Thermal Properties of Matter • Topic 3 of 3

Heat Transfer

Heat naturally flows from a region of higher temperature to one of lower temperature. There are three distinct modes of heat transfer: conduction, convection and radiation. Each operates by a different physical mechanism, and most everyday situations involve a combination of them.

Conduction is the transfer of heat through a material without any bulk movement of the material itself. Energy is passed on from molecule to molecule (and, in metals, by free electrons) as they vibrate and collide. It is the dominant mode in solids, especially metals. For steady-state conduction through a rod or slab of cross-sectional area $A$, length $L$, with a temperature difference $\Delta T$ between its ends, the rate of heat flow is:

  • $\frac{Q}{t}=\frac{kA\,\Delta T}{L}$, where $k$ is the thermal conductivity of the material (unit $\text{W}\,\text{m}^{-1}\,\text{K}^{-1}$).
  • A large $k$ means a good conductor (metals); a small $k$ means a good insulator (wood, wool, air).

Convection is the transfer of heat by the actual movement of the heated fluid (liquid or gas). When a fluid is heated it expands, becomes less dense and rises, while cooler, denser fluid sinks to take its place, setting up a convection current. Convection explains sea breezes, the circulation of air in a room heated by a radiator, and boiling water in a pan. It cannot occur in solids because the particles are not free to move.

Radiation is the transfer of heat by electromagnetic waves and requires no medium — this is how the Sun's energy reaches the Earth through empty space. Every body above absolute zero emits thermal radiation. The key results are:

  • Stefan–Boltzmann law: the energy radiated per unit area per unit time by a black body is $E=\sigma T^4$, where $\sigma=5.67\times10^{-8}\ \text{W}\,\text{m}^{-2}\,\text{K}^{-4}$ is Stefan's constant and $T$ is the absolute temperature. The total power radiated by a body of area $A$ and emissivity $e$ is $P=e\sigma A T^4$.
  • A black body is a perfect absorber and a perfect emitter of radiation (emissivity $e=1$); it is the ideal radiator against which all others are compared.
  • Wien's displacement law (intro): the wavelength $\lambda_m$ at which the radiation is most intense is inversely proportional to the absolute temperature, $\lambda_m T=b$, where $b=2.9\times10^{-3}\ \text{m}\,\text{K}$. This is why a hot iron glows first dull red, then orange, then white as its temperature rises.

Newton's law of cooling describes how a warm body loses heat to cooler surroundings. It states that the rate of loss of heat (or the rate of fall of temperature) is directly proportional to the difference in temperature between the body and its surroundings, provided this difference is small: $\frac{dT}{dt}\propto(T-T_s)$. This is why a cup of very hot tea cools quickly at first and then more slowly as it approaches room temperature.

Conduction of heat through a rod with hot and cold endsHot (T1)Cold (T2)heat flowLQ/t = k A (T1 - T2) / L
Solved Examples
1
Worked Example
A metal rod of length 0.5 m and cross-sectional area $4\times10^{-4}\ \text{m}^2$ has its ends at $100\,^\circ$C and $0\,^\circ$C. Find the rate of heat flow. ($k=200\ \text{W}\,\text{m}^{-1}\,\text{K}^{-1}$)
Solution
  1. Step 1: Use $\frac{Q}{t}=\frac{kA\,\Delta T}{L}$ with $\Delta T=100\ \text{K}$.
  2. Step 2: Substitute: $\frac{Q}{t}=\frac{200\times4\times10^{-4}\times100}{0.5}$.
  3. Step 3: Compute the numerator $=8.0$, so $\frac{Q}{t}=\frac{8.0}{0.5}=16\ \text{W}$.

Answer: $\frac{Q}{t}=16\ \text{W}$ (i.e. 16 J per second).

2
Worked Example
The thickness of a wall is doubled while everything else stays the same. How does the rate of heat conduction change?
Solution
  1. Step 1: Since $\frac{Q}{t}=\frac{kA\,\Delta T}{L}$, the rate is inversely proportional to thickness $L$.
  2. Step 2: Replacing $L$ by $2L$ gives $\frac{Q}{t}\rightarrow\frac{1}{2}\frac{Q}{t}$.
  3. Step 3: The rate of heat flow is halved.

Answer: The rate of conduction becomes one-half.

