teaching-notes — Physics (THERMAL PHYSICS)

Topic: THERMAL PHYSICS

Subtopic: Measurement of Temperature

MEANING OF TEMPERATURE

In our everyday life, we use the words "hot" and "cold" to describe the state of objects. However, in Physics, we need a precise and scientific way to measure how hot or cold an object is. This measurement is what we call temperature.

Temperature is defined as the measure of the average kinetic energy of the particles (atoms or molecules) of a substance. It determines the direction of net thermal energy (heat) flow between two bodies in thermal contact. Heat always flows spontaneously from a body at a higher temperature to a body at a lower temperature.

To understand temperature at the microscopic level, we must look at the kinetic theory of matter. All matter is made up of tiny particles that are constantly in motion. Because these particles are moving, they possess kinetic energy.

• When a substance is heated, its particles absorb thermal energy. This causes them to move faster (in liquids and gases) or vibrate more vigorously (in solids). As their speed increases, their kinetic energy increases. Consequently, the temperature of the substance rises.
• When a substance cools down, its particles lose energy and slow down. Their average kinetic energy decreases, and the temperature of the substance drops.

Therefore, temperature is directly related to the average speed of the particles. It is important to note that temperature does not depend on the number of particles in a substance, but on their average kinetic energy. For example, a small cup of boiling water at 100 °C has the same temperature as a large pot of boiling water at 100 °C, even though the large pot contains far more total thermal energy because it has more particles.

PARTICLE MOTION AND TEMPERATURE

Example:

Consider two copper blocks, Block A and Block B. Block A is at 80 °C and Block B is at 30 °C. The particles in Block A are vibrating with greater average kinetic energy than the particles in Block B. If the two blocks are placed in contact, thermal energy will flow from Block A to Block B until they reach the same temperature (thermal equilibrium).

PHYSICAL PROPERTIES THAT CHANGE WITH TEMPERATURE

To measure temperature, we make use of physical properties of substances that change continuously, measurably, and uniquely with temperature. A physical property that changes with temperature is called a thermometric property. The substance that exhibits this property is called a thermometric substance.

The following are the main physical properties that change with temperature and are used to construct different types of thermometers:

1. Volume of a Liquid (Thermal Expansion)

Most liquids expand (increase in volume) when heated and contract (decrease in volume) when cooled. This is the most common property used in school laboratories. The thermometric substances used are usually mercury or coloured alcohol. As the temperature rises, the liquid expands and rises up a narrow capillary tube.

2. Electrical Resistance of a Metal

The electrical resistance of a metallic conductor (such as a platinum wire) changes when its temperature changes. For most metals, electrical resistance increases as the temperature increases because the vibrating metal atoms offer more resistance to the flow of free electrons. This property is used in resistance thermometers, which are highly accurate over wide temperature ranges.

3. Electromotive Force (e.m.f.)

When two wires of different metals (such as copper and constantan) are joined together to form a circuit, and the two junctions are kept at different temperatures, an electrical voltage or electromotive force (e.m.f.) is generated between them. This e.m.f. changes with the temperature difference between the junctions. This device is called a thermocouple. It is used to measure rapidly changing temperatures and very high temperatures, such as inside industrial furnaces.

4. Gas Pressure at Constant Volume

When a fixed mass of gas is kept at a constant volume, its pressure increases as the temperature increases. This is because the gas particles move faster and collide with the walls of the container more frequently and with greater force. This property is used in constant-volume gas thermometers.

5. Density of a Substance

Since heating causes most substances to expand in volume while their mass remains constant, their density changes with temperature. Density is defined as mass per unit volume. As temperature increases, volume increases, which causes the density of the substance to decrease.

Formula:

ρ = mV

Where:

• ρ (rho) = density of the substance
• m = mass of the substance
• V = volume of the substance

Worked Example on Density Change:

A copper rod has a mass of 89.0 g. At 20 °C, its volume is 10.0 cm³. When heated to 300 °C, the rod expands and its volume becomes 10.2 cm³. Calculate the density of the copper rod at (a) 20 °C and (b) 300 °C, and describe how the density has changed.

