Physics

Chapter 1: Temperature and Heat

Imron Rosyadi

Chapter Overview

Temperature and Heat

1.1 Temperature and Thermal Equilibrium
1.2 Thermometers and Temperature Scales
1.3 Thermal Expansion
1.4 Heat Transfer, Specific Heat, and Calorimetry
1.5 Phase Changes
1.6 Mechanisms of Heat Transfer

1.1 Temperature and Thermal Equilibrium

Learning Objectives

By the end of this section, you will be able to:

• Define temperature and describe it qualitatively
• Explain thermal equilibrium
• Explain the zeroth law of thermodynamics

What is Temperature?

Temperature has evolved from the common concepts of hot and cold.

In physics, temperature is operationally defined as the quantity measured by a thermometer.

• More physically, temperature is proportional to the average kinetic energy of translational motion of molecules.

Differences in temperature drive heat transfer, which is the movement of energy due to a temperature difference.

Temperature is a fundamental concept in thermodynamics. Our initial understanding comes from our senses of hot and cold, but physics provides a more rigorous definition. Operationally, we define temperature by how we measure it with a thermometer. On a microscopic level, temperature is directly related to the average kinetic energy of the particles within a substance. This means that hotter objects have faster-moving particles. These temperature differences are what drive the flow of energy we call heat transfer.

Thermal Equilibrium

Thermal equilibrium occurs when two objects in close contact, allowing energy transfer, no longer have a net transfer of energy between them.

• This also means that two objects in thermal equilibrium have the same temperature.

Note

Key Idea: If objects remain in contact for a long time, they typically reach thermal equilibrium.

The Zeroth Law of Thermodynamics

The Zeroth Law of Thermodynamics states:

• If object A is in thermal equilibrium with object B,
• And object B is in thermal equilibrium with object C,
• Then object A is in thermal equilibrium with object C.

This law is fundamental for defining temperature, as it implies that if two objects are in thermal equilibrium, they share a common property: their temperature.

Figure 1.2: Illustrating the Zeroth Law of Thermodynamics.

The Zeroth Law is crucial because it allows us to use a thermometer to measure the temperature of an object. If the thermometer (A) reaches equilibrium with object B and gives a certain reading, and then it’s placed in contact with object C and gives the same reading, it means B and C are at the same temperature. This is how a thermometer can measure the temperature of “something else” and not just its own temperature. Without this law, comparing temperatures would be much more complex.

1.2 Thermometers and Temperature Scales

Learning Objectives

By the end of this section, you will be able to:

• Describe several different types of thermometers
• Convert temperatures between the Celsius, Fahrenheit, and Kelvin scales

Thermometers

Any physical property that depends consistently on temperature can be used to make a thermometer.

Common properties:

• Volume expansion: Alcohol or mercury thermometers.
• Electrical resistance: Used in electronic thermometers.
• Color/Infrared radiation emission: Pyrometers, liquid crystal strips.

Figure 1.3: Examples of different thermometer types.

Thermometers are essential tools for quantitative temperature measurement. The principle behind them is that various physical properties change predictably with temperature. For instance, most substances expand when heated, which is the basis for traditional liquid-in-glass thermometers. Electrical resistance also changes with temperature, allowing for digital thermometers. Even the color or amount of infrared radiation emitted by an object can indicate its temperature, as seen in pyrometers.

Temperature Scales

Three common temperature scales: Fahrenheit, Celsius, and Kelvin.

• Scales are defined by reproducible temperatures, typically freezing and boiling points of water.

Celsius (°C)

• Freezing point of water: 0°C
• Boiling point of water: 100°C

Fahrenheit (°F)

• Freezing point of water: 32°F
• Boiling point of water: 212°F
• A 1°C difference is 1.8 times larger than a 1°F difference (\(\Delta T_F = \frac{9}{5} \Delta T_C\)).

Kelvin (K)

• Absolute temperature scale: Zero point is absolute zero (where average kinetic energy is zero).
• Freezing point of water: 273.15 K
• Boiling point of water: 373.15 K
• Temperature differences are the same as Celsius (\(\Delta T_C = \Delta T_K\)).
• SI unit for temperature.

It’s important to understand the origins and relationships between these scales. Celsius is metric and scientifically convenient due to its 100-degree range for water’s phase changes. Fahrenheit is still used in everyday contexts in some countries. Kelvin is the absolute scale, meaning its zero point, absolute zero, is the lowest possible temperature. This scale is preferred in science because many physical laws simplify when expressed in absolute temperature.

