The Workbook

\(H\) is the heat current, the heat crossing a section per second. \(K\) is the thermal conductivity of the material, \(A\) the area of cross-section and \(\dfrac{dT}{dx}\) the temperature gradient.

Heat flows from hot to cold, that is towards decreasing \(T\). Where \(\dfrac{dT}{dx}\) is negative, \(H\) is positive along \(x\): the minus sign makes this so.

\(K = \dfrac{H}{A\,(dT/dx)}\) has the unit \(\dfrac{\text{W}}{\text{m}^2 \times \text{K/m}} = \) W/m K, and dimensions \(\dfrac{[\mathrm{ML^2T^{-3}}]}{[\mathrm{L^2}][\mathrm{KL^{-1}}]} = [\mathrm{MLT^{-3}K^{-1}}]\).

In the steady state the temperature at each point of the rod no longer changes with time, and the same heat current passes every section.

Gradient \(= \dfrac{100 - 0}{1} = 100\) K/m. At \(0.3\) m from the hot end, \(T = 100 - 100 \times 0.3 = 70^\circ\)C.

\(H = KA\dfrac{\Delta T}{L} = 400 \times 10^{-4} \times \dfrac{100}{1} = 4\) W.

\(R = \dfrac{L}{KA}\) is the temperature difference needed per unit heat current, in K/W. \(H\) plays the part of the current \(I\), \(\Delta T\) of the potential difference \(V\) and \(R\) of the electrical resistance. \(H = \dfrac{\Delta T}{R}\) is Ohm's law for heat.

\(R = \dfrac{1}{400 \times 10^{-4}} = 25\) K/W. \(H = \dfrac{100}{25} = 4\) W, as before.

1. The rod is insulated, so the same \(H\) crosses every section. With \(K\) and \(A\) uniform, \(\dfrac{dT}{dx} = -\dfrac{H}{KA}\) is the same everywhere: a constant slope.

2. At \(x = 0.3\) m the line gives \(70^\circ\)C.

3. Heat leaks out of the sides, so the current along the rod falls with \(x\) and the gradient becomes smaller further along. The graph curves, steep near the hot end and flatter near the cold end.

1. \(A\): \(\dfrac{100 - 80}{20} = 1\) \(^\circ\)C per cm. \(B\): \(\dfrac{80 - 20}{20} = 3\) \(^\circ\)C per cm.

2. \(A\). The same heat current passes through both, and \(H = KA\dfrac{dT}{dx}\). The material with the smaller gradient must have the larger \(K\).

3. \(K_A \times 1 = K_B \times 3\), so \(K_A : K_B = 3 : 1\).

The same \(H\) flows through both rods, and the temperature drops add: \(\Delta T = HR_1 + HR_2 = H(R_1 + R_2)\). So \(R = R_1 + R_2\).

\(\dfrac{2L}{KA} = \dfrac{L}{K_1A} + \dfrac{L}{K_2A}\), so \(\dfrac{2}{K} = \dfrac{1}{K_1} + \dfrac{1}{K_2}\) and \(K = \dfrac{2K_1K_2}{K_1 + K_2}\).

\(K = \dfrac{2 \times 400 \times 100}{500} = 160\) W/m K.

\(R \propto \dfrac{1}{K}\) for equal \(L\) and \(A\), so \(R_1 : R_2 = \dfrac{1}{400} : \dfrac{1}{100} = 1 : 4\).

The temperature drop is shared in the ratio of the resistances. Drop across rod 1 \(= \dfrac{1}{5} \times 100 = 20^\circ\)C, so the junction is at \(80^\circ\)C.

1. \(R = \dfrac{L}{KA}\): \(R_1 : R_2 = \dfrac{0.1}{400} : \dfrac{0.3}{300} = 2.5 \times 10^{-4} : 10^{-3} = 1 : 4\).

2. Drop across rod 1 \(= \dfrac{1}{5} \times 100 = 20^\circ\)C. Junction temperature \(= 80^\circ\)C.

3. \(R_1 = \dfrac{0.1}{400 \times 2 \times 10^{-4}} = 1.25\) K/W and \(R_2 = \dfrac{0.3}{300 \times 2 \times 10^{-4}} = 5\) K/W. \(H = \dfrac{\Delta T}{R_1 + R_2} = \dfrac{100}{6.25} = 16\) W.

