The amount of heat required to heat a liquid is formula. Quantity of heat
In practice, thermal calculations are often used. For example, when constructing buildings, it is necessary to take into account what quantity of heat The entire heating system must supply 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, it will take 2 times large quantity heat - 800 J. The temperature of the copper part must be increased not by 1 ° C, but by 260 ° C, which 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 hot water, and the amount of heat received cold water, are equal to each other. 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
- What do you need to know to calculate the amount of heat received by a body when heated?
- 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.
- Write a formula to calculate the amount of heat.
- What conclusion can be drawn from the experiment of mixing cold and hot water? Why are these energies not equal in practice?
Exercise 8
- How much heat is required to heat 0.1 kg of water by 1 °C?
- 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.
- How much heat was released when water with a volume of 20 liters cooled, if the temperature changed from 100 to 50 °C?
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 more high 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. Specific heat 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?
The concept of the amount of heat was formed on early stages development of modern physics, when there were no clear ideas about internal structure substances, what energy is, what forms of energy exist in nature and energy as a form of movement and transformation of matter.
The amount of heat is understood as a physical quantity equivalent to the energy transferred to a material body in the process of heat exchange.
The outdated unit of heat is the calorie, equal to 4.2 J; today this unit is practically not used, and the joule has taken its place.
Initially, it was assumed that the carrier of thermal energy was some completely weightless medium with the properties of a liquid. Numerous physical problems of heat transfer have been and are still being solved based on this premise. The existence of hypothetical caloric was the basis for many essentially correct constructions. It was believed that caloric is released and absorbed in the phenomena of heating and cooling, melting and crystallization. The correct equations for heat transfer processes were obtained based on incorrect physical concepts. There is a known law according to which the amount of heat is directly proportional to the mass of the body participating in heat exchange and the temperature gradient:
Where Q is the amount of heat, m body mass, and the coefficient With– a quantity called specific heat capacity. Specific heat capacity is a characteristic of a substance involved in a process.
Work in thermodynamics
As a result of thermal processes, purely mechanical work can be performed. For example, when a gas heats up, it increases its volume. Let's take a situation like the picture below:
IN in this case mechanical work will be equal to the force of gas pressure on the piston multiplied by the path traveled by the piston under pressure. Of course, this is the simplest case. But even in it one can notice one difficulty: the pressure force will depend on the volume of the gas, which means that we are not dealing with constants, but with variable quantities. Since all three variables: pressure, temperature and volume are related to each other, calculating work becomes significantly more complicated. There are some ideal, infinitely slow processes: isobaric, isothermal, adiabatic and isochoric - for which such calculations can be performed relatively simply. A graph of pressure versus volume is plotted and the work is calculated as an integral of the form.
What will heat up faster on the stove - a kettle or a bucket of water? The answer is obvious - a teapot. Then the second question is why?
The answer is no less obvious - because the mass of water in the kettle is less. Great. And now you can do the real thing yourself physical experience at home. To do this you will need two identical small saucepans, an equal amount of water and vegetable oil, for example, half a liter and a stove. Place saucepans with oil and water on the same heat. Now just watch what will heat up faster. If you have a thermometer for liquids, you can use it; if not, you can simply test the temperature with your finger from time to time, just be careful not to get burned. In any case, you will soon see that the oil heats up significantly faster than water. And one more question, which can also be implemented in the form of experience. Which will boil faster - warm water or cold? Everything is obvious again - the warm one will be first at the finish line. Why all these strange questions and experiments? To determine physical quantity, called the “amount of heat”.
Quantity of heat
The amount of heat is the energy that a body loses or gains during heat transfer. This is clear from the name. When cooling, the body will lose a certain amount of heat, and when heating, it will absorb. And the answers to our questions showed us What does the amount of heat depend on? Firstly, the greater the mass of a body, the greater the amount of heat that must be expended to change its temperature by one degree. Secondly, the amount of heat required to heat a body depends on the substance of which it consists, that is, on the type of substance. And thirdly, the difference in body temperature before and after heat transfer is also important for our calculations. Based on the above, we can determine the amount of heat using the formula:
Q=cm(t_2-t_1) ,
where Q is the amount of heat,
m - body weight,
(t_2-t_1) - difference between initial and final body temperatures,
c is the specific heat capacity of the substance, found from the corresponding tables.
Using this formula, you can calculate the amount of heat that is necessary to heat any body or that this body will release when cooling.
