Thermal Properties of Matter • Topic 1 of 3

Temperature & Thermal Expansion

Temperature is a measure of the degree of hotness or coldness of a body. More precisely, it tells us the average kinetic energy of the random motion of the molecules in a substance. Heat, by contrast, is the energy that flows from a hotter body to a colder one because of the temperature difference between them. Two bodies are said to be in thermal equilibrium when they are at the same temperature and no net heat flows between them.

Temperature scales. A thermometer assigns numbers to temperature using fixed reference points. Three scales are commonly used:

  • Celsius ($^\circ$C): ice point at $0\,^\circ$C and steam point at $100\,^\circ$C, with 100 equal divisions between them.
  • Fahrenheit ($^\circ$F): ice point at $32\,^\circ$F and steam point at $212\,^\circ$F, giving 180 divisions.
  • Kelvin (K): the SI absolute scale, with its zero at $-273.15\,^\circ$C (absolute zero), the lowest possible temperature.

The scales are connected by simple relations. To convert between Celsius and Fahrenheit use $\frac{C}{5}=\frac{F-32}{9}$, and between Celsius and Kelvin use $T(\text{K})=t(^\circ\text{C})+273.15$. A change of $1\,^\circ$C equals a change of $1$ K, but equals only $1.8\,^\circ$F.

Thermal expansion is the increase in the dimensions of a body when its temperature rises. Heating makes the molecules vibrate more vigorously about their mean positions, so the average spacing between them grows. There are three kinds:

  • Linear expansion (change in length): $\Delta L=\alpha L\,\Delta T$, where $\alpha$ is the coefficient of linear expansion (unit $\text{K}^{-1}$).
  • Area (superficial) expansion: $\Delta A=\beta A\,\Delta T$, with $\beta$ the coefficient of area expansion.
  • Volume (cubical) expansion: $\Delta V=\gamma V\,\Delta T$, with $\gamma$ the coefficient of volume expansion.

For an isotropic solid (one that expands equally in all directions) the three coefficients are simply related: $\beta=2\alpha$ and $\gamma=3\alpha$, so $\alpha:\beta:\gamma=1:2:3$. This is why a metal ring expands enough to slip onto a slightly larger shaft when heated, and why railway tracks and bridges are built with expansion gaps.

Anomalous expansion of water. Most substances expand on heating, but water behaves strangely between $0\,^\circ$C and $4\,^\circ$C: as it is warmed from $0\,^\circ$C it actually contracts until $4\,^\circ$C, then expands beyond that. Water therefore has its maximum density at $4\,^\circ$C. This single fact explains why ice forms on the top of a pond first while the water below stays at $4\,^\circ$C, allowing fish and aquatic life to survive a freezing winter.

Thermal expansion of a rod and temperature scalesL (at T)L + dL (at T + dT)dLdL = a L dT, b = 2a, g = 3aCKF100373212027332
Solved Examples
1
Worked Example
Convert $37\,^\circ$C (normal body temperature) into the Fahrenheit and Kelvin scales.
Solution
  1. Step 1: For Fahrenheit, use $F=\frac{9}{5}C+32$.
  2. Step 2: Substitute: $F=\frac{9}{5}\times37+32=66.6+32=98.6\,^\circ\text{F}$.
  3. Step 3: For Kelvin, $T=C+273.15=37+273.15=310.15\ \text{K}$.

Answer: $37\,^\circ\text{C}=98.6\,^\circ\text{F}\approx310\ \text{K}$.

2
Worked Example
At what temperature do the Celsius and Fahrenheit scales read the same value?
Solution
  1. Step 1: Set $C=F=x$ and use $C=\frac{5}{9}(F-32)$.
  2. Step 2: Substitute: $x=\frac{5}{9}(x-32)$, so $9x=5x-160$.
  3. Step 3: Solve: $4x=-160$, giving $x=-40$.

Answer: $-40\,^\circ\text{C}=-40\,^\circ\text{F}$.

3
Worked Example
A steel rod is 1 m long at $20\,^\circ$C. Find its increase in length when heated to $120\,^\circ$C. ($\alpha_{steel}=1.2\times10^{-5}\ \text{K}^{-1}$)
Solution
  1. Step 1: Use $\Delta L=\alpha L\,\Delta T$ with $\Delta T=120-20=100\ \text{K}$.
  2. Step 2: Substitute: $\Delta L=1.2\times10^{-5}\times1\times100$.
  3. Step 3: Compute: $\Delta L=1.2\times10^{-3}\ \text{m}=1.2\ \text{mm}$.

