - t 1 ) .