heat and temp W_250717_153535

Units of Heat

  • Heat = energy transferred because of a temperature difference between two objects.
  • Because heat is energy, it is measured with energy units.
    • SI: joule (J)
    • CGS: calorie (cal) with 1cal=4.184J1\,\text{cal}=4.184\,\text{J}
    • MKS: kilocalorie (kcal) with 1kcal=4184J=103cal1\,\text{kcal}=4184\,\text{J}=10^3\,\text{cal}
    • British: British thermal unit (Btu) with 1Btu=1055J1\,\text{Btu}=1055\,\text{J}
    • Mechanical (Imperial): foot-pound 1ft⋅lb=1.356J1\,\text{ft·lb}=1.356\,\text{J}
  • Definition: 1 kilocalorie is the heat required to raise the temperature of 1 kg of water by 1 K.
  • Historical constant: Robert Mayer first stated the mechanical equivalent of heat, 1cal=4.184J1\,\text{cal}=4.184\,\text{J}.

Internal Energy

  • Internal energy (thermal energy) = sum of total molecular kinetic energy + potential energy inside a substance.
    • Kinetic part is mainly translational; directly linked to temperature.
    • Temperature rise ⇒ increase in average molecular kinetic energy ⇒ increase in internal energy.
    • Potential part comes from intermolecular forces; grows with intermolecular separation.
  • Modes of molecular motion:
    • Vibrational, Rotational, Translational.
  • Heat = transfer of internal (thermal) energy from one body to another.

Thermal Capacity

  • When two objects at different temperatures contact, heat flows from hot → cold until thermal equilibrium.
  • Thermal (heat) capacity CC of an object:
    C=ΔQΔTC=\frac{\Delta Q}{\Delta T}
    ΔQ\Delta Q: heat added • ΔT=T<em>2T</em>1\Delta T=T<em>2-T</em>1
  • SI unit: J K1\text{J K}^{-1}.
  • Dependence: C=cmC=c m (directly proportional to mass and the materialʼs specific heat capacity $c$).
  • Consequences / applications of high or low $C$:
    • Small $C$ ⇒ object heats/cools easily.
    • Water has very high $C$ (≈4× that of aluminium for equal mass), so it is used for cooling engines, solar-heating reservoirs, and hot-water bags; moderates regional climate.
  • Molar thermal capacity (per mole):
    Cmolar=1nΔQΔTC_{\text{molar}}=\frac{1}{n}\frac{\Delta Q}{\Delta T}
    Unit: J mol1K1\text{J mol}^{-1}\,\text{K}^{-1}.

Specific Heat Capacity

  • Specific heat capacity cc: heat needed to raise 1 kg of a substance by 1 K.
    c=Cm=ΔQmΔTc=\frac{C}{m}=\frac{\Delta Q}{m\,\Delta T}
  • Heat required for any mass $m$ over $\Delta T$:
    ΔQ=mcΔT\Delta Q = m c \Delta T
  • SI unit: J kg1K1\text{J kg}^{-1}\,\text{K}^{-1} (others: kcal kg1C1\text{kcal kg}^{-1}\,^{\circ}\text{C}^{-1}, cal g1C1\text{cal g}^{-1}\,^{\circ}\text{C}^{-1}).
  • Representative $c$ values (J kg⁻¹ K⁻¹):
    • Water 4184 • Ice 2089 • Aluminium 898 • Glass 837 • Copper 385 • Lead 130 • Helium gas 5180 • Hydrogen gas 14 250.
  • Climatic moderation: Large lakes absorb huge heat in summer and release it in winter due to waterʼs high $c$.

