Kiến thức về phương trình cân bằng nhiệt Vật lý 8 (có bài tập thực hành)

The heat balance equation is one of the important parts of knowledge in Physics 8, often appearing in exercises and exam questions in the heat section. Therefore, this article helps you understand the nature of the heat balance equation and guides you in solving exercises most effectively.

Principle of heat transfer

There are three characteristics to learn about the principle of heat transfer. Based on the phenomena you observe in life, nature, technology…, the moment two objects exchange heat with each other:

• Heat flows from an object with a higher temperature to an object with a lower temperature

• Heat transfer occurs until the temperatures of the two objects are equal and then stops

• The heat released by one object is equal to the heat absorbed by another object

Understanding this principle, you will no longer wonder whether when you add a drop of boiling water to a bucket of hot water, the drop of water transfers heat to the bucket of water or the bucket of water transfers heat to the drop of water. Also rely on this to easily identify the heat balance point.

What is heat and what is the formula for calculating heat?

To supplement our knowledge of the heat balance equation, we need to review heat and its calculation formula.

Heat is the amount of thermal energy that an object adds or loses during the heat transfer process. Heat gain depends on three factors: Mass of object, increase in temperature of object & substance that makes up object.

=> Formula for heat absorbed by the object:

In there

• Q: Heat (J)

• m: Mass of object (kg)

• ∆t: Object temperature increase (Celsius or K)

• c: Specific heat capacity of material (J/kg.K)

Heat balance equation

The heat balance formula would be:

Q absorbed is the heat absorbed by the object as explained above with the formula Q absorbed = mc∆t

=> Q radiated = mc∆t

Note: These two formulas have the same calculation, but differ in the temperature change

• With Q collected, ∆t = t2 – t1 (t1 is the initial temperature, t2 is the final temperature)

• With Q released, ∆t = t1 – t2 (t1 is the initial temperature, t2 is the final temperature)

Method for solving heat balance exercises

In order for you to solve the heat balance exercises, we follow the steps below:

Step 1: Need to determine which object gives off heat and which object absorbs heat?

Step 2: Write down the formula to calculate the heat emitted by an object

Step 3: Write the formula to calculate the heat absorbed by an object

Step 4: Write the heat balance equation => the quantity to find.

Example: Drop a 0.15 kg aluminum sphere heated to 100°C into a glass of water 20 °C. After some time, the temperature of the sphere and the water are both 25 °C. Calculate the volume of water? Suppose only the ball and the water transfer heat to each other?

Solution instructions: The heat emitted by the sphere is: Qradia = mc∆t = 0.15.880. (100- 25) = 9900 (J)

The heat absorbed by the water is: Qthu = m(water).c(water).∆t = m(water).4200.(25-20) = 21000m

We have Qto = Qcollect => m(water) = Qto/21000 0.5 kg

Calorimetry exercises for grade 8 with answers

Lesson 1: If two objects with different temperatures are placed in contact with each other then:

A. The heat transfer process stops when the temperature of the two objects is the same.

B. The heat transfer process stops when the temperature of an object reaches 0°C.

C. The heat transfer process continues until the thermal energy of the two objects is the same.

D. Heat transfer process until the specific heat capacity of the two objects is the same.

Answer instructions: Heat transfer occurs until the temperature of the two objects is the same and then stops

⇒ Stonep ahnA

Lesson 2: Which of the following is a heat balance equation?

Answer instructions: Heat balance equation: Qradia = Qthu

⇒ Stonep ahnB

Lesson 3: Pour 5 liters of water at 20°C into 3 liters of water at 45°C. The temperature at equilibrium is:

Answer instructions:

Convert: m1 = 5 liters of water = 5 kg, m2 = 3 liters of water = 3 kg, t1 = 20°C, t2 = 45°C

– Call the temperature at equilibrium t

– The heat absorbed by 5 liters of water is: Q1 = m1c.(t – t1)

– The heat absorbed by 3 liters of water is: Q2 = m2c.(t2 – t)

– Applying the heat balance equation we have:

Q1 = Q2 ⇔ m1c.(t – t1) = m2c.(t2 – t)

⇔ m1.(t – t1) = m2.(t2 – t)

⇔ 5.(t – 20) = 3.(45 – t)

⇔ t = 29.375 ≈ 29.4°C

⇒ Stonep ahnD

Lesson 4: Which of the following is true about the principle of heat transfer:

A. Heat spontaneously transfers from an object with a lower temperature to an object with a higher temperature.

B. Heat spontaneously transfers from an object with a higher temperature to an object with a lower temperature.

C. Heat flows from an object with a higher specific heat to an object with a lower specific heat.

D. Heat flows from an object with a lower specific heat to an object with a higher specific heat.

Answer instructions: Heat flows from an object with a higher temperature to an object with a lower temperature.

⇒ Stonep ahnB

Lesson 5: Drop a 2 kg piece of steel at a temperature of 345°C into a jar containing 3 liters of water. After equilibration the final temperature is 30°C. Ignore heat loss through the environment. Know that the specific heat capacities of steel and water are 460 J/kg.K, 4200 J/kg.K, respectively. The initial temperature of the water is:

Answer instructions

Conversion: 3 liters of water = 3 kg

Let the initial temperature of water be t0

– The heat emitted by the steel piece is:

Q1 = m1c1Δt1 = 2.460.(345 – 30) = 289800 J

– The heat absorbed by the water is:

Q2 = m2c2Δt2 = 3.4200.(30 – t0)

– Applying the heat balance equation, we have:

Q1 = Q2 ⇔ 289900 = 3.4200.(30 – t0)

⇒ t0 = 7°C

⇒ Stonep ahnA

Problem 6: Drop a 0.15 kg aluminum sphere heated to 100°C into a beaker of water at 20°C. After a while, the temperature of the sphere and the water are both 25°C. Consider that the sphere and water only transfer heat to each other. We know that the specific heat capacity of aluminum and water is 800 J/kg.K, 4200 J/kg.K. The mass of water is:

Answer instructions

We have:

Aluminum m1 = 0.15kg, c1 = 880J/kg.K, t1 = 1000C

Water: m2 = ?, c2 = 4200J/kg.K, t2 = 200C

Equilibrium temperature t = 25°C

The heat emitted by the aluminum sphere is: Q1 = m1c1(t1 – t)

The heat received by the water is: Q2 = m2c2(t – t2)

Applying the heat balance equation, we have:

Q1 = Q2 ⇔ m1c1(t1 – t) = m2c2(t – t2)

⇔ 0.15.880.(100 – 25) = m2.4200.(25 – 20)

⇔ m2 = 0.471 kg

⇒ Stonep ahnB

Lesson 7: People want to mix bath water with a temperature of 38°C. How many liters of boiling water must be added to 15 liters of cold water at 24°C?

Answer instructions

Conversion: 15 liters of water = 15 kg

The equilibrium temperature of the mixing water is t = 38°C

The heat released by boiling water is: Q1 = m1c(t1 – t)

The heat received by 15 liters of cold water is: Q2 = m2c(t – t2)

Applying the heat balance equation, we have:

Q1 = Q2 ⇔ m1c(t1 – t) = m2c(t – t2)

⇔ m1(t1 – t) = m2(t – t2)

⇔ m1.(100 – 38) = 15.(38 – 24)

⇔ m1 = 3.38 kg

⇒ Stonep ahnB

Lesson 8: People drop three pieces of copper, aluminum, and lead of the same mass into a cup of hot water. Compare the final temperatures of the three metal pieces above:A. The temperature of the lead piece is highest, then the copper piece, then the aluminum piece.

B. The temperature of the copper piece is highest, then the aluminum piece and the lead piece.

C. The temperature of the aluminum piece is highest, then the copper piece and the lead piece.

D. The temperature of the three pieces is equal.

⇒ Stonep ahnD

Problem 9: People drop a piece of copper weighing 0.5 kg into 500 g of water. The copper piece cools from 80°C to 20°C. How many degrees does the water heat up? We know that the specific heat capacity of copper is 380 J/kg.K, and that of water is 4200 J/kg.K.