Temperature Conversions

Relationships between temperature scales:

Figure 1.4: Comparing Fahrenheit, Celsius, and Kelvin scales.

To convert from…	Use this equation…
Celsius to Fahrenheit	\(T_F = \frac{9}{5} T_C + 32\)
Fahrenheit to Celsius	\(T_C = \frac{5}{9} (T_F - 32)\)
Celsius to Kelvin	\(T_K = T_C + 273.15\)
Kelvin to Celsius	\(T_C = T_K - 273.15\)
Fahrenheit to Kelvin	\(T_K = \frac{5}{9} (T_F - 32) + 273.15\)
Kelvin to Fahrenheit	\(T_F = \frac{9}{5} (T_K - 273.15) + 32\)

These equations are essential for converting between the different temperature scales. Remember that converting Fahrenheit to Kelvin often involves an intermediate step of converting to Celsius first. Understanding the relationship between the scales, especially how temperature differences are related, is key.

Example: Room Temperature Conversion

Problem: “Room temperature” is generally defined as 25°C.

1. What is room temperature in °F?
2. What is it in K?

Solution:

1. Celsius to Fahrenheit:

\(T_F = \frac{9}{5} T_C + 32 = \frac{9}{5}(25^\circ\text{C}) + 32 = 45 + 32 = 77^\circ\text{F}\).

2. Celsius to Kelvin:

\(T_K = T_C + 273.15 = 25^\circ\text{C} + 273.15 = 298 \text{ K}\).

Constant-Volume Gas Thermometer

• Used as a standard for precise temperature measurement.

• Temperature is proportional to the pressure of a gas held at constant volume. \[T = \frac{P}{P_{TP}} T_{TP}\]

• Where \(T_{TP}\) is the triple-point temperature (273.16 K).

• Results are most accurate when extrapolated to zero gas density.

Tip

The triple point of water (ice, liquid water, and water vapor coexist stably) provides a highly accurate and reproducible reference temperature.

While liquid-in-glass thermometers are common, for highly accurate scientific measurements, the constant-volume gas thermometer is often used as a standard. Its principle relies on the linear relationship between the pressure of an ideal gas and its absolute temperature at constant volume. The triple point of water is a crucial fixed point for calibrating these thermometers, offering a more precise reference than just freezing or boiling points, which depend on pressure.

1.3 Thermal Expansion

Learning Objectives

By the end of this section, you will be able to:

• Answer qualitative questions about the effects of thermal expansion
• Solve problems involving thermal expansion, including those involving thermal stress

What is Thermal Expansion?

Thermal expansion is the change in size or volume of a system as its temperature changes.

• Most substances expand when heated and contract when cooled.
• This is due to increased kinetic energy of atoms/molecules, leading to increased average distance between them.

Figure 1.5: Expansion joints in bridges accommodate thermal expansion.

Thermal expansion is a very common phenomenon, from rising hot air balloons to the creaking sounds of houses at night. It’s important for engineers to account for thermal expansion when designing structures like bridges and railways, which is why expansion joints are installed. The underlying cause is the increased vibrational energy of particles at higher temperatures, which effectively pushes them further apart.

Linear Thermal Expansion

For a given material, the change in length (\(\Delta L\)) is proportional to the original length (\(L\)) and the change in temperature (\(\Delta T\)).

\[\Delta L = \alpha L \Delta T\]

• \(\alpha\) is the coefficient of linear expansion, a material property (units: 1/°C or 1/K).
• \(\Delta T\) is the change in temperature.

Note

\(\alpha\) is typically very small and is approximated as constant for most practical purposes.

Material	\(\alpha\) (1/°C)
Aluminum	\(25 \times 10^{-6}\)
Steel	\(12 \times 10^{-6}\)
Pyrex® glass	\(3 \times 10^{-6}\)

Table 1.2 (partial): Coefficients of Linear Expansion.

This equation describes how a material’s length changes with temperature. It’s a linear approximation, meaning it works well for relatively small temperature changes. The coefficient of linear expansion, alpha, is unique to each material and phase. Understanding these coefficients allows us to predict and account for changes in dimensions due to temperature fluctuations.

Example: Golden Gate Bridge Expansion

Problem: The main span of the Golden Gate Bridge is 1275 m long at its coldest (–15°C). What is its change in length when it warms to 40°C? Assume it’s made entirely of steel (\(\alpha = 12 \times 10^{-6} /^\circ\text{C}\)).

Strategy: Use \(\Delta L = \alpha L \Delta T\).

