Temperature and Heat

1.1 Temperature and Thermal Equilibrium

Definition of Thermal Contact: Two systems are considered to be in thermal contact if energy can be exchanged between them through the process of heat.
Definition of Thermal Equilibrium: Two systems are in thermal equilibrium if they are in thermal contact and there is no net exchange of energy between them.
Measurement of Temperature: Temperature is defined as a measure of the hotness or coldness of an object.
Definition of Heat: When two systems at different initial temperatures are placed in contact, they exchange energy until reaching a common intermediate temperature. This exchange of energy, driven specifically by temperature differences, is called heat.

Zeroth Law of Thermodynamics and Heat Transfer

Zeroth Law of Thermodynamics: If systems $A$ and $B$ are separately in thermal equilibrium with a third system $C$ , then systems $A$ and $B$ are in thermal equilibrium with each other.
Direction of Energy Transfer (Quick Quiz 10.1): When two objects of different sizes, masses, and temperatures are placed in thermal contact, energy travels from the object at the higher temperature to the object at the lower temperature. * The direction of energy transfer depends solely on temperature. * Transfer direction is independent of the size of the object or which object possesses more mass.

10.2 Thermometers and Temperature Scales

Thermometer Principles: All thermometers operate on the principle that specific physical properties of a system change as its temperature fluctuates. These properties include: 1. The volume of a liquid. 2. The dimensions of a solid. 3. The pressure of a gas maintained at a constant volume. 4. The volume of a gas maintained at a constant pressure. 5. The electric resistance of a conductor. 6. The color of an object.
Common Liquid-in-Glass Thermometers: These consist of a mass of liquid (typically mercury or alcohol) that expands into a glass capillary tube when heated. * Physical property: Volume of a liquid. * Operation: Any temperature change within the thermometer's range is defined as proportional to the change in length of the liquid column.
Operational Limits: * Mercury thermometer: Cannot be used below the freezing point of mercury, which is $-39^{\circ}C$ . * Alcohol thermometer: Cannot be used above the boiling point of alcohol, which is $85^{\circ}C$ .
Universal Thermodynamics Measurement: A gas thermometer approaches the requirement for a universal thermometer where readings are independent of the substance used.
Constant-Volume Gas Thermometer: This measures the variation of pressure of a fixed volume of gas with temperature. The pressure is given by: $P = P_0 + \rho gh$

Temperature Conversion and the Kelvin Scale

The Kelvin Scale: This is the absolute temperature scale based on the definition of absolute zero at $-273.15^{\circ}C$ . * The zero point of the Kelvin scale is $0\,K$ . * Relationship to Celsius: $T = T_C + 273.15$ .
Absolute Temperature Benchmarks: * Hydrogen bomb: $10^9\,K$ * Interior of the Sun: $10^7\,K$ * Solar corona: $10^6\,K$ * Surface of the Sun: $10^4\,K$ * Copper melts: $1.36 \times 10^3\,K$ * Water freezes: $2.73 \times 10^2\,K$ * Liquid nitrogen: $77\,K$ * Liquid hydrogen: $20\,K$ * Liquid helium: $4\,K$ * Lowest temperature achieved experimentally: $\approx 10^{-9}\,K$
The Fahrenheit Scale: Commonly used in the United States. * Ice point: $32^{\circ}F$ * Steam point: $212^{\circ}F$
Conversion Formulas: * Celsius to Fahrenheit: $T_F = \frac{9}{5} T_C + 32$ * Celsius to Kelvin: $T = T_C + 273.15$ * Temperature Differences: $\Delta T_C = \Delta T = \frac{5}{9} \Delta T_F$
Concept of "Twice as Hot" (Quick Quiz 10.2): This refers to the ratio of absolute temperatures ( $K$ ). * Example: An ice cube at $-20^{\circ}C$ ( $253.15\,K$ ) and flames at $233^{\circ}C$ ( $506.15\,K$ ) represent a pair where one is approximately twice as hot as the other because $\frac{506.15}{253.15} \approx 2$ .