The amount of heat is measured in joules (1 J), like any type of energy. However, this value was introduced not so long ago, and people began measuring the amount of heat much earlier. And they used a unit that is widely used in our time - calorie (1 cal). 1 calorie is the amount of heat required to heat 1 gram of water by 1 degree Celsius. Guided by these data, those who like to count calories in the food they eat can, just for fun, calculate how many liters of water can be boiled with the energy they consume with food during the day.
Exercise 81.
Calculate the amount of heat that will be released during the reduction of Fe 2 O 3 metallic aluminum if 335.1 g of iron was obtained. Answer: 2543.1 kJ.
Solution:
Reaction equation:
= (Al 2 O 3) - (Fe 2 O 3) = -1669.8 -(-822.1) = -847.7 kJ
Calculation of the amount of heat that is released when receiving 335.1 g of iron is made from the proportion:
(2 . 55,85) : -847,7 = 335,1 : X; x = (0847.7 . 335,1)/ (2 . 55.85) = 2543.1 kJ,
where 55.85 atomic mass of iron.
Answer: 2543.1 kJ.
Thermal effect of reaction
Task 82.
Gaseous ethanol C2H5OH can be obtained by the interaction of ethylene C 2 H 4 (g) and water vapor. Write the thermochemical equation for this reaction, having first calculated its thermal effect. Answer: -45.76 kJ.
Solution:
The reaction equation is:
C 2 H 4 (g) + H 2 O (g) = C2H 5 OH (g); = ?
The values of standard heats of formation of substances are given in special tables. Considering that the heats of formation of simple substances are conventionally assumed to be zero. Let's calculate the thermal effect of the reaction using a consequence of Hess's law, we get:
= (C 2 H 5 OH) – [ (C 2 H 4) + (H 2 O)] =
= -235.1 -[(52.28) + (-241.83)] = - 45.76 kJ
Reaction equations in which about the symbols chemical compounds their states of aggregation or crystalline modification are indicated, as well as the numerical value of thermal effects, called thermochemical. IN thermochemical equations, unless specifically stated, the values of thermal effects at constant pressure Q p are indicated equal to the change in enthalpy of the system. The value is usually given on the right side of the equation, separated by a comma or semicolon. The following abbreviated designations for the state of aggregation of a substance are accepted: G- gaseous, and- liquid, To
If heat is released as a result of a reaction, then< О. Учитывая сказанное, составляем термохимическое уравнение данной в примере реакции:
C 2 H 4 (g) + H 2 O (g) = C 2 H 5 OH (g); = - 45.76 kJ.
Answer:- 45.76 kJ.
Task 83.
Calculate the thermal effect of the reduction reaction of iron (II) oxide with hydrogen based on the following thermochemical equations:
a) EO (k) + CO (g) = Fe (k) + CO 2 (g); = -13.18 kJ;
b) CO (g) + 1/2O 2 (g) = CO 2 (g); = -283.0 kJ;
c) H 2 (g) + 1/2O 2 (g) = H 2 O (g); = -241.83 kJ.
Answer: +27.99 kJ.
Solution:
The reaction equation for the reduction of iron (II) oxide with hydrogen has the form:
EeO (k) + H 2 (g) = Fe (k) + H 2 O (g); = ?
= (H2O) – [ (FeO)
The heat of formation of water is given by the equation
H 2 (g) + 1/2O 2 (g) = H 2 O (g); = -241.83 kJ,
and the heat of formation of iron (II) oxide can be calculated by subtracting equation (a) from equation (b).
=(c) - (b) - (a) = -241.83 – [-283.o – (-13.18)] = +27.99 kJ.
Answer:+27.99 kJ.
Task 84.
When gaseous hydrogen sulfide and carbon dioxide interact, water vapor and carbon disulfide CS 2 (g) are formed. Write the thermochemical equation for this reaction and first calculate its thermal effect. Answer: +65.43 kJ.
Solution:
G- gaseous, and- liquid, To-- crystalline. These characters are omitted if state of aggregation substances are obvious, for example, O 2, H 2, etc.
The reaction equation is:
2H 2 S (g) + CO 2 (g) = 2H 2 O (g) + CS 2 (g); = ?
The values of standard heats of formation of substances are given in special tables. Considering that the heats of formation of simple substances are conventionally assumed to be zero. The thermal effect of a reaction can be calculated using a corollary of Hess's law:
= (H 2 O) + (СS 2) – [(H 2 S) + (СO 2)];
= 2(-241.83) + 115.28 – = +65.43 kJ.