Solution:

1. Calculate \(\Delta T = T_{final} - T_{initial} = 40^\circ\text{C} - (-15^\circ\text{C}) = 55^\circ\text{C}\).

2. Substitute values:

   \(\Delta L = (12 \times 10^{-6} /^\circ\text{C})(1275 \text{ m})(55^\circ\text{C})\)

   \(\Delta L = 0.84 \text{ m}\)

Significance: A change of 0.84 meters (about 33 inches) is significant for a bridge, highlighting the need for expansion joints.

Thermal Expansion in 2D and 3D

Objects expand in all dimensions: length, area, and volume.

• Holes in materials also expand with temperature.

Figure 1.7: Expansion in two and three dimensions.

Area Expansion:

For small \(\Delta T\): \(\Delta A = 2 \alpha A \Delta T\)

Volume Expansion:

\[\Delta V = \beta V \Delta T\] - \(\beta\) is the coefficient of volume expansion. - For isotropic solids, \(\beta \approx 3\alpha\).

It’s important to realize that thermal expansion isn’t limited to just length. An entire object, including any internal voids or holes, will expand proportionally in all dimensions. The relationship \(\beta \approx 3\alpha\) for solids simplifies calculations, showing that volume expansion is roughly three times the linear expansion. For liquids and gases, only volume expansion is typically defined, as their shapes aren’t fixed.

Anomalous Expansion of Water

Water is an important exception to typical thermal expansion behavior.

• Most substances are densest as solids.
• Water is densest at 4°C.
• It expands (density decreases) both above 4°C and below 4°C (down to 0°C).

Figure 1.8: Density of water vs. temperature.

Important

This unusual property allows aquatic life to survive in cold climates. Water at 4°C sinks to the bottom, and ice (less dense) forms on the surface, insulating the water below.

Thermal Stress

Thermal stress occurs when an object’s temperature changes but it is prevented from expanding or contracting freely.

• This stress can be very large and cause damage.
• Engineers design components with expansion joints or use materials with similar expansion coefficients (e.g., steel in concrete).

To calculate thermal stress:

• Imagine the object expanding/contracting freely (\(\Delta L = \alpha L_0 \Delta T\)).
• Calculate the force needed to return it to its original length (\(F/A = Y (\Delta L / L_0)\)).
• Combining these: \(F/A = Y \alpha \Delta T\). (\(Y\) is Young’s modulus).

Thermal stress is a critical consideration in many engineering applications. Imagine a concrete bridge without expansion joints – the expansion on a hot day could cause it to buckle, or contraction on a cold day could lead to cracks. By understanding and calculating thermal stress, engineers can choose appropriate materials and designs to prevent such failures.

1.4 Heat Transfer, Specific Heat, and Calorimetry

Learning Objectives

By the end of this section, you will be able to:

• Explain phenomena involving heat as a form of energy transfer
• Solve problems involving heat transfer

Internal Energy and Heat

Internal energy is the sum of the microscopic energies of a system’s molecules.

• Includes thermal energy, proportional to temperature.

Heat is energy transferred spontaneously due to a temperature difference.

• It’s a form of energy flow, not a property an object “has.”
• SI unit: Joules (J).
• Other common unit: calorie (cal) or kilocalorie (kcal).

Figure 1.9: Heat transfer from a soda to ice to reach thermal equilibrium.

It’s vital to distinguish between internal energy and heat. Internal energy is a property of a system, representing the total energy of its molecules. Heat, on the other hand, is the transfer of energy between systems due to a temperature difference. A common misconception is to say an object “has heat”; rather, it has internal energy, and heat is the process by which that internal energy can change.

Mechanical Equivalent of Heat

James Prescott Joule’s experiments established that work and heat can produce the same effects.

• Mechanical equivalent of heat: The amount of work needed to produce the same effects as heat transfer.
• \(1.000 \text{ kcal} = 4186 \text{ J}\)

Figure 1.10: Joule’s experiment, showing conversion of gravitational potential energy into increased internal energy (heat).

Tip

Heat and work are not state variables. Internal energy and temperature are state variables.

Joule’s work was groundbreaking because it helped consolidate the idea that heat is a form of energy, not a separate fluid (caloric). His experiments demonstrated that mechanical work could directly lead to an increase in a substance’s temperature, quantifying the relationship between mechanical energy and thermal energy. This was a crucial step in developing the law of conservation of energy and the field of thermodynamics.

Temperature Change and Heat Capacity

When heat is transferred to a substance (without phase change or work), its temperature changes.

