understand that the lowest possible temperature is zero kelvin on the thermodynamic temperature scale and that this is known as absolute zero

Temperature Scales and Thermal Concepts

Learning Objectives

• Define thermal equilibrium and state the direction of spontaneous heat flow.
• Identify the Celsius and Kelvin temperature scales, convert between them, and state that 0 K is absolute zero.
• Use the Kelvin scale in thermodynamic calculations (e.g. Carnot efficiency, low‑temperature phenomena).
• Define specific heat capacity and latent heat and apply the relevant equations.
• Explain the significance of absolute zero and how it is experimentally approached.

14.1 Thermal Equilibrium

Thermal equilibrium is a *state* in which two or more bodies are at the same temperature, so that no net heat flows between them. Heat always flows spontaneously from the hotter body to the colder body until this state is reached, in accordance with the first law of thermodynamics (energy is conserved).

Typical experiment: A hot metal rod is placed in a beaker of water at room temperature. The rod cools and the water warms; after a few minutes the temperatures are equal and the system has reached thermal equilibrium.

14.2 Temperature Scales

Why Different Scales Exist

• Celsius (°C) – defined by the freezing (0 °C) and boiling (100 °C) points of water at 1 atm. It is the scale used for everyday measurements.
• Kelvin (K) – the thermodynamic (absolute) scale. Its zero point, absolute zero, is the temperature at which a perfect crystal would have minimum possible entropy. The scale is defined independently of any material; the SI definition fixes the size of the kelvin so that 0 K = –273.15 °C.
• Fahrenheit (°F) – mentioned only for historical context (see footnote).

Kelvin–Celsius Relationship

Conversion Formulas (required for the syllabus)

\[

T{\text{K}} = T{\text{°C}} + 273.15

\]

\[

T{\text{°C}} = T{\text{K}} - 273.15

\]

Optional (for broader understanding):

\[

T{\text{K}} = \frac{5}{9}\,(T{\text{°F}}-32) + 273.15

\qquad

T{\text{°F}} = \frac{9}{5}\,T{\text{K}} - 459.67

\]

Worked Example 1 – Converting 350 K to °C (AO2)

1. Apply the Kelvin–Celsius formula:
   

   \(T_{\text{°C}} = 350\;\text{K} - 273.15 = 76.85\;°\text{C}\)

Therefore, 350 K = 76.9 °C (to three significant figures).

Worked Example 2 – Converting 350 K to °F (optional, AO2)

1. First convert to °C (as above): 76.85 °C.
2. Then convert to °F:
   

   \(T_{\text{°F}} = \frac{9}{5}\times 76.85 + 32 \approx 138.3\;°\text{F}\)

Result: 350 K ≈ 138 °F.

Worked Example 3 – Carnot efficiency between 500 K and 300 K (AO2)

\[

\eta = 1 - \frac{T{\text{cold}}}{T{\text{hot}}}

= 1 - \frac{300\;\text{K}}{500\;\text{K}}

= 0.40\;(\text{40 %})

\]

The maximum theoretical efficiency of a heat engine operating between these temperatures is 40 %.

14.3 Specific Heat Capacity and Latent Heat

Definitions

• Specific heat capacity (c) – energy required to raise the temperature of 1 kg of a substance by 1 K (or 1 °C). Units: J kg⁻¹ K⁻¹.
• Latent heat (L) – energy required to change the phase of 1 kg of a substance at constant temperature. Common forms:

  • Fusion (melting) – \(L_f\)
  • Vapourisation (boiling) – \(L_v\)

Key Equations

\[

Q = m\,c\,\Delta T \qquad\text{(sensible heating)}

\]

\[

Q = m\,L \qquad\text{(phase change)}

\]

Example 1 – Heating water

Calculate the heat needed to raise 200 g of water from 20 °C to 80 °C. (\(c_{\text{water}} = 4180\;\text{J kg}^{-1}\text{K}^{-1}\))

\[

m = 0.200\;\text{kg},\quad \Delta T = 80 - 20 = 60\;\text{K}

\]

\[

Q = (0.200)(4180)(60) = 5.0\times10^{4}\;\text{J}

\]

≈ 5.0 × 10⁴ J of energy is required.

Example 2 – Melting ice

How much heat is required to melt 50 g of ice at 0 °C? (\(L_f = 3.34\times10^{5}\;\text{J kg}^{-1}\))

\[

m = 0.050\;\text{kg}

\]

\[

Q = m\,L_f = (0.050)(3.34\times10^{5}) = 1.7\times10^{4}\;\text{J}

\]

≈ 1.7 × 10⁴ J of heat must be supplied.

14.4 Absolute Zero

At absolute zero (0 K) the thermal motion of particles reaches its minimum possible value. Quantum mechanics tells us that particles still possess zero‑point energy, but no further thermal energy can be removed.

Experimental evidence

• Extrapolation of the ideal‑gas law \(PV = nRT\) to \(T = 0\) K predicts zero pressure for a fixed amount of gas.
• Specific‑heat measurements tend to zero as \(T \rightarrow 0\) K (consistent with the third law of thermodynamics).
• Observation of phenomena such as superconductivity, superfluidity and Bose–Einstein condensation only at temperatures very close to 0 K.

14.5 Practical Implications

• The Kelvin scale is the only temperature scale that can be used directly in thermodynamic equations.
• Absolute zero provides the reference point for calculating Carnot efficiencies and other theoretical limits.
• Understanding low‑temperature behaviour underpins modern technologies such as MRI, particle accelerators and quantum computers.

14.6 Experimental Determination of Absolute Zero (AO3)

A classic laboratory method:

1. Place a fixed amount of gas in a sealed container.
2. Measure its pressure at several temperatures (e.g. using a water bath). Record \(P\) versus the corresponding temperature in °C.
3. Plot the data; the points should lie on a straight line according to the ideal‑gas law.
4. Extrapolate the line to the temperature at which \(P = 0\). The intercept gives an experimental value for absolute zero (≈ –273 °C).

This activity demonstrates the use of the ideal‑gas law, the concept of a limit that cannot be reached experimentally, and the link between pressure, temperature and absolute zero.

Footnote: The Fahrenheit scale is retained here only for optional enrichment. The Cambridge AS & A Level syllabus requires knowledge of the Celsius and Kelvin scales; detailed Fahrenheit work is not examined.