2H 2 S (g) + CO 2 (g) = 2H 2 O (g) + CS 2 (g); = +65.43 kJ.
Answer:+65.43 kJ.
Thermochemical reaction equation
Task 85.
Write the thermochemical equation for the reaction between CO (g) and hydrogen, as a result of which CH 4 (g) and H 2 O (g) are formed. How much heat will be released during this reaction if 67.2 liters of methane were obtained in terms of normal conditions? Answer: 618.48 kJ.
Solution:
Reaction equations in which their state of aggregation or crystal modification, as well as the numerical value of thermal effects are indicated next to the symbols of chemical compounds, are called thermochemical. In thermochemical equations, unless specifically stated, the values of thermal effects at constant pressure Q p equal to the change in enthalpy of the system are indicated. The value is usually given on the right side of the equation, separated by a comma or semicolon. The following abbreviated designations for the state of aggregation of a substance are accepted: G- gaseous, and- something, To- crystalline. These symbols are omitted if the aggregative state of the substances is obvious, for example, O 2, H 2, etc.
The reaction equation is:
CO (g) + 3H 2 (g) = CH 4 (g) + H 2 O (g); = ?
The values of standard heats of formation of substances are given in special tables. Considering that the heats of formation of simple substances are conventionally assumed to be zero. The thermal effect of a reaction can be calculated using a corollary of Hess's law:
= (H 2 O) + (CH 4) – (CO)];
= (-241.83) + (-74.84) – (-110.52) = -206.16 kJ.
The thermochemical equation will be:
22,4 : -206,16 = 67,2 : X; x = 67.2 (-206.16)/22?4 = -618.48 kJ; Q = 618.48 kJ.
Answer: 618.48 kJ.
Heat of formation
Task 86.
The thermal effect of which reaction is equal to the heat of formation. Calculate the heat of formation of NO based on the following thermochemical equations:
a) 4NH 3 (g) + 5O 2 (g) = 4NO (g) + 6H 2 O (l); = -1168.80 kJ;
b) 4NH 3 (g) + 3O 2 (g) = 2N 2 (g) + 6H 2 O (l); = -1530.28 kJ
Answer: 90.37 kJ.
Solution:
The standard heat of formation is equal to the heat of reaction of the formation of 1 mole of this substance from simple substances at standard conditions(T = 298 K; p = 1.0325.105 Pa). The formation of NO from simple substances can be represented as follows:
1/2N 2 + 1/2O 2 = NO
Given is reaction (a), which produces 4 mol of NO, and given reaction (b), which produces 2 mol of N2. Oxygen is involved in both reactions. Therefore, to determine the standard heat of formation of NO, we compose the following Hess cycle, i.e., we need to subtract equation (a) from equation (b):
Thus, 1/2N 2 + 1/2O 2 = NO; = +90.37 kJ.
Answer: 618.48 kJ.
Task 87.
Crystalline ammonium chloride is formed by the reaction of ammonia and hydrogen chloride gases. Write the thermochemical equation for this reaction, having first calculated its thermal effect. How much heat will be released if 10 liters of ammonia were consumed in the reaction, calculated under normal conditions? Answer: 78.97 kJ.
Solution:
Reaction equations in which their state of aggregation or crystal modification, as well as the numerical value of thermal effects are indicated next to the symbols of chemical compounds, are called thermochemical. In thermochemical equations, unless specifically stated, the values of thermal effects at constant pressure Q p equal to the change in enthalpy of the system are indicated. The value is usually given on the right side of the equation, separated by a comma or semicolon. The following have been accepted: To-- crystalline. These symbols are omitted if the aggregative state of the substances is obvious, for example, O 2, H 2, etc.
The reaction equation is:
NH 3 (g) + HCl (g) = NH 4 Cl (k). ; = ?
The values of standard heats of formation of substances are given in special tables. Considering that the heats of formation of simple substances are conventionally assumed to be zero. The thermal effect of a reaction can be calculated using a corollary of Hess's law:
= (NH4Cl) – [(NH 3) + (HCl)];
= -315.39 – [-46.19 + (-92.31) = -176.85 kJ.
The thermochemical equation will be:
The heat released during the reaction of 10 liters of ammonia in this reaction is determined from the proportion:
22,4 : -176,85 = 10 : X; x = 10 (-176.85)/22.4 = -78.97 kJ; Q = 78.97 kJ.
Answer: 78.97 kJ.
