The Workbook

Every body at a temperature above absolute zero radiates heat at all times, at a rate that depends on its temperature. It also absorbs the radiation falling on it from its surroundings.

No. At the same temperature as its surroundings the body still radiates, but it absorbs exactly as much as it emits. The exchange is dynamic, and the net loss is zero.

\(a\), \(r\) and \(t\) are the fractions of the incident radiant energy that are absorbed, reflected and transmitted. Energy is conserved, so \(a + r + t = 1\).

A perfectly black body absorbs everything: \(a = 1\), \(r = t = 0\). Here \(a = 1 - 0.3 - 0.1 = 0.6\).

1. At a given temperature the ratio of emissive power to absorptive power is the same for all surfaces. It equals the emissive power of a black body. Good absorbers are good emitters.

2. The black can. A dull black surface absorbs well, so by Kirchhoff's law it emits well and radiates heat away faster. The polished surface is a poor absorber, so it is a poor emitter.

3. The black can again: a good absorber takes in the sunlight, while the polished can reflects most of it.

\(P = \sigma AT^4\) for a black body; \(P = e\sigma AT^4\) for a grey body with emissivity \(e\) between \(0\) and \(1\).

\(P = 5.67 \times 10^{-8} \times 0.1 \times 500^4 = 5.67 \times 10^{-9} \times 6.25 \times 10^{10} = 354\) W.

The body radiates \(e\sigma AT^4\) and absorbs \(e\sigma AT_0^4\) from the surroundings, so the net loss is \(P = e\sigma A(T^4 - T_0^4)\).

\(P = 0.5 \times 5.67 \times 10^{-8} \times 0.1 \times (6.25 \times 10^{10} - 0.81 \times 10^{10}) = 2.835 \times 10^{-9} \times 5.44 \times 10^{10} = 154\) W.

\(P \propto T^4\): doubling \(T\) makes \(P\) sixteen times. For \(P\) to become four times, \(T^4\) must become four times, so \(T\) rises by a factor \(4^{1/4} = \sqrt{2} = 1.41\).

\(P = \sigma A(T^4 - T_0^4) = 5.67 \times 10^{-8} \times 1.5 \times (8.1 \times 10^9 - 7.07 \times 10^9) = 8.5 \times 10^{-8} \times 1.03 \times 10^9 = 87\) W.

A resting person produces about \(100\) W, so most of it is lost as radiation.

The wavelength at which a black body radiates most strongly is inversely proportional to its absolute temperature: \(\lambda_{max}T = b\).

Sun: \(T = \dfrac{2.9 \times 10^{-3}}{5 \times 10^{-7}} = 5800\) K. Body at \(300\) K: \(\lambda_{max} = \dfrac{2.9 \times 10^{-3}}{300} = 9.7 \times 10^{-6}\) m \(= 9.7\ \mu\)m, in the infrared. That is why warm bodies do not glow visibly.

1. \(T_1\). Its peak is at a shorter wavelength (Wien's law), and its curve is higher everywhere, so it radiates more at every wavelength (Stefan's law).

2. The area under the curve is the total power radiated per unit area, \(\sigma T^4\). If \(T_1 = 2T_2\) the areas are in the ratio \(2^4 = 16 : 1\).

The peak is at \(\lambda_{max} = 1\ \mu\)m. \(T = \dfrac{b}{\lambda_{max}} = \dfrac{2.9 \times 10^{-3}}{10^{-6}} = 2900\) K, about the temperature of a bulb filament.

\(T = \dfrac{2.9 \times 10^{-3}}{2.5 \times 10^{-7}} = 11600\) K, twice the Sun's temperature.

Its peak is in the ultraviolet, so within the visible range it emits more blue than red. The star looks bluish-white; the Sun, peaking in the green-yellow, looks yellowish-white.

\(P \propto T^4\): \(\dfrac{P_1}{P_2} = \left(\dfrac{300}{600}\right)^4 = \dfrac{1}{16}\), so \(1 : 16\).

\(\lambda_{max} \propto \dfrac{1}{T}\): \(\dfrac{\lambda_1}{\lambda_2} = \dfrac{600}{300} = 2\), so \(2 : 1\).