Solution:

(a) At 20 °C:

Formula: ρ₁ = mV₁

Substitution: ρ₁ = 89.0 g10.0 cm³

Answer: ρ₁ = 8.90 g cm^-3

(b) At 300 °C:

Formula: ρ₂ = mV₂

Substitution: ρ₂ = 89.0 g10.2 cm³

Answer: ρ₂ = 8.73 g cm^-3

Observation: The density of the copper rod decreased from 8.90 g cm^-3 to 8.73 g cm^-3 as the temperature increased because of thermal expansion.

The table below summarizes the thermometric substances, their physical properties, and the thermometers that use them:

MEASUREMENT OF TEMPERATURE AND CALIBRATION OF THERMOMETERS

To measure temperature, we need a thermometer with a scale. Designing and marking a scale on a thermometer is called calibration. To calibrate a thermometer, we must choose two reference temperatures called fixed points.

A fixed point is a standard, easily reproducible temperature at which a physical state change occurs under standard atmospheric pressure (101,325 Pa or 1 atm).

The two standard fixed points used for the Celsius scale are:

1. The Lower Fixed Point (Ice Point): This is the temperature of pure melting ice at standard atmospheric pressure. On the Celsius scale, this is defined as 0 °C.
2. The Upper Fixed Point (Steam Point): This is the temperature of steam above pure boiling water at standard atmospheric pressure. On the Celsius scale, this is defined as 100 °C.

Steps for Calibrating a Liquid-in-Glass Thermometer:

• Step 1: Finding the Lower Fixed Point (0 °C): Immerse the bulb of the uncalibrated thermometer into pure melting ice. Wait for the liquid level in the capillary tube to stop moving. Mark this level on the stem as 0 °C. Let this length of the liquid column be l₀.
• Step 2: Finding the Upper Fixed Point (100 °C): Place the bulb of the thermometer in steam just above pure boiling water in a hypsometer or boiling flask. Wait for the liquid level to become steady. Mark this level on the stem as 100 °C. Let this length of the liquid column be l₁₀₀.
• Step 3: Dividing the Scale: Measure the distance between the 0 °C mark and the 100 °C mark. Divide this distance into 100 equal parts. Each part represents a temperature change of 1 degree Celsius (1 °C). Markings can also be extended below 0 °C and above 100 °C.

CALIBRATION OF A THERMOMETER

Mathematical Formula for Temperature on the Celsius Scale:

If we use any thermometric property X (such as length of liquid, resistance, or voltage), the temperature θ (theta) in °C corresponding to a value X_θ is calculated using the formula:

θ = X_θ - X₀X₁₀₀ - X₀ × 100 °C

Where:

• θ = unknown temperature to be determined
• X_θ = value of the property at the unknown temperature θ
• X₀ = value of the property at the ice point (0 °C)
• X₁₀₀ = value of the property at the steam point (100 °C)

For a liquid-in-glass thermometer, we use the length of the liquid column l:

θ = l_θ - l₀l₁₀₀ - l₀ × 100 °C

Worked Example 1:

An uncalibrated mercury thermometer has a mercury thread of length 5.0 cm when placed in pure melting ice, and 25.0 cm when placed in steam above boiling water. When placed in a warm beaker of water, the mercury thread length is 15.0 cm. Calculate the temperature of the warm water.

Solution:

Identify the given values:

• l₀ = 5.0 cm
• l₁₀₀ = 25.0 cm
• l_θ = 15.0 cm

Formula: θ = l_θ - l₀l₁₀₀ - l₀ × 100 °C

Substitution: θ = 15.0 - 5.025.0 - 5.0 × 100 °C

Working: θ = 10.020.0 × 100 °C = 0.5 × 100 °C

Answer: θ = 50 °C

Worked Example 2:

A platinum resistance thermometer has a resistance of 10.0 Ω at 0 °C and 30.0 Ω at 100 °C. When placed in a chemical reaction mixture, its resistance is measured as 26.0 Ω. Determine the temperature of the mixture.

Solution:

Identify the given values:

• R₀ = 10.0 Ω
• R₁₀₀ = 30.0 Ω
• R_θ = 26.0 Ω

Formula: θ = R_θ - R₀R₁₀₀ - R₀ × 100 °C

Substitution: θ = 26.0 - 10.030.0 - 10.0 × 100 °C

Working: θ = 16.020.0 × 100 °C = 0.8 × 100 °C

Answer: θ = 80 °C

Types of Liquid-in-Glass Thermometers:

There are two main liquid-in-glass thermometers used in daily life and science:

• Laboratory Thermometer: Usually contains alcohol or mercury. It has a wide range (typically -10 °C to 110 °C) and is used to measure temperature in laboratory experiments. It has a continuous straight capillary tube.
• Clinical Thermometer: Used to measure human body temperature. It has a narrow range (35 °C to 42 °C) because human body temperature only varies slightly around 37 °C. It features a constriction (a narrow bend) in the capillary tube just above the bulb. This constriction prevents the mercury from falling back into the bulb immediately when the thermometer is removed from the patient's mouth, allowing the temperature to be read accurately.