Temperature Conversion Examples

Example 1: Converting 50.0 degrees Fahrenheit: * To Celsius: $T_C = \frac{5}{9}(T_F - 32) = \frac{5}{9}(50 - 32) = \frac{5}{9}(18) = 10^{\circ}C$ * To Kelvin: $T = T_C + 273.15 = 10 + 273.15 = 283\,K$
Example: Warehouse Temperature Difference: * Given Difference ( $\Delta T_F$ ): $33^{\circ}F$ * Difference in Celsius: $\Delta T_C = \frac{5}{9} \Delta T_F = \frac{5}{9}(33) = 18.3^{\circ}C$ * Difference in Kelvin: $\Delta T_K = \Delta T_C = 18.3\,K$

10.3 Thermal Expansion of Solids and Liquids

General Principle: As temperature increases, volume generally increases. Engineering applications must account for this using thermal-expansion joints in bridges, highways, and buildings to prevent buckling or structural failure.
Linear Expansion in Solids: The change in length ( $\Delta L$ ) is directly proportional to the initial length ( $L_i$ ) and the change in temperature ( $\Delta T$ ). * Formula: $\Delta L = \alpha L_i \Delta T$ * Final Length: $L_f = L_i (1 + \alpha \Delta T)$ * $\alpha$ : Average coefficient of linear expansion (Unit: $(^{\circ}C)^{-1}$ ).
Area Expansion: The change in area ( $\Delta A$ ). * Formula: $\Delta A = \gamma A_i \Delta T$ * $\gamma = 2\alpha$
Volume Expansion: The change in volume ( $\Delta V$ ). * Formula: $\Delta V = \beta V_i \Delta T$ * $\beta = 3\alpha$

Average Coefficients of Expansion (Near Room Temperature)

Linear ( $\alpha, \times 10^{-6} (^{\circ}C)^{-1}$ ): * Aluminum: $24$ * Brass and Bronze: $19$ * Concrete: $12$ * Copper: $17$ * Glass (ordinary): $9$ * Glass (Pyrex): $3.2$ * Invar (Ni-Fe alloy): $0.9$ * Lead: $29$ * Steel: $11$
Volume ( $\beta, (^{\circ}C)^{-1}$ ): * Acetone: $1.5 \times 10^{-4}$ * Benzene: $1.24 \times 10^{-4}$ * Ethyl alcohol: $1.12 \times 10^{-4}$ * Gasoline: $9.6 \times 10^{-4}$ * Glycerin: $4.85 \times 10^{-4}$ * Mercury: $1.82 \times 10^{-4}$ * Turpentine: $9.0 \times 10^{-4}$ * Air (at $0^{\circ}C$ ): $3.67 \times 10^{-3}$ * Helium: $3.665 \times 10^{-3}$

Thermal Expansion Examples

Example 10.3: Expansion of a Railroad Track: * A steel track ( $30.000\,m$ ) warms from $0.0^{\circ}C$ to $40.0^{\circ}C$ . * $\Delta L = (11 \times 10^{-6} (^{\circ}C)^{-1})(30.000\,m)(40.0^{\circ}C) = 0.013\,m = 1.3\,cm$ . * Final length: $30.013\,m$ . * Thermal stress: If the track is fixed (nailed down) and cannot expand, thermal stress develops, which can lead to buckling.
Exercise: Copper Telephone Wire: * Length $35.0\,m$ ; at $-20.0^{\circ}C$ . Find length increase at $35.0^{\circ}C$ . * $\Delta L = (17 \times 10^{-6})(35)(35 - (-20)) = 0.0327\,m = 3.27\,cm$ .
Example 10.4: Fitting a Copper Ring to a Steel Rod: * Initial Ring Area: $9.980\,cm^2$ ; initial temperature $20.0^{\circ}C$ . Rod Area: $10.000\,cm^2$ . * If only the ring is heated, use $\Delta A = \gamma A_i \Delta T = (2\alpha) A_i \Delta T$ . * If both are heated: $\Delta T = \frac{A_s - A_c}{\gamma_c A_c - \gamma_s A_s}$ $\Delta T = \frac{10.000 - 9.980}{(34 \times 10^{-6})(9.980) - (22 \times 10^{-6})(10.000)} \approx 170^{\circ}C$ .