At the Earth's distance the Sun's power is spread evenly over a sphere of radius \(d\): \(P = 4\pi d^2 S = 4 \times 3.14 \times (1.5 \times 10^{11})^2 \times 1400 = 12.56 \times 2.25 \times 10^{22} \times 1400 = 3.96 \times 10^{26}\) W.

1. The Sun radiates equally in all directions. All of its power crosses the sphere of radius \(d\) centred on the Sun, whose area is \(4\pi d^2\). \(S\) is the power per unit area on that sphere, so \(P = 4\pi d^2 S\).

2. The Earth blocks a disc of area \(\pi R_E^2\) facing the Sun: \(P_E = \pi R_E^2 S = 3.14 \times (6.4 \times 10^6)^2 \times 1400 = 3.14 \times 4.1 \times 10^{13} \times 1400 = 1.8 \times 10^{17}\) W.

\(P = \sigma(4\pi R^2)T^4\), so \(T^4 = \dfrac{P}{4\pi R^2\sigma} = \dfrac{3.96 \times 10^{26}}{12.56 \times 4.9 \times 10^{17} \times 5.67 \times 10^{-8}} = \dfrac{3.96 \times 10^{26}}{3.49 \times 10^{11}} = 1.13 \times 10^{15}\).

\(T = (1.13 \times 10^{15})^{1/4} = 5800\) K, the same as the value from Wien's law. Two different laws give the same temperature, which confirms that the Sun radiates nearly as a black body.

Absorbed: \(\pi R_E^2 S\). Radiated: \(\sigma(4\pi R_E^2)T^4\). In the steady state they are equal, so \(T^4 = \dfrac{S}{4\sigma}\) and \(T = \left(\dfrac{S}{4\sigma}\right)^{1/4}\).

\(T^4 = \dfrac{1400}{4 \times 5.67 \times 10^{-8}} = 6.17 \times 10^9\), so \(T = 280\) K, about \(7^\circ\)C. The real average is about \(288\) K; the difference is the greenhouse effect of the atmosphere.

1. A ray that enters the hole strikes the inner wall, and part of it is absorbed. The rest is reflected to another part of the wall, where more is absorbed. After many reflections almost nothing finds its way back out through the small hole. So the hole absorbs nearly all the radiation that enters: \(a \approx 1\).

2. Black-body radiation at the temperature of the walls. Its spectrum depends only on the temperature, not on the material of the box.

The emissive power \(E\) is the radiant energy emitted per second per unit area of the surface, in W/m\(^2\).

Kirchhoff's law: \(\dfrac{E}{a} = E_b\), the same for every surface at a given temperature. So \(E = aE_b\): a surface that absorbs a fraction \(a\) emits the same fraction of the black-body value.

\(P = e\sigma AT^4\), so \(A = \dfrac{P}{e\sigma T^4} = \dfrac{100}{0.5 \times 5.67 \times 10^{-8} \times 8.1 \times 10^{13}} = \dfrac{100}{2.3 \times 10^6} = 4.4 \times 10^{-5}\) m\(^2 = 0.44\) cm\(^2\).

White cloth reflects most of the sunlight, so it absorbs little heat. Dark cloth absorbs most of it. A silvered surface is a poor absorber and, by Kirchhoff's law, a poor emitter. The silvering on both walls of the flask stops heat from leaving the hot liquid by radiation, or from entering a cold one.

\(T^4 - T_0^4 = (T^2 + T_0^2)(T + T_0)(T - T_0)\). For \(T\) close to \(T_0\), \(T^2 + T_0^2 \approx 2T_0^2\) and \(T + T_0 \approx 2T_0\), so \(T^4 - T_0^4 \approx 4T_0^3(T - T_0)\).

The net power is then \(4e\sigma AT_0^3(T - T_0)\), proportional to the excess temperature. This is Newton's law of cooling.

\(\sigma = \dfrac{P}{AT^4}\): unit \(\dfrac{\text{W}}{\text{m}^2\,\text{K}^4} = \) W m\(^{-2}\) K\(^{-4}\).

