Learn
Games

10.12 - EXERCISES

Interactive Audio Lesson

Listen to a student-teacher conversation explaining the topic in a relatable way.

Triple Points and Temperature Conversions

Unlock Audio Lesson

Signup and Enroll to the course for listening the Audio Lesson

Teacher
Teacher

Today we're exploring the triple points of substances like neon at 24.57 K and carbon dioxide at 216.55 K. Can anyone tell me how to convert these temperatures to Celsius and Fahrenheit?

Student 1
Student 1

I think we can use the conversion formulas for that. What are those formulas again?

Teacher
Teacher

Great question, Student_1! The formula to convert Kelvin to Celsius is T(°C) = T(K) - 273.15. For Fahrenheit, we use T(°F) = T(°C) × 9/5 + 32. So for neon, that would be -248.58°C or -415.4°F and for carbon dioxide, -56.6°C or -69.9°F.

Student 2
Student 2

Why is the triple point of water such an important reference point in thermometry?

Teacher
Teacher

Excellent point, Student_2! The triple point of water is critical because it's a fixed point that can be reproduced under controlled conditions. It’s more reliable than the melting and boiling points.

Student 3
Student 3

So, does every substance have a triple point?

Teacher
Teacher

Exactly, Student_3! Every pure substance does, and it represents the temperature and pressure at which all three phases of the substance exist in equilibrium.

Student 4
Student 4

Could you give us an example of a calculation?

Teacher
Teacher

Sure! Let's calculate the temperatures for neon and carbon dioxide together now.

Thermal Expansion Calculations

Unlock Audio Lesson

Signup and Enroll to the course for listening the Audio Lesson

Teacher
Teacher

Let's move on to thermal expansion! Why does a steel tape calibrated at 27°C measure differently on a hot day? Who remembers the coefficient of linear expansion for steel?

Student 1
Student 1

It's 1.20 × 10–5 K–1, right?

Teacher
Teacher

Exactly! Let’s calculate the actual length of a steel rod measured at 63.0 cm on a hot day at 45°C. Can anyone set up the equation for me?

Student 2
Student 2

We could use the formula ΔL = L0 * α * ΔT, where ΔL is the change in length.

Teacher
Teacher

Very good, Student_2! Now, plugging in the numbers, let's solve for the actual length on that day.

Student 3
Student 3

Does this apply to all materials?

Teacher
Teacher

This principle applies to many materials, but coefficients of expansion vary for different substances.

Student 4
Student 4

What happens if the material cools down again?

Teacher
Teacher

It will revert to its initial dimensions as it contracts, following the same formula!

Student 2
Student 2

That makes sense! Thanks!

Heat Transfer Calculations

Unlock Audio Lesson

Signup and Enroll to the course for listening the Audio Lesson

Teacher
Teacher

Let's explore heat transfer in melting ice. If we have a copper block of mass 2.5 kg at 500°C, how do we determine the maximum amount of ice it can melt?

Student 1
Student 1

Do we need to know the specific heat of copper?

Teacher
Teacher

Indeed! The specific heat of copper is 0.39 J g–1 K–1, and the heat of fusion for water is 335 J g–1. First, let's calculate the heat lost by the copper block as it cools to 0°C.

Student 2
Student 2

And then we would convert that heat into how much ice can be melted?

Teacher
Teacher

That's right, Student_2! The equation Q = mcΔT will help us find the heat transferred.

Student 3
Student 3

Is there a specific way to calculate the mass of ice melted?

Teacher
Teacher

Yes! After finding the heat transferred, we can use the heat of fusion to determine how much ice is melted. Let’s go through the calculations together!

Student 4
Student 4

Can we use that for other types of materials too?

Teacher
Teacher

Yes, the principles apply broadly, but you'll need to adjust the specific heat values for different materials.

Introduction & Overview

Read a summary of the section's main ideas. Choose from Basic, Medium, or Detailed.

Quick Overview

This section focuses on various exercises related to thermodynamic concepts, including calculations of temperature conversions, properties of specific fixed points, and practical applications of principles of heat transfer and expansion.

Standard

The exercises in this section challenge students to compute various thermodynamic values, such as temperature conversions across scales, properties of substances at different temperatures, and effects of thermal expansion. They employ mathematical reasoning to solve practical problems based on heat transfer and specific heat capacities.

Detailed

Detailed Summary

In this section, we explore a series of exercises that test understanding of thermal properties and behaviors of substances. Key exercises include:

Temperature Conversions

  • Triple Points: Students will calculate the triple points of substances like neon and carbon dioxide in Celsius and Fahrenheit scales.
  • Absolute Temperatures: The relationship between different absolute temperature scales is examined, specifically looking at the significance of the triple point of water.

Resistance and Temperature Relationships

  • The resistance of thermometers at different temperatures is analyzed, prompting calculations based on resistance values and temperature coefficients.

Thermal Properties and Thermal Expansion

  • Exercises focus on practical applications, such as measuring lengths affected by temperature changes, the expansion of materials, and the calculations of melting ice based on heat transfer from a heated object.

Energy Calculations

  • Students are tasked with calculating the maximum amount of ice that can be melted by a heated copper block, emphasizing the principles of specific heat and latent heat of fusion.

Through these exercises, students gain hands-on experience with critical thermal principles, preparing them for more advanced studies in thermodynamics.

