10.2 Thermodynamics

10.2 Thermodynamics on NMAT Physics

Within CEM NMAT Physics (30 items, ~30 minutes), Thermodynamics pairs with Mechanics as classical thermal physics: temperature, heat, energy conservation for thermal systems, ideal gases, heat transfer, and expansion. Expect definitional traps (heat vs temperature) and short numerical items (calorimetry, (PV = nRT), linear expansion).

Quick frame: Temperature is a state property. Heat (Q) is energy in transit because of a temperature difference. Internal energy (U) is energy stored in the system. The first law is bookkeeping among (Q), (W), and $\Delta U$.

Temperature scales

Common scales: Celsius (°C), Kelvin (K), Fahrenheit (°F).

• Absolute zero: $0\,\mathrm{K} = -273.15\,^\circ\mathrm{C}$.
• Conversion: $T(\mathrm{K}) = T(^\circ\mathrm{C}) + 273.15$ (NMAT often uses +273).
• Size of one degree: $1\,\mathrm{K}$ interval = $1\,^\circ\mathrm{C}$ interval (same spacing).
• Fahrenheit: $T_F = \tfrac{9}{5}T_C + 32$ (less common on NMAT but know freezes/boils: 32°F / 212°F for water at 1 atm).

Gas laws and ideal-gas formulas always need absolute temperature (kelvin). Using °C directly in (PV = nRT) is a classic wrong-answer generator.

Worked example — scale conversion. Body temperature 37°C → $37 + 273 = 310\,\mathrm{K}$.

Heat versus temperature; specific heat; latent heat

Temperature reflects how “hot” a system is (related to average translational KE of molecules in an ideal gas). Heat is energy transferred between systems because of a temperature difference. A large cold lake can hold more thermal energy than a small hot cup of coffee; the cup still has higher temperature.

Specific heat capacity (c): heat needed to raise temperature of unit mass by 1 degree (no phase change):

$$Q = mc\Delta T$$

Water has a high $c \approx 4186\,\mathrm{J/(kg\cdot ^\circ C)} \approx 1.00\,\mathrm{cal/(g\cdot ^\circ C)}$ — it resists temperature change, important for climate and physiology analogies.

Latent heat (L): heat for phase change at constant temperature:

[ Q = mL ]

• Fusion (melt/freeze): $L_f$ for water ≈ $3.34 \times 10^5\,\mathrm{J/kg}$.
• Vaporization (boil/condense): $L_v$ for water ≈ $2.26 \times 10^6\,\mathrm{J/kg}$ (order-of-magnitude larger than fusion).

During pure melting of ice at 0°C, temperature stays 0°C until all ice melts if heat is supplied slowly at equilibrium conditions — energy goes into breaking structure, not raising (T).

Calorimetry (worked problems)

Assume an isolated calorimeter: heat lost by hot parts = heat gained by cold parts (signs: write $Q_{\mathrm{hot}} + Q_{\mathrm{cold}} = 0$ or equate magnitudes carefully).

Worked example — mixture without phase change. 0.200 kg of water at 80.0°C is mixed with 0.300 kg of water at 20.0°C in an insulated cup (ignore cup). Find final temperature $T_f$. Use $c = 4186\,\mathrm{J/(kg\cdot ^\circ C)}$ (it will cancel).

• $m_h c (T_f - 80) + m_c c (T_f - 20) = 0$
• $0.200(T_f - 80) + 0.300(T_f - 20) = 0$
• $0.200 T_f - 16 + 0.300 T_f - 6 = 0$
• $0.500 T_f = 22 \Rightarrow T_f = 44.0\,^\circ\mathrm{C}$.

Weighted average intuition: more cold water pulls $T_f$ below the midpoint of 50°C — here 44°C matches.

Worked example — latent heat then warming. How much heat to turn 0.050 kg ice at 0°C into water at 0°C, then warm that water to 30°C? Use $L_f = 3.34 \times 10^5\,\mathrm{J/kg}$, $c_w = 4186\,\mathrm{J/(kg\cdot ^\circ C)}$.

• Melt: $Q_1 = mL_f = 0.050 \times 3.34 \times 10^5 = 1.67 \times 10^4\,\mathrm{J}$.
• Warm: $Q_2 = mc\Delta T = 0.050 \times 4186 \times 30 \approx 6.28 \times 10^3\,\mathrm{J}$.
• Total $Q \approx 2.30 \times 10^4\,\mathrm{J}$.

Always sequence phases: you cannot apply $Q = mc\Delta T$ through a phase change as if ice “heats” as solid past 0°C without melting when energy is still melting ice at the boundary.

First law of thermodynamics and PV work (conceptual)

First law (energy conservation for thermodynamic systems):

$$\Delta U = Q - W$$

using the common physics convention that (W) is work done by the system. Equivalent form $\Delta U = Q + W$ appears when (W) is work done on the system — read the sign convention on any formula sheet or stem carefully.