\(b = \lambda_{max}T\): unit m K.

The rate of loss of heat of a body is proportional to its excess temperature over the surroundings. So is the rate of fall of temperature: \(\dfrac{dT}{dt} = -k(T - T_0)\). \(T_0\) is the temperature of the surroundings and \(k\) is a constant that depends on the body's surface, area and heat capacity.

The law holds well when the excess temperature is small, up to a few tens of degrees. It holds best when there is forced convection, as in a steady draught.

Heat lost per second \(= hA(T - T_0)\). The temperature falls by \(dT\) when heat \(ms\,dT\) is lost: \(-ms\dfrac{dT}{dt} = hA(T - T_0)\), so \(\dfrac{dT}{dt} = -\dfrac{hA}{ms}(T - T_0)\) and \(k = \dfrac{hA}{ms}\).

The small body. For the same shape \(A \propto r^2\) and \(m \propto r^3\), so \(k \propto \dfrac{1}{r}\). A small body has more surface per unit mass and cools faster. This is why small animals lose heat quickly.

\(\dfrac{80 - 70}{5} = k(75 - 30)\), so \(2 = 45k\) and \(k = \dfrac{2}{45}\) per minute.

From \(70\) to \(60\): \(\dfrac{10}{t} = \dfrac{2}{45}(65 - 30) = \dfrac{70}{45}\), so \(t = \dfrac{450}{70} = 6.4\) minutes. The second fall takes longer because the excess is smaller.

Write \(\theta = T - T_0\): \(\dfrac{d\theta}{dt} = -k\theta\), so \(\dfrac{d\theta}{\theta} = -k\,dt\). Integrating, \(\ln\theta = -kt + \) constant, and with \(\theta = T_1 - T_0\) at \(t = 0\): \(\theta = (T_1 - T_0)e^{-kt}\).

\(\dfrac{60 - 20}{100 - 20} = \dfrac{1}{2} = e^{-10k}\), so \(k = \dfrac{\ln 2}{10} = 0.0693\) per minute.

From \(60\) to \(40\): \(\dfrac{40 - 20}{60 - 20} = \dfrac{1}{2} = e^{-kt}\), so \(t = \dfrac{\ln 2}{k} = 10\) minutes again. Each halving of the excess takes the same time.

Excess at start \(= 80^\circ\)C. After \(30\) minutes, three halvings: \(\dfrac{80}{8} = 10^\circ\)C. Temperature \(= 20 + 10 = 30^\circ\)C.

1. \(60^\circ\)C at \(10\) min and \(40^\circ\)C at \(20\) min. The excess goes \(80 \to 40 \to 20\), halving every \(10\) minutes.

2. At \(40\) min the excess has halved four times: \(\dfrac{80}{16} = 5^\circ\)C. \(T = 25^\circ\)C.

3. The rate of cooling is proportional to the excess. As the body approaches room temperature the excess shrinks, so the slope of the curve shrinks with it.

1. Taking logs of \(T - T_0 = (T_1 - T_0)e^{-kt}\): \(\ln(T - T_0) = \ln(T_1 - T_0) - kt\). This is a straight line in \(t\) with slope \(-k\) and intercept \(\ln(T_1 - T_0)\).

2. Slope \(= \dfrac{3.69 - 4.38}{10} = -0.069\) per minute, so \(k = 0.069\) per minute, in agreement with \(\dfrac{\ln 2}{10}\).

Rate \(\propto\) excess. At \(60^\circ\)C the excess is \(40^\circ\)C, at \(40^\circ\)C it is \(20^\circ\)C, half as much. The rate is \(1^\circ\)C per minute.

\(k = \dfrac{hA}{ms} \propto \dfrac{r^2}{r^3} = \dfrac{1}{r}\). With the same excess, the rate of cooling is proportional to \(k\): \(\dfrac{1}{1} : \dfrac{1}{2} = 2 : 1\). The small sphere cools twice as fast.

\(\dfrac{10}{10} = k(55 - 25)\), so \(k = \dfrac{1}{30}\) per minute.

\(\dfrac{10}{t} = \dfrac{1}{30}(45 - 25) = \dfrac{2}{3}\), so \(t = 15\) minutes.

