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- Theory Formula for Calculation of Heat
- Heat of an object

- HEAT BALANCE Equation and Equation for Calculation of THERMAL EMISSION WHEN BURNING FUEL
- The formula for calculating heat capacity
- Heat balance equation
- WHAT IS THERMAL?
- What is specific heat?
- SOME EXERCISES ON THERMAL
- Simple caloric exercise
- PRACTICE

### Theory Formula for Calculation of Heat

#### Heat of an object

A. Core knowledge:

1. The heat of an object to heat up depends on what factors? The amount of heat required by an object to heat up depends on the mass of the object’s temperature rise and the specific heat capacity of the substance.

2. Heat calculation formula:

You are viewing: Heat Calculation Formula

The formula for calculating the heat absorbed Q = m . c . t where: Q is the heat (J), m is the mass of the object (kg), ∆t is the temperature rise of the object (0C or K), c is the specific heat capacity of the substance (J/kg). .K).

In there:

Q is the amount of heat absorbed or released by the object. The unit is Jun (J).

m is the mass of the object, measured in kilograms.

c is the specific heat of the substance, measured in J/kg.K

The specific heat of a substance can indicate the amount of heat required to raise 1 kg of that substance by 1 degree Celsius.

∆t is the temperature change or in other words the temperature change (Celsius or K )

t = t2 – t1

∆t > 0 : heat-radiating object

t heat collector

- For example:

When we say that the calorific value of coal is 5,106 J/kg, it means that when completely burning 1 kg of coal, it will give off a heat amount of 5,106.

3. Specific heat capacity of a substance: Specific heat capacity of a substance indicates the amount of heat required to raise 1 kg of that substance by 10C

Note: In order to calculate the heat Q when there is a heat exchange (with an exothermic object, an endothermic object), when the conservation of energy is recognized, it is necessary to have both the quantities Qradiant and Qthu. But both of these quantities cannot be measured directly, but can only be investigated through intermediate quantities (time, fuel consumption, this knowledge after learning Jun-lenz’s law and the heat capacity of the newly learned HS fuel). The heat Q depends simultaneously on three quantities c, m, ∆t0C. When studying the change of Q, it is necessary to consider the change of each quantity when the remaining quantities are constant. In order to have a deeper understanding of the knowledge learned in everyday life such as: When boiling the same super water to a higher temperature (the greater the temperature rise), the longer it takes, that is, the amount of heat that the water needs to collect is greater. When you boil two kettles of water, you find that a full kettle gains more heat because it takes longer to cook. Heat two equal volumes of cooking oil and water to the same temperature, and the water needs to collect more heat. From that we can see, the amount of heat Q that the object takes to heat up depends on the temperature increase, the mass m and the substance of the object.

### HEAT BALANCE Equation and Equation for Calculation of THERMAL EMISSION WHEN BURNING FUEL

Heat balance equation

Q collection = Q radiated

- Q collected: is the total heat of the objects when collected.
- Q radiated: the total heat of the objects when released.

Formula for calculating the amount of heat released when fuel is burned

Q = qm

In there:

- Q: is the heat released by the object (J).
- q: is the calorific value of the fuel (J/kg)
- m: is the mass of the fuel when completely burned, in kg.

### The formula for calculating heat capacity

- The factors that determine the warming of an object when heat energy is absorbed:

- The amount of heat an object takes in to heat up depends on three factors:

– Mass of an object -> The larger the mass, the greater the amount of heat absorbed by the object.

– The temperature rise of the object -> The greater the temperature rise, the greater the amount of heat absorbed by the object.

– Substances that make up objects.

Formula for calculating heat:

Q = m . c . t

Where: Q is the heat (J),

m is the mass of the object (kg),

∆t is the temperature rise of the object (0C or K),

c is the specific heat capacity of the material (J/kg.K).

- Specific heat capacity of a substance indicates the amount of heat required to increase 1kg of that substance by 10C .

- The heat balance equation and the formula for calculating the heat released when fuel is burned:

#### Heat balance equation

Q absorbed = Q radiated Q is the total heat of the objects in. Q released is the total heat of the objects released._ The formula for calculating the heat released when burning the fuel:

Q = qm

Q is the heat released (J).q is the calorific value of the fuel (J/kg) m is the mass of the completely burned fuel in kg.

#### WHAT IS THERMAL?

Heat is a form of energy stored in matter through the chaotic thermal motion of the particles that make up matter.