3
Worked Example
A black body is at a temperature of 500 K. Find the energy radiated per unit area per second. ($\sigma=5.67\times10^{-8}\ \text{W}\,\text{m}^{-2}\,\text{K}^{-4}$)
Solution
  1. Step 1: Use the Stefan–Boltzmann law $E=\sigma T^4$.
  2. Step 2: Compute $T^4=(500)^4=6.25\times10^{10}\ \text{K}^4$.
  3. Step 3: $E=5.67\times10^{-8}\times6.25\times10^{10}=3.54\times10^{3}\ \text{W}\,\text{m}^{-2}$.

Answer: $E\approx3.54\times10^{3}\ \text{W}\,\text{m}^{-2}$.

4
Worked Example
The absolute temperature of a black body is doubled. By what factor does its radiated energy per unit area increase?
Solution
  1. Step 1: By Stefan's law $E\propto T^4$.
  2. Step 2: Doubling $T$ gives $E'\propto(2T)^4=16T^4$.
  3. Step 3: Therefore $E'=16E$.

Answer: The radiated energy increases 16 times.

5
Worked Example
The Sun's surface temperature is about 5800 K. Estimate the wavelength of maximum emission using Wien's law. ($b=2.9\times10^{-3}\ \text{m}\,\text{K}$)
Solution
  1. Step 1: Wien's law gives $\lambda_m=\frac{b}{T}$.
  2. Step 2: Substitute: $\lambda_m=\frac{2.9\times10^{-3}}{5800}$.
  3. Step 3: Compute: $\lambda_m=5.0\times10^{-7}\ \text{m}=500\ \text{nm}$ (green light, the centre of the visible band).

Answer: $\lambda_m\approx500\ \text{nm}$.

6
Worked Example
A body cools from $80\,^\circ$C to $70\,^\circ$C in 5 minutes when the surroundings are at $20\,^\circ$C. Using Newton's law of cooling, estimate the time to cool from $70\,^\circ$C to $60\,^\circ$C.
Solution
  1. Step 1: Newton's law: $\frac{\Delta T}{t}=k\left(\bar{T}-T_s\right)$, where $\bar{T}$ is the average temperature of the interval.
  2. Step 2: First interval: average $=75\,^\circ$C, excess $=55$; $\frac{10}{5}=k\times55$, so $k=\frac{2}{55}$. Second interval: average $=65\,^\circ$C, excess $=45$.
  3. Step 3: $\frac{10}{t}=\frac{2}{55}\times45$, so $t=\frac{10\times55}{2\times45}=\frac{550}{90}\approx6.1\ \text{min}$.

Answer: About $6.1$ minutes — cooling slows as the temperature difference falls.

Key Points

  • Three modes of heat transfer: conduction (through solids, no bulk motion), convection (moving fluid), radiation (EM waves, no medium needed).
  • Conduction rate: $\frac{Q}{t}=\frac{kA\,\Delta T}{L}$; $k$ is the thermal conductivity (unit $\text{W}\,\text{m}^{-1}\,\text{K}^{-1}$).
  • Stefan–Boltzmann law: a black body radiates $E=\sigma T^4$ per unit area, with $\sigma=5.67\times10^{-8}\ \text{W}\,\text{m}^{-2}\,\text{K}^{-4}$.
  • A black body is a perfect absorber and emitter ($e=1$); Wien's law $\lambda_m T=b$ shows hotter bodies peak at shorter wavelengths.
  • Newton's law of cooling: rate of cooling $\propto(T-T_s)$ for a small temperature difference.
Quick Check — 4 MCQs
Tap an option to check your answer0 / 4
Q1.The rate of heat conduction through a rod is given by:
Explanation: Conduction rate is $\frac{Q}{t}=\frac{kA\,\Delta T}{L}$ — larger area and temperature difference increase it, larger length decreases it.
Q2.Heat transfer that does not require any material medium is:
Explanation: Radiation travels as electromagnetic waves and can pass through vacuum, as the Sun's heat reaching Earth shows.
Q3.According to the Stefan–Boltzmann law, the energy radiated per unit area by a black body is proportional to:
Explanation: $E=\sigma T^4$, so the radiated energy varies as the fourth power of the absolute temperature.
Q4.Newton's law of cooling states that the rate of cooling is proportional to:
Explanation: $\frac{dT}{dt}\propto(T-T_s)$ for small temperature differences.
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