Both rods have the same \(\Delta T\) across them, so the currents add: \(H = K_1A_1\dfrac{\Delta T}{L} + K_2A_2\dfrac{\Delta T}{L} = K(A_1 + A_2)\dfrac{\Delta T}{L}\).

So \(K = \dfrac{K_1A_1 + K_2A_2}{A_1 + A_2}\). For equal areas \(K = \dfrac{400 + 100}{2} = 250\) W/m K.

1. Copper: \(H_1 = 400 \times 10^{-4} \times 100 = 4\) W. Steel: \(H_2 = 50 \times 10^{-4} \times 100 = 0.5\) W.

2. \(H = 4.5\) W. \(R = \dfrac{\Delta T}{H} = \dfrac{100}{4.5} = 22.2\) K/W. Check: \(R_1 = 25\) and \(R_2 = 200\) K/W in parallel give \(\dfrac{25 \times 200}{225} = 22.2\) K/W.

Wall alone: \(H = KA\dfrac{\Delta T}{L} = 1 \times 10 \times \dfrac{20}{0.2} = 1000\) W.

With the board: \(R_{wall} = \dfrac{0.2}{1 \times 10} = 0.02\) K/W and \(R_{board} = \dfrac{0.02}{0.05 \times 10} = 0.04\) K/W. \(H = \dfrac{20}{0.06} = 333\) W. A thin board of poor conductor cuts the loss to a third.

In the steady state no heat collects at the joint. So the current entering the joint from slab 1 equals the current leaving into slab 2: the same \(H\) passes both.

Slab 1: \(T_1 - T_j = HR_1\). Slab 2: \(T_j - T_2 = HR_2\). Adding, \(T_1 - T_2 = H(R_1 + R_2)\), so \(H = \dfrac{\Delta T}{R_1 + R_2}\).

Drop across the wall \(= HR_{wall} = 333 \times 0.02 = 6.7^\circ\)C. Interface temperature \(= 25 - 6.7 = 18.3^\circ\)C. Most of the drop, \(13.3^\circ\)C, is across the thin board.

The hand is warmer than the spoon. Metal has a large \(K\), so it carries heat away from the skin quickly and the skin cools. Wood has a small \(K\), so the skin loses heat slowly. The nerves sense the rate of heat loss, not the temperature of the spoon.

1. Gradient \(= \dfrac{50 - 80}{50 - 20} = -1\) \(^\circ\)C per cm. The temperature falls by \(1^\circ\)C for every centimetre along the rod.

2. \(T_1 = 80 + 20 \times 1 = 100^\circ\)C at \(x = 0\). \(T_2 = 50 - 50 \times 1 = 0^\circ\)C at \(x = 100\) cm.

\(H = KA\dfrac{\Delta T}{L} = 0.05 \times 1 \times \dfrac{30}{0.05} = 30\) W.

In one hour \(Q = 30 \times 3600 = 1.08 \times 10^5\) J. Mass melted \(= \dfrac{1.08 \times 10^5}{3.36 \times 10^5} = 0.32\) kg.

1. \(H \propto \Delta T\): doubles.

2. \(H \propto \dfrac{1}{L}\): halves.

3. \(H \propto A \propto r^2\): four times.

At every section the heat arriving from the hotter side each second equals the heat leaving to the colder side. No heat is stored in any part of the rod, so no part warms or cools. The rod is like a pipe with water flowing steadily through it: full of flow, but with nothing changing.

\(H = KA\dfrac{\Delta T}{L} \propto \dfrac{r^2}{L}\). \(\dfrac{H_1}{H_2} = \dfrac{1^2/1}{2^2/2} = \dfrac{1}{2}\), so \(1 : 2\).

Parallel. In series the resistance is \(2R\), so \(H_s = \dfrac{\Delta T}{2R}\). In parallel it is \(\dfrac{R}{2}\), so \(H_p = \dfrac{2\Delta T}{R}\). \(H_p : H_s = 4 : 1\).

\(K = \dfrac{HL}{A\Delta T}\): \([K] = \dfrac{[\mathrm{ML^2T^{-3}}][\mathrm{L}]}{[\mathrm{L^2}][\mathrm{K}]} = [\mathrm{MLT^{-3}K^{-1}}]\).

\(R = \dfrac{\Delta T}{H}\): \([R] = \dfrac{[\mathrm{K}]}{[\mathrm{ML^2T^{-3}}]} = [\mathrm{M^{-1}L^{-2}T^{3}K}]\).