Convection is heat transfer by the bulk movement of a fluid: the warm fluid itself moves and carries its heat with it. In natural convection the fluid moves because of the density change on heating. In forced convection a fan or pump drives it.

\(h\) is the convective heat transfer coefficient, the power transferred per unit area per kelvin of temperature difference: W/m\(^2\) K.

1. Air touching the plate is warmed by conduction. It expands, becomes less dense than the air around it, and the denser cool air pushes it up. Cool air flows in from the sides to take its place, is warmed in turn, and rises. A steady circulation results.

2. \(h\) becomes larger. The fan sweeps the warm air away as fast as it forms. So the air next to the plate is always cool, and the temperature difference across the thin layer at the surface stays large. Forced convection can raise \(h\) by ten times or more.

\(\dfrac{dQ}{dt} = hA(T_s - T_f) = 10 \times 2 \times 20 = 400\) W.

Convection: \(h(T_s - T_f) = 10 \times 20 = 200\) W/m\(^2\).

Radiation: \(\sigma(T^4 - T_0^4) = 5.67 \times 10^{-8} \times (1.049 \times 10^{10} - 0.81 \times 10^{10}) = 5.67 \times 10^{-8} \times 2.39 \times 10^9 = 135\) W/m\(^2\).

Total \(= 335\) W/m\(^2\). At ordinary temperatures the two processes are of similar size, and both are proportional to the small excess, which is why Newton's law works.

1. \(P\). It falls faster from the same start in the same room, so its rate of cooling per degree of excess is larger.

2. \(P\) reaches \(50^\circ\)C (excess halved from \(60\) to \(30\)) at \(10\) min, \(Q\) at \(20\) min. The halving time is \(\dfrac{\ln 2}{k}\), so \(k_P : k_Q = 20 : 10 = 2 : 1\).

3. \(k = \dfrac{hA}{ms}\). \(P\) could have a smaller mass, a larger surface area, or a blacker, rougher surface with a larger \(h\).

\(k = \dfrac{hA}{ms} = \dfrac{10 \times 0.05}{0.2 \times 4200} = 5.95 \times 10^{-4}\) per second \(= 0.036\) per minute.

Halving time \(= \dfrac{\ln 2}{k} = \dfrac{0.693}{0.036} = 19\) minutes. Hot tea at \(60^\circ\)C above room temperature is \(30^\circ\)C above it after about \(19\) minutes.

The fan drives forced convection. It replaces the warm air next to the skin with cooler room air, so the skin loses heat faster. It also carries away the moist air near the skin, so sweat evaporates faster and takes its latent heat from the body. The air is no cooler, but the body loses heat faster.

1. The vacuum stops conduction and convection between the walls. Both need a material medium, and there is almost no gas to carry heat.

2. The silvered surfaces stop radiation. Silver is a poor emitter, so the hot inner wall radiates little, and a poor absorber, so the outer wall reflects what does arrive.

3. The stopper stops convection and evaporation from the open top, and is a poor conductor itself.

Land has a smaller specific heat than water and warms faster in the day. The air over the land is heated, rises, and cooler air from over the sea flows in to replace it: the sea breeze.

At night the land cools faster than the sea. Now the air over the sea is warmer and rises, and air flows from the land to the sea: the land breeze.

Surface area per metre \(= 2\pi r \times 1 = 2 \times 3.14 \times 0.02 = 0.1256\) m\(^2\).

\(\dfrac{dQ}{dt} = hA\Delta T = 10 \times 0.1256 \times 50 = 62.8\) W per metre.

Conduction needs a medium and the medium does not move; heat passes from molecule to molecule. Convection needs a fluid medium that moves in bulk, carrying its heat. Radiation needs no medium at all and crosses a vacuum, as sunlight does; it is the only one of the three that can.

The true rate of cooling changes as the body cools, because the excess keeps falling. Over a short interval the excess changes little, so its average, \(\dfrac{T_1 + T_2}{2} - T_0\), is a good stand-in for the actual excess. Over a large fall the rate at the start is much larger than at the end, and the average excess underestimates the time.