• (Q > 0): heat added to the system (in the usual “into system positive for (Q)” convention).
• (W > 0) (by system): system expands and does work on surroundings → tends to decrease (U) if (Q = 0).

For an ideal gas, internal energy (U) depends only on temperature (and amount): $\Delta U = 0$ in any isothermal process. In a free expansion of an ideal gas into vacuum ((Q = 0), (W = 0)), $\Delta U = 0$ and (T) stays constant.

PV work conceptual: for a quasi-static expansion, $W = \int P\,dV$. On a (P)–(V) diagram, work by the system is the area under the path. Isobaric expansion ((P) constant): $W = P\Delta V$. Isochoric process ((V) constant): (W = 0) (no area).

Worked example — first law numbers. A gas absorbs 500 J of heat and expands doing 200 J of work. $\Delta U = Q - W = 500 - 200 = 300\,\mathrm{J}$.

Ideal gas law applications

$$PV = nRT = Nk_B T$$

• (P) absolute pressure (Pa), (V) in m³, (T) in K, (n) in moles, $R \approx 8.314\,\mathrm{J/(mol\cdot K)}$.
• Combined gas law for fixed (n): $P_1 V_1 / T_1 = P_2 V_2 / T_2$.
• Boyle (isothermal): $PV = \mathrm{const}$. Charles (isobaric): $V/T = \mathrm{const}$. Gay-Lussac (isochoric): $P/T = \mathrm{const}$.

Worked example — combined gas law. A gas at 1.0 atm and 300 K occupies 2.0 L. Pressure becomes 2.0 atm and temperature 600 K. New volume?

• $V_2 = V_1 (P_1/P_2)(T_2/T_1) = 2.0 \times (1/2) \times (600/300) = 2.0 \times 0.5 \times 2 = 2.0\,\mathrm{L}$.
• Pressure doubled but absolute temperature also doubled → volume unchanged.

Worked example — mole count. At STP approximate (1 atm, 273 K), one mole of ideal gas occupies about 22.4 L. If a problem gives (P), (V), (T), solve (n = PV/(RT)) with consistent units (often convert L·atm using $R = 0.0821\,\mathrm{L\cdot atm/(mol\cdot K)}$).

Heat transfer: conduction, convection, radiation

1. Conduction: energy transfer through a material without bulk motion — molecular collisions and free electrons in metals. Rate often $H = kA\Delta T / L$ (Fourier): high (k) (metals) conduct well; thick walls and small $\Delta T$ reduce heat flow.
2. Convection: bulk movement of fluid carries thermal energy (forced fan, or natural buoyancy of warm air). Blood circulation and wind chill are convective themes in applied settings.
3. Radiation: emission/absorption of electromagnetic waves; does not require a medium. Net power often modeled by Stefan–Boltzmann ideas ($\propto T^4$ for ideal blackbody absolute temperature). Dark surfaces absorb/emit more efficiently than shiny ones in introductory comparisons.

Exam discrimination: vacuum thermos reduces conduction and convection; silvering reduces radiation. Touching a metal vs wood at the same temperature: metal’s high conductivity drains heat from skin faster → feels colder.

Thermal expansion

Most materials expand when heated:

• Linear: $\Delta L = \alpha L_0 \Delta T$
• Area (approx): $\Delta A \approx 2\alpha A_0 \Delta T$
• Volume: $\Delta V = \beta V_0 \Delta T$ with $\beta \approx 3\alpha$ for isotropic solids

Liquids have their own $\beta$; water is anomalous near 4°C (density maximum) — useful factoid, not usually heavy calculation.

Worked example — linear expansion. A steel rod $L_0 = 2.00\,\mathrm{m}$, $\alpha = 1.2 \times 10^{-5}\,/^\circ\mathrm{C}$, heated by $50\,^\circ\mathrm{C}$:

• $\Delta L = (1.2 \times 10^{-5})(2.00)(50) = 1.2 \times 10^{-3}\,\mathrm{m} = 1.2\,\mathrm{mm}$.

Gaps in bridges and bimetallic strips exploit differential expansion ($\alpha_1 \neq \alpha_2$).

Common thermodynamics traps

1. Using °C in (PV = nRT) instead of K.
2. Confusing heat (Q) with temperature (T).
3. Applying $Q = mc\Delta T$ during a pure phase change at constant (T).
4. Wrong first-law sign convention for work.
5. Gauge vs absolute pressure in gas laws (need absolute (P)).
6. Forgetting that isothermal ideal-gas $\Delta U = 0$ even if (Q) and (W) are nonzero ((Q = W) in the “by system” convention).

Connecting to medical reasoning (light touch)

Fever is elevated temperature; antipyretics and cooling alter heat balance. Specific heat of tissue/water dominates thermal load. Evaporative cooling (sweat) uses latent heat of vaporization. None of this replaces formulas on NMAT, but it helps you remember what heat transfer mode is at work.

Thermodynamics completes the first half of NMAT Physics classical topics. Next chapters cover waves/optics, electricity & magnetism, and modern physics — still at introductory college application depth, still unit-sensitive, still diagram-friendly.