Heat conducted up through the ice (area \(A\)) in time \(dt\): \(dQ = KA\dfrac{\theta}{x}dt\). This heat comes from the latent heat of the layer that freezes: \(dQ = (\rho A\,dx)L\).

Equating: \(\rho L\,dx = \dfrac{K\theta}{x}dt\), so \(x\,dx = \dfrac{K\theta}{\rho L}dt\).

Integrate from \(x_1\) to \(x_2\): \(\dfrac{x_2^2 - x_1^2}{2} = \dfrac{K\theta}{\rho L}t\), so \(t = \dfrac{\rho L}{2K\theta}(x_2^2 - x_1^2)\).

\(t = \dfrac{\rho L}{2K\theta}x^2 = \dfrac{900 \times 3.36 \times 10^5}{2 \times 2 \times 20} \times (0.1)^2 = \dfrac{3.024 \times 10^8}{80} \times 0.01 = 37800\) s.

That is \(\dfrac{37800}{3600} = 10.5\) hours.

\(t = \dfrac{\rho L}{2K\theta}(x_2^2 - x_1^2) = \dfrac{3.024 \times 10^8}{80} \times (0.04 - 0.01) = 3.78 \times 10^6 \times 0.03 = 113400\) s \(= 31.5\) h.

The second \(10\) cm takes three times as long as the first, although it is the same thickness.

\(t \propto x_2^2 - x_1^2\). From \(0\) to \(x\): \(x^2\). From \(x\) to \(2x\): \(4x^2 - x^2 = 3x^2\). Ratio \(1 : 3\).

1. Upward, from the water side at \(0^\circ\)C to the air side at \(-\theta\). Heat always flows from the warmer face to the colder face.

2. At the lower face, where water turns to ice. The latent heat released there is exactly the heat conducted up through the ice to the cold air. That is why the ice grows only as fast as the ice can conduct.

3. Water at \(0^\circ\)C touching ice at \(0^\circ\)C is at the freezing point. Any water that cools further freezes, joining the ice. Below the ice the water is at \(0^\circ\)C at the top and up to \(4^\circ\)C at the bottom, where the densest water sinks.

\(\dfrac{dx}{dt} = \dfrac{K\theta}{\rho L x} = \dfrac{2 \times 20}{900 \times 3.36 \times 10^5 \times 0.05} = \dfrac{40}{1.512 \times 10^7} = 2.65 \times 10^{-6}\) m/s.

In one hour: \(2.65 \times 10^{-6} \times 3600 = 9.5 \times 10^{-3}\) m \(= 9.5\) mm per hour.

The latent heat released at the bottom must be conducted up through the whole thickness of ice. The temperature difference across the ice stays at \(\theta\), so the heat current \(\dfrac{K\theta}{x}\) falls as \(x\) grows. Less heat removed per second means less ice frozen per second: \(\dfrac{dx}{dt} \propto \dfrac{1}{x}\).

1. From \(x\,dx = \dfrac{K\theta}{\rho L}dt\) with \(x = 0\) at \(t = 0\): \(x^2 = \dfrac{2K\theta}{\rho L}t\). So \(x^2\) is proportional to \(t\): a straight line through the origin.

2. \(x = 10\) cm means \(x^2 = 100\) cm\(^2\): \(t = 10.5\) h. \(x = 20\) cm means \(x^2 = 400\) cm\(^2\): \(t = 42\) h.

3. Slope \(= \dfrac{100 \text{ cm}^2}{10.5 \text{ h}} = \dfrac{10^{-2}}{37800} = 2.65 \times 10^{-7}\) m\(^2\)/s. \(\dfrac{2K\theta}{\rho L} = \dfrac{80}{3.024 \times 10^8} = 2.65 \times 10^{-7}\) m\(^2\)/s. They agree.

The thin shell has area \(2\pi r l\), so \(H = -K(2\pi r l)\dfrac{dT}{dr}\). In the steady state \(H\) is the same for every \(r\).

Separate: \(dT = -\dfrac{H}{2\pi K l}\dfrac{dr}{r}\). Integrate from \(r_1\) to \(r_2\): \(T_2 - T_1 = -\dfrac{H}{2\pi K l}\ln\dfrac{r_2}{r_1}\), so \(H = \dfrac{2\pi K l(T_1 - T_2)}{\ln(r_2/r_1)}\).