Volume Expansion Examples

Example 10.5: Global Warming and Oceans: * Fractional change in ocean volume for $1^{\circ}C$ rise: $\frac{\Delta V}{V_0} = \beta \Delta T = (2.07 \times 10^{-4})(1) = 2 \times 10^{-4}$ . * Change in depth for ocean (average depth $4000\,m$ ): $\Delta L = \alpha L_i \Delta T = (\frac{\beta}{3}) L_i \Delta T = (6.90 \times 10^{-5})(4000)(1) \approx 0.3\,m$ .
Exercise 10.5: Gasoline Spillage from Aluminum Cylinder: * Aluminum cylinder ( $1.00\,L$ ) and gasoline both heated from $5.00^{\circ}C$ to $65.0^{\circ}C$ . * $\Delta V_{gas} = (9.6 \times 10^{-4})(1.00)(60) = 0.0576\,L$ . * $\Delta V_{Al} = (3 \times 24 \times 10^{-6})(1.00)(60) = 0.00432\,L$ . * Volume spilled: $V_{spilled} = \Delta V_{gas} - \Delta V_{Al} = 57.6\,cm^3 - 4.32\,cm^3 = 53.3\,cm^3$ .
Example: Underground Gasoline Tank: * Filling tank with $1.20 \times 10^3\,gallons$ at $90^{\circ}F$ which cools to $52^{\circ}F$ . * Temperature change in Celsius: $\Delta T_C = \frac{5}{9}(52 - 90) = -21.1^{\circ}C$ . * Volume transferred according to tanker gauge ( $V_i$ ): $V_i = \frac{V_f}{1 + \beta \Delta T} = \frac{1200}{1 + (9.6 \times 10^{-4})(-21.1)} = 1225\,gallons$ .

11.1 Heat and Internal Energy

Internal Energy ( $U$ ): The energy associated with the atoms and molecules of the system. It encompasses: * Kinetic energy from random translational, rotational, and vibrational motion. * Potential energy associated with the bonds between particles.
Heat Transfer: Energy moving between system and environment due to temperature differences.
Units of Heat: * calorie (cal): Energy to raise $1\,g$ of water from $14.5^{\circ}C$ to $15.5^{\circ}C$ . * Calorie (Cal): Equal to $1000\,calories$ or $1\,kilocalorie$ . Used for food energy. * British thermal unit (Btu): Energy to raise $1\,lb$ of water from $63^{\circ}F$ to $64^{\circ}F$ . * Joule (J): Standard SI unit. * Conversion: $1\,cal = 4.186\,J$ ; $1\,Cal = 4186\,J$ ; $1\,Btu = 1.055 \times 10^3\,J$ .
Joule’s Experiment: Demonstrated the conversion between mechanical work and thermal energy. Falling blocks rotating paddles in water cause temperature increase. This established the mechanical equivalent of heat.

11.2 Specific Heat

Principle: Adding energy to a system (without change in state) usually causes a rise in temperature. The amount of energy required to raise the temperature of a given mass varies by substance.
Specific Heat Capacity ( $c$ ): The amount of energy ( $Q$ ) transferred to a sample of mass ( $m$ ) to produce a change in temperature ( $\Delta T$ ). * Formula: $Q = mc \Delta T$ * $c = \frac{Q}{m \Delta T}$ * Specific Heat of Water: $4186\,J/kg \cdot ^{\circ}C$ * Specific Heat of Copper: $387\,J/kg \cdot ^{\circ}C$
Sign Convention: * $+Q$ : Energy flows into the system ( $\Delta T$ is positive). * $-Q$ : Energy flows out of the system ( $\Delta T$ is negative).

Specific Heat and Calorimetry Examples

Example 11.1: Working Off Breakfast: * Breakfast energy: $3.20 \times 10^2\,Cal = 3.20 \times 10^2 \times 4186 = 1.34 \times 10^6\,J$ . * Lifting a $25.0\,kg$ barbell vertical distance $0.400\,m$ . * Energy expended per repetition (raising and lowering): $2mgh = 2(25.0)(9.80)(0.400) = 196\,J$ . * Number of reps: $n = \frac{1.34 \times 10^6}{196} = 6.84 \times 10^3\,times$ .
Example 11.2: Stressing a Strut: * Steel strut ( $L = 2.00\,m, m = 1.57\,kg$ ) absorbs $2.50 \times 10^5\,J$ of thermal energy. * $\Delta T = \frac{Q}{mc} = \frac{2.50 \times 10^5}{(1.57)(448)} = 355^{\circ}C$ . * $\Delta L = \alpha L_i \Delta T = (11 \times 10^{-6})(2.00)(355) = 7.8 \times 10^{-3}\,m$ .
Example 11.3: Finding Specific Heat: * $125\,g$ block at $90.0^{\circ}C$ placed in $0.326\,kg$ water at $20.0^{\circ}C$ . Final temperature $T_f = 22.4^{\circ}C$ . * Energy conservation: $Q_{cold} = -Q_{hot}$ . * $m_w c_w (T_f - T_{wi}) = -m_x c_x (T_f - T_{xi})$ . * $c_x = \frac{m_w c_w (T_f - T_{wi})}{m_x (T_{xi} - T_f)} = \frac{(0.326)(4190)(2.4)}{(0.125)(67.6)} \approx 390\,J/kg \cdot ^{\circ}C$ .