The amount of heat (\(Q\)) transferred is approximately:

\[Q = mc\Delta T\]

• \(m\) is the mass of the substance.
• \(\Delta T\) is the change in temperature.
• \(c\) is the specific heat (or specific heat capacity).
  • Specific heat depends on the material and its phase (solid, liquid, gas).
  • SI units: J/(kg·K) or J/(kg·°C).

Note

Water has one of the largest specific heats, which helps stabilize Earth’s climate.

This equation is fundamental for calculating temperature changes due to heat transfer. Specific heat is a crucial property, as it tells us how much energy is needed to change the temperature of a unit mass of a substance by one degree. Water’s high specific heat is why it takes a long time to boil and also why large bodies of water moderate local climates. The specific heat can vary with temperature, but for many problems, we assume it’s constant over the given temperature range.

Specific Heats of Various Substances

Substances	Specific Heat (\(c\)) J/(kg·°C)
Solids
Aluminum	900
Copper	387
Glass	840
Ice (average, –50°C to 0°C)	2090
Iron, steel	452
Liquids
Ethanol	2450
Mercury	139
Water (15.0°C)	4186
Gases
Air (dry)	721 (1015)
Steam (100°C)	1520 (2020)

Table 1.3 (partial): Specific Heats of Various Substances. (Values in parentheses for gases are at constant pressure.)

Example: Heating a Pan of Water

Problem: A 0.500-kg aluminum pan and 0.250 L of water in it are heated from 20.0°C to 80.0°C.

1. How much heat is required?
2. What percentage of the heat is used to raise the temperature of the pan and the water?

Knowns: - \(m_{Al} = 0.500 \text{ kg}\), \(c_{Al} = 900 \text{ J/(kg}\cdot^\circ\text{C})\) - \(V_W = 0.250 \text{ L} \implies m_W = 0.250 \text{ kg}\) (since density of water is 1 kg/L) - \(c_W = 4186 \text{ J/(kg}\cdot^\circ\text{C})\) - \(\Delta T = 80.0^\circ\text{C} - 20.0^\circ\text{C} = 60.0^\circ\text{C}\)

Solution:

1. Heat to water: \(Q_W = m_W c_W \Delta T = (0.250 \text{ kg})(4186 \text{ J/(kg}\cdot^\circ\text{C}))(60.0^\circ\text{C}) = 62.8 \text{ kJ}\).
2. Heat to aluminum: \(Q_{Al} = m_{Al} c_{Al} \Delta T = (0.500 \text{ kg})(900 \text{ J/(kg}\cdot^\circ\text{C}))(60.0^\circ\text{C}) = 27.0 \text{ kJ}\).
3. Total heat: \(Q_{Total} = Q_W + Q_{Al} = 62.8 \text{ kJ} + 27.0 \text{ kJ} = 89.8 \text{ kJ}\).

Percentages:

2. Pan: \((27.0 \text{ kJ} / 89.8 \text{ kJ}) \times 100\% = 30.1\%\).
3. Water: \((62.8 \text{ kJ} / 89.8 \text{ kJ}) \times 100\% = 69.9\%\).

Calorimetry

Calorimetry is the use of a calorimeter (a container preventing heat transfer with surroundings) to measure heat.

For objects thermally isolated from their surroundings: - Heat gained by colder objects = Heat lost by hotter objects. - This is based on the conservation of energy.

\[Q_{hot} + Q_{cold} = 0\]

Tip

When solving calorimetry problems, set the sum of all heat transfers within the isolated system to zero. Remember that heat gained is positive, and heat lost is negative.

1.5 Phase Changes

Learning Objectives

By the end of this section, you will be able to:

• Describe phase transitions and equilibrium between phases
• Solve problems involving latent heat
• Solve calorimetry problems involving phase changes

Phase Transitions

A phase transition is a change from one state of matter to another.

• Melting (fusion): Solid \(\to\) Liquid (opposite: freezing)
• Evaporation (vaporization): Liquid \(\to\) Gas (opposite: condensation)
• Sublimation: Solid \(\to\) Gas (opposite: deposition)

Figure 1.13: Dry ice subliming (left) and frost forming (right).

Phase transitions are common and crucial processes. They occur at specific temperatures and pressures, known as melting points and boiling points, respectively. These points can vary with pressure. Sublimation is a less common but interesting phase change where a solid directly turns into a gas, like dry ice.

Phase Diagrams

A phase diagram is a pT graph showing the phase of a substance at different pressures and temperatures.

