Chapter 14 – Temperature, Heat & Phase Change
Learning Objectives
- Explain thermal equilibrium and the zeroth law of thermodynamics.
- Distinguish thermal contact from thermal isolation in experimental set‑ups.
- Convert between the Celsius and Kelvin temperature scales (Fahrenheit is shown only as a historical note).
- Derive and apply the specific‑heat‑capacity relation \(Q = mc\Delta T\).
- State the difference between specific heat capacity (\(c\)) and molar heat capacity (\(C\)).
- Recognise the distinction between \(c{p}\) (constant‑pressure) and \(c{v}\) (constant‑volume) and when they are effectively equal.
- Use the latent‑heat equations \(Q = mL\) for fusion and vaporisation.
- Link heat transfer to internal‑energy change (\(\Delta U = mc\Delta T\)) and the first law of thermodynamics.
- Assess and propagate uncertainties in experimental determinations of \(c\) and \(L\).
14.1 Thermal Equilibrium
Zeroth Law of Thermodynamics
- If two systems are each in thermal equilibrium with a third system, they are in thermal equilibrium with each other.
- Practical consequence: a thermometer in contact with a system reads the same temperature as any other thermometer in contact with the same system.
Thermal Contact vs Thermal Isolation
- Thermal contact: objects are allowed to exchange energy (e.g. a metal block placed in water).
- Thermal isolation: the exchange of heat with the surroundings is minimised (e.g. a calorimeter surrounded by a foam jacket).
- In exam questions the phrase “insulated container” always implies thermal isolation from the laboratory.
Experimental Indicators of Equilibrium
- No net heat flow – the temperatures of the bodies no longer change when they remain in contact.
- Identical thermometer readings after a sufficient period of contact.
- Energy balance: the heat lost by the warmer body equals the heat gained by the cooler body (plus any heat absorbed by the calorimeter).
14.2 Temperature Scales
Celsius (°C)
Based on the freezing point (0 °C) and boiling point (100 °C) of liquid water at 1 atm.
Kelvin (K)
SI unit of thermodynamic temperature. Absolute zero (0 K) is the theoretical point at which molecular motion ceases. The size of one kelvin is identical to one degree Celsius, so temperature differences are the same in K and °C.
Historical Note – Fahrenheit (°F)
Fahrenheit is not required for the Cambridge 9702 exam, but is retained here for completeness:
- Freezing point of water = 32 °F, boiling point = 212 °F.
- Conversion formulas are shown in the table below but will not be marked in the exam.
Conversion Formulas
| From → To | Formula |
|---|
| Celsius → Kelvin | \(K = ^\circ\!C + 273.15\) |
| Kelvin → Celsius | \(^\circ\!C = K - 273.15\) |
| Celsius → Fahrenheit (historical) | \(^\circ\!F = \dfrac{9}{5}\,^\circ\!C + 32\) |
| Fahrenheit → Celsius (historical) | \(^\circ\!C = \dfrac{5}{9}\,(\,^\circ\!F - 32\,)\) |
Worked Conversion
Convert \(25^\circ\!C\) to Kelvin and Fahrenheit:
- K = \(25 + 273.15 = 298.15\ \text{K}\)
- \(^\circ\!F = \dfrac{9}{5}\times25 + 32 = 77^\circ\!F\) (historical only)
14.3 Specific Heat Capacity
Definition
The specific heat capacity \(c\) of a substance is the amount of heat energy required to raise the temperature of 1 kg of that substance by 1 K (or 1 °C). SI unit: \(\text{J kg}^{-1}\text{K}^{-1}\).
Constant‑Pressure vs. Constant‑Volume
- \(c_{p}\): heat capacity when the process occurs at constant pressure (the usual situation for liquids and solids in calorimetry).
- \(c_{v}\): heat capacity when the volume is held constant (important for gases).
- For solids and liquids \(c{p}\approx c{v}\) because the work term \(p\Delta V\) is negligible.
- For ideal gases \(c{p}=c{v}+R\) (where \(R\) is the gas constant).
Derivation from the First Law
For a closed system with only pressure‑volume work:
\[
\Delta U = Q - W \qquad\text{where } W = p\Delta V .
\]
If the process is at constant volume (\(\Delta V=0\)) or the work term is negligible (as in most calorimetry experiments), then
\[
\Delta U = Q = mc\Delta T .
\]
Thus the heat required to change the temperature of a mass \(m\) by \(\Delta T\) is
\[
Q = mc\Delta T .
\]
Specific vs. Molar Heat Capacity
Typical Specific Heat Capacities
| Substance | Specific Heat, \(c\) (J kg⁻¹ K⁻¹) |
|---|
| Water (liquid, 0 °C–100 °C) | 4186 |
| Ice (solid) | 2100 |
| Aluminium | 900 |
| Iron | 450 |
| Air (1 atm, 20 °C) | 1005 |
Worked Example – Heating Water
How much heat is required to raise the temperature of \(2.5\ \text{kg}\) of water from \(20^\circ\!C\) to \(80^\circ\!C\)?
- Data: \(m = 2.5\ \text{kg}\), \(c = 4186\ \text{J kg}^{-1}\text{K}^{-1}\), \(\Delta T = 80-20 = 60\ \text{K}\).
- Apply \(Q = mc\Delta T\):
\[
Q = (2.5)(4186)(60) = 6.28\times10^{5}\ \text{J}.
\]
- Result: \(\boxed{6.3\times10^{5}\ \text{J}}\) of heat must be supplied.
Uncertainty Propagation for \(Q = mc\Delta T\)
If the uncertainties are \(\delta m,\ \delta c,\ \delta\Delta T\), the relative uncertainty in \(Q\) is
\[
\frac{\delta Q}{Q}= \sqrt{\left(\frac{\delta m}{m}\right)^{2}
+ \left(\frac{\delta c}{c}\right)^{2}
+ \left(\frac{\delta\Delta T}{\Delta T}\right)^{2}} .
