Chapter 18: Temperature, Heat, and the First Law of Thermodynamics Notes

Thermodynamics

Thermodynamics is the study of thermal energy, also known as internal energy, within systems. Temperature is a fundamental concept in thermodynamics.

Temperature

Our perception of temperature is not always reliable. For example, metal and wood at the same temperature may feel different to the touch because metal conducts heat away from our fingers more quickly than wood.

The Zeroth Law of Thermodynamics

Every object possesses a property called temperature. When two objects are in thermal equilibrium, their temperatures are equal, and vice versa.

Thermal Equilibrium

Thermal equilibrium is achieved when two bodies in contact reach the same temperature.

Celsius and Fahrenheit Scales

Celsius Scale

The Celsius scale is defined as: $T_c = T - 273.150$, where T is in Kelvin.

• The Celsius degree has the same magnitude as the Kelvin.

• The zero point of the Celsius scale is shifted for convenience relative to absolute zero.

Key Temperatures

• Absolute zero: 0 K or -273.15°C

• Water freezing temperature: 273.15 K or 0°C

Kelvin Scale

Scientists use the Kelvin scale to measure temperature. Zero Kelvin is the lowest attainable temperature.

Relationship between Celsius and Kelvin

• $1 \text{ } C^0 = 1 \text{ } K$

• $0^0C = 273.15 \text{ } K$

Heat

Heat is the energy transferred between a system and its environment due to a temperature difference.

Units of Heat

• SI unit: joule (J)

• Other units:

  • calorie (cal): $1 \text{ cal} = 4.1860 \text{ J}$

  • British thermal unit (Btu): $1 \text{ cal} = 3.969 × 10^{-3} \text{ Btu}$

Work

Work is the energy transferred when a force causes a displacement.

Work Done by Gas

For a small change in volume, the work done by a gas is: $dW = F ds = P A ds = P dV$, where P is pressure and V is volume.

Total Work

The total work done during a thermodynamic process is: $W = \int<em>{V</em>i}^{V<em>f} P dV$, where $V</em>i$ and $V_f$ are the initial and final volumes, respectively.

Work as Area Under the Curve

In a P-V diagram, the work done by the system is the area under the curve.

Positive and Negative Work

• Positive work: The final volume is greater than the initial volume (expansion).

• Negative work: The final volume is smaller than the initial volume (compression).

Thermodynamic Cycles

• Work in a cycle is the area of the loop in a P-V diagram.

  • Positive loop area: Positive work.

  • Negative loop area: Negative work.

Path Dependence of Work

Work depends on the path taken in a thermodynamic process.

Illustration

• Process i to a: Constant pressure, increase temperature, heat flows into the system, $W = \text{area under the curve}$.

• Process a to f: Constant volume, decrease temperature, heat flows out of the system, $W = 0$.

Work done by the system and heat flow are path-dependent.

First Law of Thermodynamics

The change in internal energy of a system is given by: $\Delta E_{int} = Q - W$, where Q is the heat transferred to the system and W is the work done by the system.

• $\Delta E<em>{int} = E</em>{int,f} - E_{int,i}$

• W and Q are path-dependent, but $Q - W$ is independent of the path.

Conservation of Energy

The first law of thermodynamics is a statement of the conservation of energy. It assumes that changes in kinetic and potential energies of the system are negligible.

Special Cases of the First Law of Thermodynamics

Adiabatic Processes

No heat transferred: $Q = 0$. This occurs when the system is well insulated or the process is rapid. Then, $\Delta E_{int} = -W$.

Constant-Volume Processes

$W = \int<em>V^V P dV = 0$. Thus, $\Delta E</em>{int} = Q$.

Cyclical Processes

Final state = Initial state, so $\Delta E_{int} = 0$ and $Q = W$.

Free Expansion Processes

$Q = 0$, $W = 0$, so $\Delta E_{int} = 0$.

Free Expansions

• System is insulated, thus $Q = 0$.

• $W = \int<em>{V</em>i}^{V<em>f} P dV = \int</em>{V<em>i}^{V</em>f} 0 dV = 0$

• Thus, $\Delta E_{int} = 0$.

Examples

Example 1

Steam to Water:

• Initial state: $P<em>i = 1 \text{ Atm}$, $V</em>i = 1 × 10^{-3} m^3$, $T_i = 100^\circ C$, $m = 1 \text{ kg}$

• Final state: $P<em>f = 1 \text{ Atm}$, $V</em>f = 1.671 m^3$, $T_i = 100^\circ C$, $m = 1 \text{ kg}$

Calculations:

• $W = \int<em>{V</em>i}^{V<em>f} P dv = P \int</em>{V<em>i}^{V</em>f} dv = P(V<em>f - V</em>i) = 1.01 × 10^5 Pa (1.671 m^3 - 1 × 10^{-3} m^3) = 169 \text{ kJ}$

• $Q = L_V m = 2256 \text{ kJ/kg} × 1 \text{ kg} = 2256 \text{ kJ}$

• $\Delta E_{int} = Q - W = 2256 \text{ kJ} - 169 \text{ kJ} = 2090 \text{ kJ}$

Example 2

Ideal gas expanding at constant pressure:

• $Q = 500 \text{ J}$

• $V<em>i = 0.2 m^3$, $V</em>f = 0.3 m^3$, $P = 4.0 × 10^3 Pa$

Calculations:

• Isobaric expansion: $W = \int<em>{V</em>i}^{V<em>f} P dv = P \int</em>{V<em>i}^{V</em>f} dv = P \Delta V = P(V<em>f - V</em>i) = 4.0 × 10^3 Pa (0.3 m^3 - 0.2 m^3) = 400 \text{ J}$