• Triple point: Unique combination of temperature and pressure where solid, liquid, and gas phases coexist in equilibrium.
• Critical point: Temperature and pressure above which liquid and gas phases cannot be distinguished (supercritical fluid).

Figure 1.12: Phase diagram for water.

Phase diagrams are powerful tools for visualizing the conditions under which a substance exists in different phases. For water, the phase diagram shows its unique negative slope for the melting curve, meaning increasing pressure can melt ice. The triple point is a very precise and stable reference point for temperature calibration. Beyond the critical point, the distinction between liquid and gas vanishes, leading to exotic properties of supercritical fluids.

Equilibrium during Phase Changes

At a phase transition temperature (e.g., melting point, boiling point), phases coexist in equilibrium.

• Adding heat causes more of one phase; removing heat causes more of the other.
• Vapor pressure: The equilibrium pressure of a gas above its liquid at a given temperature in a closed container.
  • Boiling occurs when vapor pressure equals ambient pressure.

Figure 1.14: Liquid-gas equilibrium at different boiling points.

During a phase change, the system is in equilibrium, meaning the rates of forward and reverse processes are equal. For example, at the boiling point, molecules are evaporating and condensing at the same rate. This is why temperature remains constant during a phase change even as heat is added or removed. The concept of vapor pressure is crucial for understanding evaporation and boiling.

Latent Heat

During a phase change, heat transfer occurs without changing the temperature.

• This energy is used to break or form molecular bonds.
• Latent heat: The energy per unit mass involved in a phase change.

\[Q = mL_f \quad \text{(melting/freezing)}\] \[Q = mL_v \quad \text{(vaporization/condensation)}\]

• \(L_f\) is the latent heat of fusion.
• \(L_v\) is the latent heat of vaporization.

Figure 1.15: Temperature vs. heat added for water, showing latent heat regions.

Latent heats are “hidden” because they don’t cause a temperature change. Instead, they facilitate the rearrangement of molecules, either breaking them apart (melting, vaporization) or allowing them to come closer together (freezing, condensation). These values are specific to each substance and are typically much larger than specific heats, indicating that significant energy is involved in changing phases. This property is exploited in applications like preventing frost damage to fruit by spraying water.

Heats of Fusion and Vaporization

Substance	Melting Point (°C)	\(L_f\) (kJ/kg)	Boiling Point (°C)	\(L_v\) (kJ/kg)
Water	0.00	334	100.0	2256
Aluminum	660	380	2450	11400
Ethanol	–114	104	78.3	854
Mercury	–38.9	11.8	357	272

Table 1.4 (partial): Heats of Fusion and Vaporization.

Note

Notice that \(L_v\) is generally much larger than \(L_f\), meaning more energy is required to turn a liquid into a gas than a solid into a liquid. This is because molecules are completely separated in the gas phase.

Example: Chilling Soda with Ice

Problem: Three ice cubes (6.0 g each, at 0°C) are added to 0.250 kg of 20°C soda. Assuming no heat loss to surroundings and soda has water’s specific heat. Find the final temperature when all ice has melted.

Knowns:

• \(m_{ice} = 3 \times 0.006 \text{ kg} = 0.018 \text{ kg}\)
• \(T_{ice, initial} = 0^\circ\text{C}\)
• \(L_f = 334 \text{ kJ/kg} = 334 \times 10^3 \text{ J/kg}\) (for water/ice)
• \(m_{soda} = 0.250 \text{ kg}\)
• \(T_{soda, initial} = 20^\circ\text{C}\)
• \(c_W = 4186 \text{ J/(kg}\cdot^\circ\text{C})\) (for soda)

Strategy: Use calorimetry: \(Q_{ice} + Q_{soda} = 0\).

\(Q_{ice} = \text{heat to melt ice} + \text{heat to warm melted ice}\)

\(Q_{ice} = m_{ice} L_f + m_{ice} c_W (T_f - 0^\circ\text{C})\)

\(Q_{soda} = m_{soda} c_W (T_f - 20^\circ\text{C})\)

Solution:

\(m_{ice} L_f + m_{ice} c_W T_f + m_{soda} c_W (T_f - 20^\circ\text{C}) = 0\)

Solve for \(T_f\):

\(T_f = \frac{m_{soda} c_W (20^\circ\text{C}) - m_{ice} L_f}{(m_{soda} + m_{ice}) c_W}\)