\]
Common Mistakes
- Using temperature differences in °F – only K or °C may be used in the equation.
- Applying the specific heat of one phase to another (e.g. using liquid‑water \(c\) for ice).
- Neglecting the heat capacity of the calorimeter, container or thermometer.
- Omitting the required uncertainty analysis.
- Confusing \(c{p}\) and \(c{v}\) for gases; remember that the exam usually deals with liquids/solids where \(c{p}\approx c{v}\).
14.4 Specific Latent Heat
Definitions
- Specific latent heat of fusion \(L_{f}\): heat required to melt 1 kg of a solid at its melting point (J kg⁻¹).
- Specific latent heat of vaporisation \(L_{v}\): heat required to vapour‑ise 1 kg of a liquid at its boiling point (J kg⁻¹).
Equations
\[
Q{\text{fusion}} = mL{f}, \qquad Q{\text{vaporisation}} = mL{v}.
\]
Typical Values
| Substance | \(L{f}\) (J kg⁻¹) | \(L{v}\) (J kg⁻¹) |
|---|
| Water | 3.34 × 10⁵ | 2.26 × 10⁶ |
| Aluminium | 3.97 × 10⁵ | — |
Worked Example – Melting Ice
Calculate the heat required to melt \(0.500\ \text{kg}\) of ice at \(0^\circ\!C\) to water at the same temperature.
\[
Q = mL_{f} = (0.500\ \text{kg})(3.34\times10^{5}\ \text{J kg}^{-1})
= 1.67\times10^{5}\ \text{J}.
\]
Link to Internal Energy
During a phase change the temperature remains constant, so \(\Delta T = 0\) and the supplied heat goes entirely into changing the internal energy associated with molecular arrangement:
\[
\Delta U = Q = mL .
\]
14.5 Practical Calorimetry
Typical Setup
- Insulated container (calorimeter) to minimise heat loss to the surroundings.
- Thermometer or temperature probe with known uncertainty (usually ±0.1 °C).
- Stirring bar or magnetic stirrer to ensure uniform temperature.
- Balance for mass measurements (uncertainty ±0.01 g).
- Optional: known calorimeter heat capacity \(C_{\text{cal}}\) (J K⁻¹).
Data‑Analysis Checklist
- Convert all temperature readings to °C or K before calculating \(\Delta T\) (the numerical value is the same for K and °C).
- Apply the appropriate equation:
- Temperature change only: \(Q = mc\Delta T\).
- Phase change: \(Q = mL\).
- If the calorimeter’s heat capacity is given: \(Q{\text{total}} = (mc + C{\text{cal}})\Delta T\).
- Propagate uncertainties using the relative‑uncertainty formula shown earlier.
- Calculate the experimental value of \(c\) or \(L\) and compare with the literature value.
- Discuss possible sources of error (heat loss, incomplete mixing, thermometer lag, heat absorbed by the container, assumptions about constant \(c\), etc.).
Uncertainty Checklist (Exam‑style)
- Identify uncertainties in each measured quantity (mass, temperature, \(c\) or \(C_{\text{cal}}\)).
- Use the relative‑uncertainty formula to obtain \(\delta Q\) or \(\delta c\).
- Report the final answer with the correct number of significant figures and the uncertainty (e.g. \(c = 4180 \pm 30\ \text{J kg}^{-1}\text{K}^{-1}\)).
14.6 Practice Questions
- Convert \(350\ \text{K}\) to Celsius and (historical) Fahrenheit.
- A \(0.75\ \text{kg}\) aluminium block is heated from \(25^\circ\!C\) to \(150^\circ\!C\). Calculate the heat supplied.
- Two substances are mixed in an insulated container:
- Substance A: \(mA = 1.00\ \text{kg},\ cA = 2000\ \text{J kg}^{-1}\text{K}^{-1},\ T_A = 30^\circ\!C\).
- Substance B: \(mB = 2.00\ \text{kg},\ cB = 1000\ \text{J kg}^{-1}\text{K}^{-1},\ T_B = 80^\circ\!C\).
Assuming no heat loss, find the final equilibrium temperature.
- Determine the heat required to melt \(250\ \text{g}\) of ice at \(-5^\circ\!C\) and then raise the resulting water to \(20^\circ\!C\). (Use \(c{\text{ice}} = 2100\ \text{J kg}^{-1}\text{K}^{-1}\), \(Lf = 3.34\times10^{5}\ \text{J kg}^{-1}\), \(c_{\text{water}} = 4186\ \text{J kg}^{-1}\text{K}^{-1}\).)
- In a calorimetry experiment a metal sample of mass \(0.120\ \text{kg}\) at \(95^\circ\!C\) is placed in \(0.250\ \text{kg}\) of water initially at \(20^\circ\!C\). The highest temperature recorded is \(23.5^\circ\!C\). Calculate the specific heat capacity of the metal, assuming the calorimeter’s heat capacity is negligible.
Summary Checklist
- Thermal equilibrium ⇒ no net heat flow; the zeroth law guarantees a single temperature reading.
- Use Kelvin for thermodynamic equations; temperature differences are the same in K and °C.
- \(Q = mc\Delta T\) – ensure \(c\) matches the phase and that \(\Delta T\) is in K or °C.
- \(Q = mL\) for phase changes – temperature stays constant.
- Distinguish specific heat (\(c\)) from molar heat capacity (\(C = cM\)).
- Remember the small difference between \(c{p}\) and \(c{v}\) for solids/liquids; for gases the distinction matters.
- Always propagate uncertainties and comment on experimental limitations.