Law of Heat Exchange & Calorimetry

  • Law: In an isolated system, total heat lost = total heat gained.
    ΔQ<em>lost=ΔQ</em>gained\Delta Q<em>{\text{lost}} = \Delta Q</em>{\text{gained}}
    For two bodies A (hot) and B (cold):
    m<em>Ac</em>A(T<em>AT</em>f)=m<em>Bc</em>B(T<em>fT</em>B)m<em>A c</em>A (T<em>A-T</em>f)=m<em>B c</em>B (T<em>f-T</em>B)
  • Energy-conservation statement for heat processes.
  • Calorimeter: insulated vessel (often Cu/Al) + stirrer + thermometer; measures heat exchange.
  • Determining an unknown $c$ with calorimeter mass m<em>cm<em>c and its $cc$:
    c=ΔQm<em>cc</em>cΔTmΔTc=\frac{\Delta Q - m<em>c c</em>c \Delta T}{m\,\Delta T}
  • Typical problem types: mixing liquids, condensation of steam, cooling/heating times, etc.

Change of State & Latent Heat

  • Phase changes occur abruptly at definite temperatures that depend on pressure.
  • Latent heat: energy absorbed/released during phase change at constant T.
    • Specific latent heat $L$ (fusion $Lf$ or vaporization $Lv$): ΔQ=mL\Delta Q = m L ; unit J kg1\text{J kg}^{-1}.
  • Water heating curve (ice −10 °C → steam 120 °C): heat portions
    1. Warm ice: mciceΔTm c_{\text{ice}} \Delta T
    2. Melt: mLfm L_f (0 °C plateau)
    3. Warm water: mcwΔTm c_w \Delta T
    4. Vaporize: mLvm L_v (100 °C plateau)
    5. Superheat steam: mcsteamΔTm c_{\text{steam}} \Delta T
  • Sublimation: solid → gas directly (e.g., freeze-drying).

Fusion (Melting)

  • Melting point = temperature where solid ↔ liquid at given pressure (ice: 0 °C at 1 atm).
  • Water/ice: Lf=3.335×105J kg1L_f = 3.335\times10^5\,\text{J kg}^{-1}.
  • Example: heat to melt 5 kg ice at 0 °C: ΔQ=5Lf=1.67×106J\Delta Q = 5 L_f =1.67\times10^6\,\text{J}.
  • Mechanical-heat equivalence example: Lead bullet melts on impact. Entire kinetic energy converts to mL<em>fm L<em>f leading to v=2L</em>fv=\sqrt{2L</em>f} (≈221 m s⁻¹).

Vaporization (Boiling / Condensation)

  • Boiling point = temperature where liquid ↔ vapour at given pressure (water: 100 °C at 1 atm).
  • Water: Lv=2.255×106J kg1L_v = 2.255\times10^6\,\text{J kg}^{-1}.
  • Steam burns worse than boiling water: extra energy $L_v$ released on condensation.
  • Sample: 5 kg water → steam needs 1.13×107J1.13\times10^7\,\text{J}.
  • Condensation liberates same amount of heat; exploited in steam heating systems.

Worked Mixed-Phase Example

  • Ice cubes (0.045 kg at −10 °C) in 0.3 kg tea (30 °C): sequential heats (warm ice, melt ice, warm melt) balanced with tea cooling. Outcome: final equilibrium T ≈ 15 °C; all ice melts (since T>0 °C).

Pressure Dependence of Melting & Boiling Points

  • Melting point of ice decreases with increased pressure (slope negative on P-T diagram).
    • Ice skating: blade pressure slightly melts surface, forming lubricating water film; refreezes afterward.
  • Boiling point increases with pressure.
    • Pressure cooker operates at ≈2 atm so water boils at ≈120 °C ⇒ faster cooking.
  • Phase diagram salient points (water):
    • Triple point: all 3 phases coexist.
    • Critical point: liquid & gas densities equal; beyond this water becomes supercritical fluid.

Summary of Key Equations

  • Heat-temperature relation: ΔQ=mcΔT\Delta Q = m c \Delta T.
  • Thermal capacity: C=ΔQΔT=cmC = \frac{\Delta Q}{\Delta T} = c m.
  • Molar thermal capacity: Cm=1nΔQΔTC_{\text{m}} = \frac{1}{n}\frac{\Delta Q}{\Delta T}