\(T_f = \frac{(0.250 \text{ kg})(4186 \text{ J/kg}\cdot^\circ\text{C})(20^\circ\text{C}) - (0.018 \text{ kg})(334 \times 10^3 \text{ J/kg})}{(0.250 \text{ kg} + 0.018 \text{ kg})(4186 \text{ J/kg}\cdot^\circ\text{C})}\)

\(T_f = \frac{20930 \text{ J} - 6012 \text{ J}}{(0.268 \text{ kg})(4186 \text{ J/kg}\cdot^\circ\text{C})} = \frac{14918 \text{ J}}{1122.08 \text{ J}/^\circ\text{C}} \approx 13^\circ\text{C}\)

1.6 Mechanisms of Heat Transfer

Learning Objectives

By the end of this section, you will be able to:

• Explain some phenomena that involve conductive, convective, and radiative heat transfer
• Solve problems on the relationships between heat transfer, time, and rate of heat transfer
• Solve problems using the formulas for conduction and radiation

Three Mechanisms of Heat Transfer

Whenever there is a temperature difference, heat transfer occurs by one or more of these methods:

1. Conduction: Heat transfer through stationary matter by physical contact.
2. Convection: Heat transfer by the macroscopic movement of a fluid (liquid or gas).
3. Radiation: Heat transfer by emission or absorption of electromagnetic radiation.

Figure 1.19: All three heat transfer mechanisms around a fireplace.

These three mechanisms cover all possible ways heat can transfer. It’s rare for only one mechanism to be active in a real-world scenario; usually, they act in combination. Understanding each mechanism individually is key to analyzing complex heat transfer situations. All three are driven by temperature differences, with faster transfer for larger differences.

Conduction

Heat transfer through direct physical contact.

• Occurs due to collisions between molecules, transferring kinetic energy.
• Faster in materials with free electrons (metals).
• Insulators are poor conductors (e.g., polystyrene foam, trapped air).

Rate of conductive heat transfer (\(P\)): \[P = \frac{kA(T_h - T_c)}{d}\]

• \(k\): Thermal conductivity of the material (W/(m·°C)).
• \(A\): Cross-sectional area.
• \(T_h - T_c\): Temperature difference across the material.
• \(d\): Thickness of the material.

Note

The ratio \(d/k\) is called the R factor, a measure of insulating ability. A larger R factor means better insulation.

Conduction is the primary way heat moves through solids. Think about touching a hot metal spoon versus a wooden spoon. The metal spoon conducts heat much faster because of its free electrons. Thermal conductivity, ‘k’, is a measure of how good a material is at conducting heat. Insulators, like foam or air, have low ‘k’ values.

Thermal Conductivities

Substance	Thermal Conductivity \(k\) (W/m·°C)
Diamond	2000
Silver	420
Copper	390
Aluminum	220
Steel	80
Ice	2.2
Glass (average)	0.84
Water	0.6
Polystyrene foam	0.035
Air	0.023

Table 1.5 (partial): Thermal Conductivities of Common Substances.

Example: Ice Melting in an Icebox

Problem: A polystyrene foam icebox (total area \(0.950 \text{ m}^2\), thickness \(2.50 \text{ cm}\)) contains ice, water, and canned beverages at 0°C. How much ice melts in one day if the icebox is in a car trunk at 35.0°C?

Knowns:

• \(k = 0.035 \text{ W/(m}\cdot^\circ\text{C})\) (polystyrene foam)
• \(A = 0.950 \text{ m}^2\)
• \(d = 2.50 \text{ cm} = 0.0250 \text{ m}\)
• \(T_c = 0^\circ\text{C}\), \(T_h = 35.0^\circ\text{C}\)
• \(t = 1 \text{ day} = 86,400 \text{ s}\)
• \(L_f = 334 \times 10^3 \text{ J/kg}\) (for ice)

Strategy: 1. Calculate the rate of heat transfer (\(P\)) by conduction. 2. Calculate the total heat transferred (\(Q = P \times t\)). 3. Calculate the mass of ice melted (\(m = Q / L_f\)).

Solution:

1. \(P = \frac{kA(T_h - T_c)}{d} = \frac{(0.035 \text{ W/m}\cdot^\circ\text{C})(0.950 \text{ m}^2)(35.0^\circ\text{C})}{0.0250 \text{ m}} = 46.6 \text{ W}\).
2. \(Q = (46.6 \text{ J/s})(86,400 \text{ s}) = 4.03 \times 10^6 \text{ J}\).
3. \(m = \frac{4.03 \times 10^6 \text{ J}}{334 \times 10^3 \text{ J/kg}} = 12.1 \text{ kg}\).