Answer: $\Delta L=1.2\ \text{mm}$.

4
Worked Example
A metal sheet has an area of $0.5\ \text{m}^2$ at $0\,^\circ$C. Find the increase in area when its temperature rises to $50\,^\circ$C. ($\alpha=2\times10^{-5}\ \text{K}^{-1}$)
Solution
  1. Step 1: For area expansion $\beta=2\alpha=4\times10^{-5}\ \text{K}^{-1}$ and $\Delta A=\beta A\,\Delta T$.
  2. Step 2: Substitute: $\Delta A=4\times10^{-5}\times0.5\times50$.
  3. Step 3: Compute: $\Delta A=1.0\times10^{-3}\ \text{m}^2$.

Answer: $\Delta A=1.0\times10^{-3}\ \text{m}^2$.

5
Worked Example
A glass flask of volume 1000 mL is completely filled with mercury at $0\,^\circ$C. How much mercury overflows when heated to $100\,^\circ$C? ($\gamma_{mercury}=1.8\times10^{-4}\ \text{K}^{-1}$, $\gamma_{glass}=2.7\times10^{-5}\ \text{K}^{-1}$)
Solution
  1. Step 1: Overflow corresponds to the apparent expansion, using $\gamma_{app}=\gamma_{mercury}-\gamma_{glass}$.
  2. Step 2: Compute $\gamma_{app}=1.8\times10^{-4}-2.7\times10^{-5}=1.53\times10^{-4}\ \text{K}^{-1}$.
  3. Step 3: Overflow $=\gamma_{app}V\,\Delta T=1.53\times10^{-4}\times1000\times100=15.3\ \text{mL}$.

Answer: About $15.3\ \text{mL}$ of mercury overflows.

6
Worked Example
A copper wire is heated so that its length increases by 0.2%. If $\alpha_{copper}=1.7\times10^{-5}\ \text{K}^{-1}$, find the rise in temperature.
Solution
  1. Step 1: Fractional increase $\frac{\Delta L}{L}=\alpha\,\Delta T$, and $0.2\%=2\times10^{-3}$.
  2. Step 2: Rearrange: $\Delta T=\frac{\Delta L/L}{\alpha}=\frac{2\times10^{-3}}{1.7\times10^{-5}}$.
  3. Step 3: Compute: $\Delta T\approx117.6\ \text{K}$.

Answer: $\Delta T\approx118\,^\circ\text{C}$ (or K).

Key Points

  • Temperature measures the average kinetic energy of molecules; heat is energy flowing due to a temperature difference.
  • Scale conversions: $\frac{C}{5}=\frac{F-32}{9}$ and $T(\text{K})=t(^\circ\text{C})+273.15$; absolute zero is $-273.15\,^\circ$C.
  • Linear expansion: $\Delta L=\alpha L\,\Delta T$; area: $\Delta A=\beta A\,\Delta T$; volume: $\Delta V=\gamma V\,\Delta T$.
  • For an isotropic solid $\beta=2\alpha$ and $\gamma=3\alpha$, so $\alpha:\beta:\gamma=1:2:3$.
  • Water shows anomalous expansion: it contracts from $0\,^\circ$C to $4\,^\circ$C and has maximum density at $4\,^\circ$C.
Quick Check — 4 MCQs
Tap an option to check your answer0 / 4
Q1.The relation between the coefficients of linear and volume expansion for an isotropic solid is:
Explanation: For an isotropic solid the volume coefficient is three times the linear one: $\gamma=3\alpha$.
Q2.Water has its maximum density at a temperature of:
Explanation: Because of anomalous expansion, water is densest at $4\,^\circ$C.
Q3.A temperature change of $1$ K is equal to a change of:
Explanation: The Kelvin and Celsius degrees are the same size, so $\Delta T$ of $1$ K equals $1\,^\circ$C.
Q4.The increase in length of a rod on heating is given by:
Explanation: Linear expansion follows $\Delta L=\alpha L\,\Delta T$.
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