\(H = \dfrac{2\pi K l(T_1 - T_2)}{\ln(r_2/r_1)} = \dfrac{2 \times 3.14 \times 0.1 \times 1 \times 70}{\ln 2} = \dfrac{43.96}{0.693} = 63.4\) W per metre.

1. \(R = \dfrac{\ln(r_2/r_1)}{2\pi K l} = \dfrac{0.693}{2 \times 3.14 \times 0.1 \times 1} = \dfrac{0.693}{0.628} = 1.10\) K/W.

2. \(H = \dfrac{\Delta T}{R} = \dfrac{70}{1.10} = 63.4\) W, as in Question 10.

3. \(H \propto \dfrac{1}{\ln(r_2/r_1)}\). The ratio \(r_2/r_1\) goes from \(2\) to \(4\), and \(\ln 4 = 2\ln 2\), so \(H\) falls to half. Doubling the thickness of insulation does not halve the loss unless the ratio of radii doubles.

A thin shell of radius \(r\) has area \(4\pi r^2\): \(H = -4\pi K r^2\dfrac{dT}{dr}\), the same for every \(r\).

\(dT = -\dfrac{H}{4\pi K}\dfrac{dr}{r^2}\). Integrating from \(r_1\) to \(r_2\): \(T_2 - T_1 = -\dfrac{H}{4\pi K}\left(\dfrac{1}{r_1} - \dfrac{1}{r_2}\right) = -\dfrac{H}{4\pi K}\dfrac{r_2 - r_1}{r_1 r_2}\).

So \(H = \dfrac{4\pi K r_1 r_2(T_1 - T_2)}{r_2 - r_1}\).

1. \(H = \dfrac{4 \times 3.14 \times 0.5 \times 0.1 \times 0.2 \times 50}{0.1} = 62.8\) W.

2. \(\dfrac{1/r - 1/r_2}{1/r_1 - 1/r_2} = \dfrac{1/15 - 1/20}{1/10 - 1/20} = \dfrac{1/60}{1/20} = \dfrac{1}{3}\). \(T = 50 + \dfrac{1}{3} \times 50 = 66.7^\circ\)C. The midpoint is much nearer the cold temperature, because most of the drop happens near the inner surface.

1. The same heat current passes through every cylindrical surface, but the area \(2\pi r l\) grows with \(r\). So the gradient \(\dfrac{dT}{dr} = -\dfrac{H}{2\pi K l r}\) falls as \(\dfrac{1}{r}\), and \(T\) falls logarithmically, not linearly.

2. At the inner surface, where \(r\) is smallest and the area is least. The same heat must squeeze through a smaller area, which needs a steeper gradient.

\(H = -4\pi K r^2\dfrac{dT}{dr}\) is the same at every \(r\), so \(\dfrac{dT}{dr} \propto \dfrac{1}{r^2}\).

\(\dfrac{(dT/dr)_{r_1}}{(dT/dr)_{r_2}} = \dfrac{r_2^2}{r_1^2} = \left(\dfrac{r_2}{r_1}\right)^2\).

With \(r_2 - r_1 = d\) and \(r_1 r_2 \approx r_1^2\): \(H \approx \dfrac{4\pi K r_1^2\,\Delta T}{d} = KA\dfrac{\Delta T}{d}\), the slab formula with \(A = 4\pi r_1^2\).

Exact: \(r_1 r_2 = 0.1 \times 0.105 = 0.0105\). Approximate: \(r_1^2 = 0.01\). The approximation is low by \(\dfrac{0.0005}{0.0105} \approx 5\%\).

\(t \propto x^2\) from the start. \(4\) cm takes \(4 \times 4 = 16\) h in all, so the second \(2\) cm takes \(16 - 4 = 12\) h.

\(x^2 \propto \theta t\). For the same \(t\), \(x \propto \sqrt{\theta}\): \(x = 5 \times \sqrt{2} = 7.07\) cm.

\(H \propto \dfrac{1}{\ln(r_2/r_1)}\). \(\dfrac{H_2}{H_1} = \dfrac{\ln 2}{\ln 4} = \dfrac{1}{2}\).

\(t \propto x^2\): \(t = 1 \times 5^2 = 25\) h.

As \(r_2 \to \infty\), \(\dfrac{r_1 r_2}{r_2 - r_1} \to r_1\), so \(H \to 4\pi K r_1(T_1 - T_2)\), which is not zero.

The extra insulation far from the sphere has a very large area, so it adds very little resistance. Almost all the resistance lies in the first few radii around the inner surface.