• $\Delta E_{int} = Q - W = 500 \text{ J} - 400 \text{ J} = 100 \text{ J}$

Example 3

Calculate work done by expanding 1 mole of gas:

• $P<em>i = 4.0 × 10^3 Pa$, $V</em>i = 0.2 m^3$, $T_i = 96.2 K$

Two processes:

1. Isobaric expansion to $0.3 m^3$, $T_f = 144.3 K$

2. Isothermal expansion to $0.3 m^3$

Calculations:

1. Isobaric expansion:

• $W = \int<em>{V</em>i}^{V<em>f} P dv = P \int</em>{V<em>i}^{V</em>f} dv = P \Delta V = P(V<em>f - V</em>i) = 4.0 × 10^3 Pa (0.3 m^3 - 0.2 m^3) = 400 \text{ J}$

• $\frac{T<em>f}{T</em>i} = \frac{V<em>f}{V</em>i} = \frac{3}{2}$

1. Isothermal expansion:

• $W = nRT \cdot \ln{\frac{V<em>f}{V</em>i}} = P<em>i V</em>i \cdot \ln{\frac{V<em>f}{V</em>i}} = 4.0 × 10^3 Pa × 0.2 m^3 × \ln{\frac{0.3 m^3}{0.2 m^3}} = 324 \text{ J}$

• $\frac{p<em>f}{p</em>i} = \frac{V<em>i}{V</em>f} = \frac{2}{3}$

Example 4

Heat absorbed by 3 moles of helium during isothermal expansion:

• $V<em>i = 10 \text{ L}$, $V</em>f = 20 \text{ L}$, $T = 350 \text{ K}$

Calculations:

• For an ideal gas, isothermal means $\Delta E_{int} = 0$

• $pV = nRT$

• $Q = -W_{on}$

• $W<em>{on} = -nRT \cdot \ln{\frac{V</em>f}{V_i}}$

• $p<em>i = \frac{nRT}{V</em>i} = 8.72 × 10^5 Pa$

• $p<em>f = \frac{p</em>i}{2} = 4.36 × 10^5 Pa$

• $\frac{V<em>f}{V</em>i} = \frac{20 \text{ L}}{10 \text{ L}} = 2$

• $Q = -(-6048 \text{ J}) = 6048 \text{ J}$

Heat Transfer Mechanisms

1. Conduction

2. Convection

3. Radiation

Conduction

Collisions between adjacent atoms transfer heat along the material.

Formula - Single Slab

Conduction rate: $P<em>{con} = \frac{Q}{t} = kA \frac{T</em>H - T<em>C}{L}$, where k is thermal conductivity, A is the face area, $T</em>H$ is the hot reservoir temperature, $T_C$ is the cold reservoir temperature, and L is the thickness.

Thermal Resistance (R-Value)

$R = \frac{L}{k}$. A high R-value indicates a good thermal insulator.

Thermal Conductivity

A material with low thermal conductivity is a good thermal insulator.

Formula - Two Slabs

$P<em>{con} = \frac{k</em>1 A(T<em>H - T</em>X)}{L} = \frac{k<em>2 A(T</em>X - T_C)}{L}$

$P<em>{con} = A \frac{T</em>H - T<em>C}{\frac{L</em>1}{k<em>1} + \frac{L</em>2}{k<em>2}} = A \frac{T</em>H - T<em>C}{R</em>1 + R_2}$

Formula - Many Slabs

$P<em>{con} = A \frac{T</em>H - T<em>C}{\frac{L</em>1}{k<em>1} + \frac{L</em>2}{k<em>2} + \cdots + \frac{L</em>n}{k_n}}$

Checkpoint 4 Solution

In steady state, conduction rates through slabs are equal. $k \Delta T$ is constant for slabs with the same length and cross-sectional area. Smaller temperature difference corresponds to greater thermal conductivity.

Convection

Expansion of fluid when heated leads to lower density, and buoyant forces cause it to rise.

Radiation

Thermal radiation consists of electromagnetic waves that do not require a medium for transfer (travel through vacuum).

Formula

• Rate of thermal radiation emission: $P_{rad} = \sigma \varepsilon A T^4$, where $\sigma$ is the Stefan-Boltzmann constant ($5.67 × 10^{-8} W/m^2 \cdot K^4$), $\varepsilon$ is emissivity, A is surface area, and T is temperature in Kelvins.

• Rate of thermal radiation absorption: $P<em>{abs} = \sigma \varepsilon A T</em>{env}^4$, where $T_{env}$ is the environment temperature in Kelvins.

• Net rate of energy exchange: $P<em>{net} = P</em>{abs} - P<em>{rad} = \sigma \varepsilon A (T</em>{env}^4 - T^4)$

Emissivity

• Value from 0 to 1

• Depends on the composition of the surface

• Black body radiator (ideal): $\varepsilon = 1$

• White Shiny: $\varepsilon = 0$

Example 5

Four-Layer Wall

• Steady state: conduction rates through all layers are the same

• $k<em>d = 5k</em>a$

• $L<em>d = 2L</em>a$

• $T<em>1 = 25^\circ C$, $T</em>2 = 20^\circ C$, $T_5 = -100^\circ C$

Solution:

$\frac{k<em>a A(T</em>1 - T<em>2)}{L</em>a} = \frac{k<em>d A(T</em>4 - T<em>5)}{L</em>d}$

$P<em>a = P</em>d$

$\frac{k<em>a A(T</em>1 - T<em>2)}{L</em>a} = \frac{5k<em>a A(T</em>4 - T<em>5)}{2L</em>a}$

$T<em>1 - T</em>2 = \frac{5}{2} (T<em>4 - T</em>5)$

$25^\circ C - 20^\circ C = \frac{5}{2} (T_4 - (-10^\circ C))$

$T_4 = -8^\circ C$