Significance: About 12.1 kg (26.7 lbs) of ice melts in a day, which is a significant amount and shows the effectiveness of polystyrene foam as an insulator.

Convection

Heat transfer by the macroscopic movement of a fluid.

• Forced convection: Fluid movement driven by external means (fans, pumps).
  • E.g., car cooling system, blowing on hot food.
• Natural (free) convection: Fluid movement driven by buoyant forces (hot fluid rises, cold fluid sinks).
  • E.g., hot air rising in a room, ocean currents.

Figure 1.25: Natural convection in a pot of boiling water.

Tip

Convection is typically more effective than conduction for heat transfer in fluids over macroscopic distances.

Convection is how heat moves in liquids and gases. It’s responsible for many everyday phenomena and plays a huge role in weather patterns and ocean currents. The density changes due to temperature are what drive natural convection. Think of a radiator in a room: it heats the air, which rises, cools, and sinks, creating a circulation pattern.

Convection with Phase Change

Convection can be combined with phase changes for significant cooling or heating effects.

• Evaporative cooling: When sweat evaporates from skin, it removes a large amount of heat (\(Q = mL_v\)) from the body.
  • This is why sweating cools us down, even if the air temperature is high.
• Cloud formation: Water vapor condensing into liquid droplets in the atmosphere releases latent heat, driving powerful weather systems like thunderheads and hurricanes.

Figure 1.27: Cumulus clouds formed by convection and condensation.

Radiation

Heat transfer by emission or absorption of electromagnetic waves.

• No medium is required for radiation to propagate (e.g., Sun warms Earth through vacuum).
• Hotter objects radiate more energy, especially at shorter wavelengths (e.g., from red to white hot).
• Emissivity (\(e\)): A measure of an object’s ability to emit and absorb thermal radiation (0 to 1).
  • Black objects: good absorbers and emitters (\(e \approx 1\)).
  • White/shiny objects: poor absorbers and emitters (\(e \approx 0\)).

Stefan-Boltzmann Law of Radiation: The net rate of heat transfer by radiation between an object (temperature \(T_1\), emissivity \(e\)) and its surroundings (temperature \(T_2\)) is:

\[P_{net} = \sigma A e (T_2^4 - T_1^4)\]

• \(\sigma = 5.67 \times 10^{-8} \text{ J/(s}\cdot\text{m}^2\cdot\text{K}^4)\) (Stefan-Boltzmann constant).
• \(A\): Surface area of the object.
• \(T_1, T_2\): Temperatures in Kelvins.

Radiation is unique because it doesn’t require direct contact or a fluid medium. It’s how the sun heats us. The Stefan-Boltzmann law shows a very strong dependence on temperature (to the fourth power!), which means even small temperature differences can lead to significant radiative heat transfer. Emissivity ‘e’ is also important: black surfaces absorb and emit radiation much more effectively than shiny or white surfaces.

Example: Net Heat Transfer of a Person

Problem: What is the rate of heat transfer by radiation of an unclothed person (\(A = 1.50 \text{ m}^2\), \(T_1 = 33.0^\circ\text{C}\)) standing in a dark room (\(T_2 = 22.0^\circ\text{C}\))? The emissivity of skin is \(e = 0.97\).

Strategy: Use the Stefan-Boltzmann law: \(P_{net} = \sigma A e (T_2^4 - T_1^4)\).

Remember to convert temperatures to Kelvin!

Knowns: - \(\sigma = 5.67 \times 10^{-8} \text{ J/(s}\cdot\text{m}^2\cdot\text{K}^4)\) - \(A = 1.50 \text{ m}^2\) - \(e = 0.97\) - \(T_1 = 33.0^\circ\text{C} + 273.15 = 306.15 \text{ K}\) - \(T_2 = 22.0^\circ\text{C} + 273.15 = 295.15 \text{ K}\)

Solution:

\(P_{net} = (5.67 \times 10^{-8})(1.50)(0.97) ((295.15)^4 - (306.15)^4)\)

\(P_{net} = (8.24 \times 10^{-8}) (7.59 \times 10^9 - 8.76 \times 10^9)\)

\(P_{net} = (8.24 \times 10^{-8}) (-1.17 \times 10^9)\)

\(P_{net} \approx -96.3 \text{ W}\)

Significance: The person loses about 96.3 W of heat to the environment through radiation. The negative sign indicates heat is leaving the person. This is a significant amount, suggesting the person would feel cold.