The molecules that make up matter are in constant chaotic motion, and thus they have kinetic energy.

Heat energy of an object is the sum of kinetic energies: kinetic energy of motion of the center of mass of the molecule + kinetic energy of oscillations of the atoms that make up the molecule around the common center of mass + rotational kinetic energy of the molecule around the center of mass.

Thermal energy has a close relationship with temperature. Accordingly, the higher the temperature of the object, the greater the thermal energy of the object because the molecules that make up the object move faster.

Heat can be exchanged through processes of radiation, conduction, and convection.

WHAT IS THERMAL?

Heat is understood as the amount of heat energy that an object gains or loses in the process of heat transfer.

The amount of heat absorbed by an object to heat up depends on three factors:

- Mass of an object: The greater the mass of an object, the greater the heat absorbed by the object.
- Temperature rise: If the temperature rise of an object is greater, the amount of heat absorbed by the object is also greater.
- Substances that make up objects.

#### What is specific heat?

Specific heat capacity of a substance plays a very important role in the formula for calculating specific heat & heat.

Specific heat can be understood as the heat required to provide a unit of measurement for that substance. Specifically when used to measure the mass or number of molecules (such as Mol, …). In terms of the international system of units of measurement of physics, then:

- The unit used to measure specific heat is:

Joule/Kilogram/Kelvin

or

Joule/Mol/Kelvin

nice:

J-Kg-1; J/(kg-K).

Specific heat capacity is often used in calorimetry calculations in the process of joining construction materials and for the selection of materials in heat strokes.

For heat: The amount of heat an object needs to collect to serve the heating process depends entirely on the mass of the object, the temperature rise of the object and the specific heat capacity of the material making the object.

+ With the mass of the object: The larger the mass of the object, the greater the heat absorbed by the object (proportional).

+ With the increase of the object: The higher the increase, the greater the amount of heat absorbed by the object (proportional).

In addition, the heat of a substance also depends on the composition of the material…

Table of specific heat capacity of a few substances commonly encountered in student life can be referred to:

Matter (J/kg.K) Matter (J/kg.K) Water 4200,4186,4190 Land 800 Alcohol 2500 Steel 460 Ice 1800 Copper 380 Aluminum 880 Lead 130For example: To heat 1kg of alcohol by 1 degree Celsius, it takes 2500J, then call 2500J the specific heat of alcohol.

#### SOME EXERCISES ON THERMAL

Application exercises

Exercise 1: Calculate the heat required to heat 4 kg of water from 15 degrees Celsius to 100 degrees Celsius in an iron barrel with a mass of 2 kg. The specific heat capacity of water is 4200 J/kg and the specific heat of iron is 460 J/kg.

Exercise 2: There is a 1.8 kg aluminum vessel containing 3 kg of water at a temperature of 30 degrees Celsius. After that, a piece of iron with mass 0.3 kg has been heated to 400 degrees C. Determine the temperature of the water at the beginning of thermal equilibrium. Knowing the specific heat of aluminum is 896 J/kg.K ; of water is 4.18.10^3 J/kg.K ; of iron is 0.46.10^3 J/kg.K.

Exercise 3: Use a charcoal stove to boil 3 liters of water with an initial temperature of 30 degrees Celsius in an aluminum kettle with a mass of 500g. Knowing, the efficiency of the coal stove is 35%, the specific heat of aluminum is 880 J/kg.K ; of water is 4200 J/kg.K ; The heat capacity of coal is 27.10^6 J/kg. Calculate the amount of coal required.

Exercise 4: To determine the temperature of a kiln, a piece of iron is inserted into the furnace with a mass of 22.4g. When the above iron has the same temperature as the furnace, people take it out and immediately drop into it a calorimeter with a mass of 300g containing 450g of water at a temperature of 20 degrees Celsius, the temperature of the water in the calorimeter increase to 23 degrees C. Know the specific heat of iron is 478 J/kg.K ; of the calorimeter is 418 J/kg.K ; of water is 4.18.10^3 J/kg.K. Determine the oven temperature.

Exercise 5: There is a car that can run 200km with a pulling force of 800 N and consume 10 liters of gasoline. The heat capacity of gasoline is known as q = 46.106 J/kg. Calculate the efficiency of the car engine.