Youtube Videos

Class 11th Physics Chapter 10 | Exercise Questions (10.1 to 10.20) | Thermal Properties of Matter
Class 11th Physics Chapter 10 | Exercise Questions (10.1 to 10.20) | Thermal Properties of Matter
11th Physics NCERT Solutions Oneshot | Chapter 11 Thermal Properties of Matter | Vikrant Kirar
11th Physics NCERT Solutions Oneshot | Chapter 11 Thermal Properties of Matter | Vikrant Kirar
Thermal Properties of Matter Class 11 Physics | Revised NCERT Solutions | Chapter 10 Questions 1-20
Thermal Properties of Matter Class 11 Physics | Revised NCERT Solutions | Chapter 10 Questions 1-20
Thermal Properties of Matter - NCERT Solutions (Q. 1 to 10) | Class 11 Physics Ch 10 | CBSE 2024-25
Thermal Properties of Matter - NCERT Solutions (Q. 1 to 10) | Class 11 Physics Ch 10 | CBSE 2024-25
Thermal Properties of Matter - NCERT Solutions (Q. 11 to 20) | Class 11 Physics Ch 10 | CBSE 2024-25
Thermal Properties of Matter - NCERT Solutions (Q. 11 to 20) | Class 11 Physics Ch 10 | CBSE 2024-25
class 11 Thermal properties of matter. part 1
class 11 Thermal properties of matter. part 1
Thermal Properties of Matter | NCERT Exercise | Physics | Class 11 #thermalpropertiesofmatter
Thermal Properties of Matter | NCERT Exercise | Physics | Class 11 #thermalpropertiesofmatter

Audio Book

Dive deep into the subject with an immersive audiobook experience.

Heating the Copper Block

Unlock Audio Book

Signup and Enroll to the course for listening the Audio Book

A copper block of mass 2.5 kg is heated in a furnace to a temperature of 500 °C and then placed on a large ice block.

Detailed Explanation

In this chunk, we start by understanding the scenario where a copper block, which has been heated, is placed on ice. We need to analyze how much heat will be transferred from the copper block to the ice, which can lead to the melting of some ice. The specific heat of copper, represented in joules per gram per degree Celsius (J/g°C), is an important factor in this process because it determines how much energy the copper block has when at the given temperature. We will calculate the energy lost by the copper block as it cools down, and this energy will be used to determine how much ice melts.

Examples & Analogies

Imagine you have a hot cup of coffee. When you place it next to a block of ice, the heat from the coffee will gradually transfer to the ice, causing some of it to melt. Similarly, the hot copper block will transfer heat to the ice, and by calculating this transfer of energy, we can figure out how much ice can actually melt.

Calculating Ice Melting

Unlock Audio Book

Signup and Enroll to the course for listening the Audio Book

What is the maximum amount of ice that can melt? (Specific heat of copper = 0.39 J g–1 K–1; heat of fusion of water = 335 J g–1).

Detailed Explanation

To find out how much ice can melt, we use the formula Q = mcΔT for the copper block, where Q is the heat lost, m is the mass, c is the specific heat, and ΔT is the change in temperature. The change in temperature is the difference between the initial temperature of the copper block (500 °C) and the final temperature (0 °C) when it has transferred all its heat to the ice. Using the heat of fusion of water, which is the amount of heat energy required to melt ice, we calculate how much of the ice melts when it absorbs the heat from the copper.

Examples & Analogies

Think of a campfire where hot coals are placed nearby a snow block. The heat from the coals will cause some of the snow to melt. The hotter the coals, and the longer they are allowed to sit next to the snow, the more snow will melt. This is similar to our copper block; its high temperature allows it to melt a maximum amount of ice, much like the coals with the snow.

Definitions & Key Concepts

Learn essential terms and foundational ideas that form the basis of the topic.

Key Concepts

  • Triple Point: A crucial point in thermodynamics where a substance can exist in solid, liquid, and gas phases simultaneously.

  • Specific Heat: Defines how much heat energy a substance can store, vital for understanding heat transfer.

  • Thermal Expansion: Indicates how materials change in size with temperature, affecting many engineering and scientific applications.

  • Heat Transfer: The movement of thermal energy from one object to another, essential for understanding processes like melting and boiling.

Examples & Real-Life Applications

See how the concepts apply in real-world scenarios to understand their practical implications.

Examples

  • Calculating the temperature of the triple point of water in Celsius and Fahrenheit yields 0 °C and 32 °F respectively.

  • Finding the actual length of a steel rod that is different from its calibrated size due to temperature changes, e.g., measuring 63.0 cm at 45 °C while knowing it was calibrated at 27 °C.

Memory Aids

Use mnemonics, acronyms, or visual cues to help remember key information more easily.

🎵 Rhymes Time

  • In the hot sun, materials expand, warmth flows through this land.

📖 Fascinating Stories

  • Once, a copper pot sat on a warm stove. As it heated, it expanded, just like how a balloon grows when filled with air!

🧠 Other Memory Gems

  • S.H.E. for Specific Heat, Heat of fusion, and Expansion coefficients to remember key thermal concepts!

🎯 Super Acronyms

T.C.E. for Triple Points, Coefficient of Expansion, and Energy Transfer.

Flash Cards

Review key concepts with flashcards.

Glossary of Terms

Review the Definitions for terms.

  • Term: Triple Point

    Definition:

    The unique set of conditions at which a substance can coexist in all three phases: solid, liquid, and gas.

  • Term: Specific Heat

    Definition:

    The amount of heat required to raise the temperature of one gram of a substance by one degree Celsius.

  • Term: Coefficient of Linear Expansion

    Definition:

    A measure of how much a material expands per unit of temperature increase.

  • Term: Heat of Fusion

    Definition:

    The amount of energy required to change a substance from solid to liquid at its melting point.

  • Term: Thermal Equilibrium

    Definition:

    The state in which two bodies in contact do not exchange heat, implying they are at the same temperature.