Key Takeaways

1. Temperature and Heat: Temperature is a measure of average molecular kinetic energy, while heat is the transfer of thermal energy due to a temperature difference.
2. Thermal Equilibrium & Zeroth Law: Objects in thermal equilibrium have the same temperature, and the Zeroth Law allows for consistent temperature measurement.
3. Temperature Scales: Fahrenheit, Celsius, and Kelvin (absolute scale) are used, with specific conversion formulas.
4. Thermal Expansion: Most materials expand with increasing temperature (\(\Delta L = \alpha L \Delta T\), \(\Delta V = \beta V \Delta T\)), with water being a notable exception (densest at 4°C).
5. Heat Capacity & Calorimetry: Specific heat (\(Q = mc\Delta T\)) quantifies heat transfer for temperature changes; calorimetry uses energy conservation (\(Q_{hot} + Q_{cold} = 0\)).
6. Phase Changes & Latent Heat: Heat transfer during phase changes occurs at constant temperature, using latent heats of fusion (\(L_f\)) or vaporization (\(L_v\)) (\(Q = mL\)).
7. Heat Transfer Mechanisms:
   • Conduction: Direct contact (\(P = \frac{kA\Delta T}{d}\)).
   • Convection: Fluid movement.
   • Radiation: Electromagnetic waves (\(P_{net} = \sigma A e (T_2^4 - T_1^4)\)).

Key Equations

Equation	Description
\(T_F = \frac{9}{5} T_C + 32\)	Celsius to Fahrenheit conversion
\(T_C = T_K - 273.15\)	Kelvin to Celsius conversion
\(\Delta L = \alpha L \Delta T\)	Linear thermal expansion
\(\Delta V = \beta V \Delta T\)	Volume thermal expansion
\(F/A = Y \alpha \Delta T\)	Thermal stress (fixed ends)
\(Q = mc\Delta T\)	Heat transfer for temperature change
\(Q_{hot} + Q_{cold} = 0\)	Conservation of energy in calorimetry
\(Q = mL_f\)	Heat transfer for melting/freezing
\(Q = mL_v\)	Heat transfer for vaporization/condensation
\(P = \frac{kA(T_h - T_c)}{d}\)	Rate of heat transfer by conduction
\(P_{net} = \sigma A e (T_2^4 - T_1^4)\)	Net rate of heat transfer by radiation

Key Terms

Term	Definition
Temperature	Quantity measured by a thermometer; proportional to average kinetic energy of molecular translation.
Heat Transfer	Movement of energy due to a difference in temperature.
Thermal Equilibrium	State where no net energy is transferred between objects in contact.
Zeroth Law of Thermodynamics	If A is in equilibrium with B, and B with C, then A is in equilibrium with C.
Absolute Zero	Lowest possible temperature where molecular kinetic energy is zero (0 K).
Thermal Expansion	Change in size or volume of a system due to temperature change.
Coefficient of Linear Expansion (\(\alpha\))	Material property describing fractional change in length per degree Celsius or Kelvin.
Coefficient of Volume Expansion (\(\beta\))	Material property describing fractional change in volume per degree Celsius or Kelvin.
Thermal Stress	Stress created when an object is constrained from expanding or contracting due to temperature changes.
Internal Energy	Sum of microscopic energies of a system.
Heat	Energy transferred spontaneously due to a temperature difference.
Specific Heat (\(c\))	Amount of heat needed to change the temperature of 1.00 kg of mass by 1.00°C (or 1.00 K).
Calorimetry	Measurement of heat transfer using a calorimeter.
Phase Transition	Change of a substance from one state (solid, liquid, gas) to another.
Phase Diagram	A pT graph showing the phases of a substance at different pressures and temperatures.
Triple Point	Unique temperature and pressure where solid, liquid, and gas phases of a substance coexist in equilibrium.
Latent Heat (\(L\))	Energy per unit mass required to change the phase of a substance at constant temperature.
Latent Heat of Fusion (\(L_f\))	Energy per unit mass for melting/freezing.
Latent Heat of Vaporization (\(L_v\))	Energy per unit mass for vaporization/condensation.
Conduction	Heat transfer through stationary matter by physical contact.
Thermal Conductivity (\(k\))	Material property indicating its ability to conduct heat.
Convection	Heat transfer by the macroscopic movement of a fluid.
Radiation	Heat transfer by the emission or absorption of electromagnetic waves.
Emissivity (\(e\))	A measure of how effectively an object emits and absorbs thermal radiation (0 to 1).
Stefan-Boltzmann Constant (\(\sigma\))	Fundamental constant used in the Stefan-Boltzmann law of radiation.