Exercise 6: If 100g of lead is transferred with a heat of 270J, the temperature will be increased from 20 degrees Celsius to 30 degrees Celsius. Calculate the heat capacity and specific heat of lead.

#### Simple caloric exercise

Exercise 1: Boil aluminum water with a mass of 0.5 kg, which contains 2 liters of water at 25°C. If you want to boil a kettle of water, how much heat is needed. Knowing that the specific heat of aluminum is 880J/kg.degrees, water is 4200J/kg.degrees (Answer 1: 663 kJ).

Lesson 2: Find the heat required to transfer 5 kg of copper to help change the temperature from 20 degrees Celsius to 50 degrees Celsius.

Solution: Based on the formula for calculating heat Q=mCΔt

Q=5,380.(50−20)=57000(J).

From there, calculate the amount of heat that needs to be transferred to 5kg of copper to increase the temperature from 20 degrees Celsius to 50 degrees Celsius will be:

Q = 57000 (J) Q=57000(J)

Exercise 3: There are 10 liters of water and heat energy of 840J is transferred into the water. So the question is how many degrees Celsius increase in water?

Solution: From the formula for calculating the calorific value Q=mcΔt, t=Q/m/c = 840000/10/4200= 20 degrees Celsius. So transferring 840J into 10 liters of water will increase the water to 20 degrees Celsius.

Lesson 4: There is a kettle made of aluminum with 1 liter of water inside with a water temperature of 20 degrees C. Find the heat required to boil the amount of water in the jar?

Solution: The amount of heat required to boil water is Q = Q warm + Q water = 0.4. 880.80 + 1.4200.80 = 28160 + 336000 = 364160J.

Lesson 5: Find the name of the metal knowing that it must provide 5kg of metal with a heat source of 59kJ to increase this metal from 20 degrees Celsius to 50°c. What is that metal?

Solution: The specific heat of metal is calculated by the formula: c = Q/mt= 59000/5(50-20) = 393J/kg.K. Based on the table of specific heat, we will know that the metal in question is copper.

Exercise 6: Calculate the amount of heat needed to boil 5 kg of water with a temperature change of 15 degrees C to 100 degrees C with an iron barrel of mass 1.5 kg. Knowing the specific heat of water is 4200 J/kg.K, the specific heat of iron is 460 J/kg.K.

Solution: Q = (m1c1 + m2c2)(t2 – t1) = 1843650 J.

Lesson 7: Infuse 100g of lead with a heat source of 260J, then the lead will change its temperature from 15 degrees Celsius to 35 degrees Celsius. Calculate the specific heat capacity of lead?

Solution: Q = mc(t2-t1) = C.(t2 – t1) From this it follows that C = 13J/K and c = 130J/kg.K

#### PRACTICE

Exercise 1: Given an object X of mass m(kg), knowing that the specific heat capacity of the object is C (J/kg.0C) to raise the temperature from t01C−t02C. Calculate the amount of heat to be transferred and the amount of heat released.

Instruct

– Apply the formula: Q=mCΔt

Calculate the amount of heat to be transferred.

– Apply the heat balance equation: Qthu=Qradi to deduce the heat released.

Exercise 2: Given 5kg of copper, calculate the amount of heat transferred by the balance so that the temperature can increase from 200C−500C.

Instruct

Apply: Q=mCΔtQ=mCΔt

Change numbers:

- m = 5kg
- C = 380
- Δt=50−20Δt=50−20

Output: 57000 (J)

Exercise 3: An aluminum super water has a mass of 0.5kg, inside it contains 2kg of water at 250C250C. How much heat is needed to boil the kettle to 750C750C?

Instruct

Calculate the heat transferred to the aluminum to heat up to 750C: Q=mCΔt

Calculate the heat transferred to the hot water to 750C: Q=mCΔt

Calculate the heat required to heat up the whole super water to 750C: Q=Q1+Q2

Exercise 4: Calculate the heat dissipated on the resistor within 30s.

Instruct

Apply the formula: Q=RI2t

[1] How much heat is input to heat 1 aluminum kettle with 1 liter of water with temperature from 300 C to 1000 C. (Use the formula Formula: Q = mc∆T)

[2]If a 200cm3 teapot with a temperature of 950 C is poured into a 250-g porcelain cup at an initial temperature of 250 C. What is the equilibrium temperature of the tea cup after pouring?

Posted by: Le Hong Phong High School

Category: Education

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