Answer instructions

The heat that the copper piece radiates is:

Q1 = mcuccu(80 – 20) = 0.5.380.(80 – 20) = 11400 J

The heat received by the water is:

Q2 = mwatercwaterΔt

According to the heat balance equation, we have:

Q1 = Q2 = 11400 J

⇒ Δt = Q2 \ (mwater. cwater ) = 11400 / (0.5.4200) = 5.430C

So the hot water adds 5.43°C

Lesson 10: When you drop an aluminum ball with a mass of 500g into 2kg of water at 25 degrees Celsius, its temperature after thermal equilibrium is 30 degrees Celsius. What is the initial temperature of the aluminum ball? Knowing that the heat loss in this case is 20% of the heat absorbed by the water. Knowing that the specific heat capacity of aluminum is 880 J/kg.K, the specific heat capacity of water is 4200 J/kg.K

Answer instructions

Heat absorbed by water:

Q2 = m2.c2. (t – t2) = 2.4200.(30 – 25) = 42000J

Heat loss:

Qhp= 20%.Q2= 20%.42000 = 8400J

Applying the heat balance equation results in Q1 = Q2 + Qhp

⇔m1.c1.(t1 − t) = 8400 + 42000

⇔0.5.880.(t1− 30)= 50400

Lesson 11: People drop three pieces of copper and lead of the same mass into a cup of hot water. Compare the final temperatures of the three pieces of metal above.

Answer: The temperatures of the three pieces are equal because when three pieces of metal of the same mass are dropped into a cup of hot water, the higher temperature of the cup of water will be transferred to the three pieces of metal. And finally when the temperatures of the three pieces are equal, the heat transfer process will stop.

Lesson 12: A calorimeter contains 2 liters of water at a temperature of 15°C. How many degrees will the water heat up if a 500g brass ball is placed in the calorimeter and heated to 100°C?

Take the specific heat capacity of brass to be 368J/kgK, and that of water to be 4186J/kgK. Ignore the heat transferred to the calorimeter and the outside environment.

Answer: The heat emitted by the copper sphere is:

Q2 = m2.c2.(t2 – t) = 0.5.368.(100 – t)

The heat absorbed by the water is:

Q1 = m1.c1.(t – t1) = 2.4186.(t – 15)

Because the heat released is equal to the heat absorbed, so:

Qthu = Qtoa ↔ Q2 = Q1

↔ 0.5.368.(100 – t) = 2.4186.(t – 15)

Deduce t = 16.83°C

Lesson 13: To have 100 liters of water at a temperature of 35°C, how many liters of boiling water must be poured into how many liters of water at a temperature of 15°C. Take the specific heat capacity of water as 4190J/kg.K.

Answer: Let m1 be the mass of water at 15°C and m2 be the mass of boiling water.

We have: m1 + m2 = 100kg (1)

The heat released in m2 kg of boiling water is:

Q2 = m2.c.(t2 – t) = m2.4190.(100 – 35)

The amount of heat required in m1 kg of water at 15°C to heat up to 35°C is:

Q1 = m1.c.(t – t1) = m1.4190.(100 – 35)

Because the heat released is equal to the heat absorbed, so: Q2 = Q1

m2.4190.(100 – 35) = m1.4190.(100 – 35) (2)

Solving the system of equations between (1) and (2) we get:

m1 = 76.5 kg and m2 = 23.5 kg.

Thus, 23.5 liters of boiling water must be poured into 76.5 liters of water at 15°C to get 100 liters of water at 35°C.

Conclude

The heat balance equation that we know is easy to remember, isn't it? You just need to remember Q emitted = Q received and the formula to calculate Qto & Q collected. However, some exercises do not provide us with all the elements to immediately apply the formula, which requires students to be more flexible in calculations. So we need to spend time doing many different exercises, it will definitely no longer be difficult. timhieulichsuquancaugiay.edu.vn thanks you for following the article and wishes you well in studying this subject.

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