A rod between hot and cold
Take a copper rod 1 m long, first at 0 °C. Hold its left end against a body at 100 °C and its right end in melting ice at 0 °C. Wrap its sides in lagging, a poor conductor such as glass wool, so that heat can leave only through the ends. Now watch what happens along the rod.
At first the hot end pours heat into the cold rod, more than 20 W, while almost none reaches the ice. The heat that enters is used to warm the rod itself. Slowly each part of the rod warms, and the heat current leaving the cold end grows. After about an hour the two currents are equal. Every slice now passes on exactly the heat it receives, so no slice warms any more.
A body conducting heat is in the steady state when the temperature at every point no longer changes with time. Each part then gives out as much heat as it takes in, so the same heat current crosses every section of a lagged rod.
This is how a rod can be full of flowing heat while nothing in it changes. Water flowing steadily through a pipe is a good picture: the pipe is full of flow, but the level at each point stays the same. In the steady state the rod stores no more heat; it only passes heat on.
The temperature gradient
The steady profile of Fig 2.1 is a straight line from 100 °C down to 0 °C. The fall of temperature per metre along the rod is its temperature gradient, \(dT/dx\). Here the temperature falls 100 K over 1 m, so \(dT/dx = -100\) K/m. The minus sign says the temperature falls as \(x\) grows. At \(0.3\) m from the hot end, the rod is at \(100 - 100 \times 0.3 = 70\) °C.
Why a straight line? The same heat current crosses every section of the lagged rod. A uniform rod needs the same gradient to push the same current everywhere. So the gradient is constant, and a constant slope is a straight line. On a bare rod, heat also leaks out of the sides. Less heat reaches the far part, which needs a smaller gradient, so the profile sags into a curve.
A lagged uniform rod \(2\,\text{m}\) long has its ends at \(80\,°\text{C}\) and \(20\,°\text{C}\) in the steady state. What is the temperature \(0.5\,\text{m}\) from the hot end?
In the steady state, heat keeps flowing through a rod, yet the temperature at every point stays constant. Explain how both can be true.
A point's temperature rises only if heat is stored there, that is, if more heat arrives than leaves. In the steady state every slice of the rod passes on to its colder side exactly the heat it receives from its hotter side. No heat is stored anywhere, so no temperature changes, although heat flows through the whole rod all the time.
Fourier's law
How large is the heat current? Call the heat that crosses a section per second the heat current, \(H = dQ/dt\), measured in watts. Experiment shows three things. \(H\) doubles when the area of cross-section doubles. It doubles when the temperature difference doubles. It halves when the rod is twice as long. A fourth factor is the material, measured by its thermal conductivity \(k\).
The heat current through a small slab of area \(A\) is proportional to the area and to the temperature gradient across it: \(H = -kA\,dT/dx\). The constant \(k\) is the thermal conductivity of the material.
The minus sign has a job. Heat flows towards lower temperature. Where the temperature falls along \(+x\), \(dT/dx\) is negative, and the minus sign makes \(H\) positive along \(+x\). For the copper rod, \(dT/dx = -100\) K/m, and \(H\) comes out positive, flowing from hot to cold. The unit of \(k\) follows from the law: W/(m K). Its dimensions are \([\text{M}\,\text{L}\,\text{T}^{-3}\,\text{K}^{-1}]\).
The copper rod of Fig 2.1 is \(1\,\text{m}\) long with a cross-section of \(1\,\text{cm}^2\). Its ends are at \(100\,°\text{C}\) and \(0\,°\text{C}\) in the steady state. Take \(k = 400\,\text{W/(m K)}\) for copper. Find the heat current.
Convert the area: \(1\,\text{cm}^2 = 10^{-4}\,\text{m}^2\).
\(H = kA\,\Delta T/L = 400 \times 10^{-4} \times 100/1 = 4\,\text{W}\).
So 4 J cross every section of the rod each second. They enter at the hot end and leave into the ice, as Fig 2.1 shows at the end.
An aluminium rod is \(50\,\text{cm}\) long and \(2\,\text{cm}^2\) in cross-section, with its sides lagged. Its ends are kept \(100\,\text{K}\) apart. Take \(k = 205\,\text{W/(m K)}\). Find the heat current in the steady state.
A lagged rod carries heat in the steady state. Its radius is doubled, with its length and the temperatures of its ends unchanged. The heat current becomes
A lagged rod is made of two materials A and B of equal length and equal cross-section, joined end to end. In the steady state the temperature falls from \(100\,°\text{C}\) to \(80\,°\text{C}\) across A and from \(80\,°\text{C}\) to \(20\,°\text{C}\) across B. Find the ratio \(k_A/k_B\).
Joseph Fourier was a mathematician and a prefect under Napoleon. In 1807 he wrote the law that the heat current is proportional to the temperature gradient. From it he found an equation for how temperature spreads through a body. To solve it he added up many sine waves, just as the curve of Fig 2.1 is drawn from six of them. The judges of the Paris Academy doubted his method, but it grew into Fourier series, now used in every branch of science.
From copper to still air
Fig 2.2 shows the thermal conductivities of some common materials. The scale goes up ten times at each step, because the range is enormous. Metals, with their free electrons, sit at the top. Non-metals such as glass, brick and water are hundreds of times lower. Still air is lowest of all, about 17 000 times below copper.
A material full of small pockets of still air takes on the low conductivity of air. The air cannot flow in the tiny pockets, so it cannot carry heat by convection. Wool, cotton wadding, feathers and thermocol all work this way. Their conductivity is close to 0.04 W/(m K). This book takes 0.04 W/(m K) for lagging.
A kulfi-wala's box has walls of foam \(5\,\text{cm}\) thick, of total area \(1\,\text{m}^2\), with \(k = 0.04\,\text{W/(m K)}\). It holds ice at \(0\,°\text{C}\) on a street at \(30\,°\text{C}\). Find the heat flowing in each second and the mass of ice that melts in one hour. Take \(L_f = 3.34 \times 10^5\,\text{J/kg}\).
\(H = kA\,\Delta T/L = 0.04 \times 1 \times 30/0.05 = 24\,\text{W}\).
In one hour, \(Q = 24 \times 3600 = 86\,400\,\text{J}\).
Ice melted: \(Q/L_f = 86\,400/(3.34 \times 10^5) \approx 0.26\,\text{kg}\). The foam lets the box keep its ice for a whole working day.
On a winter night in Lucknow, a sleeper's skin is at about \(33\,°\text{C}\) and the room at \(8\,°\text{C}\). A quilt of area \(2\,\text{m}^2\) and thickness \(4\,\text{cm}\) covers him, with \(k = 0.04\,\text{W/(m K)}\). Estimate the heat escaping through the quilt, in watts.
Two tawas are alike except for the handle, one of steel (\(k = 50\)) and one of wood (\(k = 0.12\,\text{W/(m K)}\)), of the same size. About how many times more heat flows along the steel handle?
In 1789 Jan Ingenhousz coated rods of different metals with wax and dipped one end of each into hot water. In the steady state the wax melted farther along the better conductors. The melted length grows as the square root of \(k\). So along copper (400) the wax melts about \(\sqrt{400/50} \approx 2.8\) times as far as along steel (50). The same experiment is still done in school laboratories today.
- Using \(H = kA\,\Delta T/L\) while the rod is still warming up.That form holds only in the steady state of a lagged uniform rod; before that the profile is not straight.
- Leaving the area in cm² or the length in cm.Put \(A\) in m² and \(L\) in m before using \(k\) in W/(m K).
- Thinking a larger temperature gradient means a better conductor.For the same heat current, the better conductor has the smaller gradient.