11.4 Latent Heat and Phase Change

Phase Transitions: A transfer of energy can result in a change of physical characteristics (solid to liquid, liquid to gas) without changing the temperature.
Internal Energy Change: During a phase change, added energy goes toward breaking intermolecular bonds, increasing potential energy rather than kinetic energy.
Latent Heat Formula: $Q = \pm mL$ * $+Q$ : Energy enters (melting/vaporization). * $-Q$ : Energy removed (freezing/condensation).
Types of Latent Heat: * Latent Heat of Fusion ( $L_f$ ): Energy for solid-to-liquid transition. For water: $3.33 \times 10^5\,J/kg$ . * Latent Heat of Vaporization ( $L_v$ ): Energy for liquid-to-gas transition. For water: $2.26 \times 10^6\,J/kg$ .

Conversion Process: Ice to Steam (Example Case)

To convert $1.00\,g$ of ice at $-30.0^{\circ}C$ to steam at $120.0^{\circ}C$ requires five distinct steps:

Warm Ice to $0^{\circ}C$ : $Q_1 = mc_{ice} \Delta T = (1.00 \times 10^{-3})(2090)(30) = 62.7\,J$ .
Melt Ice to Water at $0^{\circ}C$ : $Q_2 = mL_f = (1.00 \times 10^{-3})(3.33 \times 10^5) = 333\,J$ .
Warm Water to $100^{\circ}C$ : $Q_3 = mc_w \Delta T = (1.00 \times 10^{-3})(4186)(100) = 419\,J$ .
Vaporize Water to Steam at $100^{\circ}C$ : $Q_4 = mL_v = (1.00 \times 10^{-3})(2.26 \times 10^6) = 2260\,J$ .
Warm Steam to $120^{\circ}C$ : $Q_5 = mc_{steam} \Delta T = (1.00 \times 10^{-3})(2010)(20) = 40.2\,J$ .

Total Energy Needed: $Q_{total} = 3115\,J \approx 3.11 \times 10^3\,J$ .
Practical Consequence: Burns from steam at $100^{\circ}C$ are more severe than from liquid water at $100^{\circ}C$ because of the large amount of latent heat released during condensation on the skin.

Comprehensive Physics Problems

Example 11.5: Ice Water Equilibrium: * $6.00\,kg$ of ice at $-5.00^{\circ}C$ added to $30.0\,liters$ ( $30.0\,kg$ ) of water at $20.0^{\circ}C$ . * Calculation finds the mass of ice can melt completely. * Energy Balance: $Q_{ice} + Q_{melt} + Q_{ice-water} + Q_{water} = 0$ . * Result: Equilibrium temperature $T_f = 3.03^{\circ}C$ .
Exercise 11.5: Whale Tank Cooling: * Tank holds $1.20 \times 10^3\,m^3$ of water ( $1.2 \times 10^6\,kg$ ). * Cool from $20.0^{\circ}C$ to $10.0^{\circ}C$ using ice at $-10.0^{\circ}C$ . * Energy required from water: $Q = mc \Delta T = (1.2 \times 10^6)(4186)(10) = 5.0232 \times 10^{10}\,J$ . * Mass of ice required: $1.27 \times 10^5\,kg$ .

Questions & Discussion

Q: Why does the Zeroth Law mention a third system?
A: The third system (system C) acts as a measuring device (thermometer). If the thermometer shows the same reading for A and B, they must have the same temperature and be in equilibrium with each other even without being in direct contact.
Q: Why is the heat transfer direction independent of mass?
A: Heat flows based on the gradient of average kinetic energy per molecule (temperature), not total internal energy. A small, high-temperature object will still transfer energy to a massive, lower-temperature object until temperatures equalize.
Q: Is the expansion of a hollow object different than a solid object?
A: No. A hole in a material expands at the same rate as the material itself would if it filled the hole. A ring of copper expands as if it were a solid copper disk.