CLINICAL VS LABORATORY THERMOMETER

SUMMARY

• Temperature is a measure of the average kinetic energy of the particles of a substance.
• Heat flows from a region of higher temperature to a region of lower temperature.
• A thermometric property is a physical property that changes continuously and predictably with temperature.
• Examples of thermometric properties include volume of a liquid, electrical resistance, e.m.f., and gas pressure.
• Density decreases as temperature increases because substances expand in volume while keeping a constant mass.
• To calibrate a thermometer, two fixed points are needed: the lower fixed point (ice point, 0 °C) and the upper fixed point (steam point, 100 °C).
• The formula for calculating temperature on the Celsius scale is: θ = X_θ - X₀X₁₀₀ - X₀ × 100 °C.
• A clinical thermometer has a narrow range (35 °C to 42 °C) and a constriction to prevent the mercury from flowing back before a reading is taken.

KEY TERMS

REVISION QUESTIONS

1. Explain the difference between temperature and heat in terms of the kinetic theory of matter.
2. What happens to the average kinetic energy of the particles of a metal spoon when it is dipped into a hot cup of tea?
3. Define a thermometric property and list three examples of such properties.
4. Explain why the density of most liquids decreases when they are heated.
5. What is meant by the terms "lower fixed point" and "upper fixed point" on the Celsius scale?
6. Describe how you would determine the 100 °C mark on an unmarked mercury-in-glass thermometer in a school laboratory.
7. State two differences between a clinical thermometer and a laboratory thermometer.
8. Explain the importance of the constriction in a clinical thermometer. Why is there no constriction in a laboratory thermometer?

PRACTICE EXERCISE

1. A solid block of aluminium of mass 540 g has a volume of 200 cm³ at 25 °C. When heated to 400 °C, its volume increases to 206 cm³.
   a) Calculate the density of the block at 25 °C.
   b) Calculate the density of the block at 400 °C.
   c) State the relationship between temperature and density shown by these results.
2. A mercury-in-glass thermometer has a mercury column of length 3.0 cm at 0 °C and 23.0 cm at 100 °C.
   a) Find the length of the mercury column at 45 °C.
   b) Find the temperature of a liquid when the mercury column is 18.0 cm long.
3. The resistance of a platinum wire is 8.0 Ω in pure melting ice and 20.0 Ω in steam at standard atmospheric pressure.
   a) Calculate the temperature when its resistance is 14.0 Ω.
   b) Calculate the resistance of the wire at a temperature of 75 °C.
4. The electromotive force (e.m.f.) of a thermocouple is 0.0 mV at 0 °C and 16.0 mV at 100 °C.
   a) Find the temperature of a liquid when the thermocouple reads 12.0 mV.
   b) What is the e.m.f. at 35 °C?
5. An uncalibrated liquid-in-glass thermometer has a liquid column of length 4.0 cm at the ice point and 24.0 cm at the steam point.
   a) What is the temperature when the length of the liquid column is 9.0 cm?
   b) If the thermometer is placed in a freezer at -15 °C, what will be the length of the liquid column?
6. A constant-volume gas thermometer has a gas pressure of 1.20 × 10⁵ Pa at 0 °C and 1.64 × 10⁵ Pa at 100 °C. Calculate the temperature of an oven when the gas pressure is 1.42 × 10⁵ Pa.

ANSWERS TO PRACTICE EXERCISE

Question 1:

a) At 25 °C:

Formula: ρ = mV

Substitution: ρ = 540 g200 cm³

Answer: ρ = 2.70 g cm^-3

b) At 400 °C:

Formula: ρ = mV

Substitution: ρ = 540 g206 cm³

Answer: ρ = 2.62 g cm^-3

c) As temperature increases, density decreases. This is because the volume increases due to thermal expansion while the mass remains constant.

Question 2:

Given: l₀ = 3.0 cm, l₁₀₀ = 23.0 cm

a) Find l_θ at θ = 45 °C:

Formula: θ = l_θ - l₀l₁₀₀ - l₀ × 100 °C

Substitution: 45 = l₄₅ - 3.023.0 - 3.0 × 100

Working: 45 = l₄₅ - 3.020.0 × 100

45 = (l₄₅ - 3.0) × 5

455 = l₄₅ - 3.0

9 = l₄₅ - 3.0

l₄₅ = 9 + 3.0

Answer: l₄₅ = 12.0 cm

b) Find θ when l_θ = 18.0 cm:

Formula: θ = l_θ - l₀l₁₀₀ - l₀ × 100 °C

Substitution: θ = 18.0 - 3.023.0 - 3.0 × 100 °C

Working: θ = 15.020.0 × 100 °C = 0.75 × 100 °C

Answer: θ = 75 °C

Question 3:

Given: R₀ = 8.0 Ω, R₁₀₀ = 20.0 Ω

a) Find θ when R_θ = 14.0 Ω:

Formula: θ = R_θ - R₀R₁₀₀ - R₀ × 100 °C

Substitution: θ = 14.0 - 8.020.0 - 8.0 × 100 °C

Working: θ = 6.012.0 × 100 °C = 0.5 × 100 °C

Answer: θ = 50 °C

b) Find R₇₅ when θ = 75 °C:

Formula: θ = R_θ - R₀R₁₀₀ - R₀ × 100

Substitution: 75 = R₇₅ - 8.020.0 - 8.0 × 100

Working: 75 = R₇₅ - 8.012.0 × 100

75 = (R₇₅ - 8.0) × 8.333

Alternatively: 75100 = R₇₅ - 8.012.0

0.75 = R₇₅ - 8.012.0

0.75 × 12.0 = R₇₅ - 8.0

9.0 = R₇₅ - 8.0

R₇₅ = 9.0 + 8.0

Answer: R₇₅ = 17.0 Ω

Question 4:

Given: E₀ = 0.0 mV, E₁₀₀ = 16.0 mV

a) Find θ when E_θ = 12.0 mV:

Formula: θ = E_θ - E₀E₁₀₀ - E₀ × 100 °C

Substitution: θ = 12.0 - 0.016.0 - 0.0 × 100 °C

Working: θ = 12.016.0 × 100 °C = 0.75 × 100 °C

Answer: θ = 75 °C

b) Find E₃₅ at 35 °C:

Formula: θ = E_θ - E₀E₁₀₀ - E₀ × 100

Substitution: 35 = E₃₅ - 0.016.0 - 0.0 × 100

Working: 35 = E₃₅16.0 × 100

0.35 = E₃₅16.0

E₃₅ = 0.35 × 16.0

Answer: E₃₅ = 5.6 mV

Question 5:

Given: l₀ = 4.0 cm, l₁₀₀ = 24.0 cm

a) Find θ when l_θ = 9.0 cm:

Formula: θ = l_θ - l₀l₁₀₀ - l₀ × 100 °C

Substitution: θ = 9.0 - 4.024.0 - 4.0 × 100 °C

Working: θ = 5.020.0 × 100 °C = 0.25 × 100 °C

Answer: θ = 25 °C

b) Find l at -15 °C:

Formula: θ = l_θ - l₀l₁₀₀ - l₀ × 100

Substitution: -15 = l_-15 - 4.024.0 - 4.0 × 100

Working: -15 = l_-15 - 4.020.0 × 100

-15 = (l_-15 - 4.0) × 5

-155 = l_-15 - 4.0

-3 = l_-15 - 4.0

l_-15 = -3 + 4.0

Answer: l_-15 = 1.0 cm

Question 6:

Given: P₀ = 1.20 × 10⁵ Pa, P₁₀₀ = 1.64 × 10⁵ Pa, P_θ = 1.42 × 10⁵ Pa

Formula: θ = P_θ - P₀P₁₀₀ - P₀ × 100 °C

Substitution: θ = (1.42 × 10⁵) - (1.20 × 10⁵)(1.64 × 10⁵) - (1.20 × 10⁵) × 100 °C

Working: θ = 0.22 × 10⁵0.44 × 10⁵ × 100 °C = 0.220.44 × 100 °C = 0.5 × 100 °C

Answer: θ = 50 °C