How much a black body radiates
The last lesson showed that every body radiates and that a black body radiates the most. This lesson asks how much. The answer depends on one thing only, the absolute temperature, and it depends on it very strongly. This lesson uses the Stefan-Boltzmann constant \(\sigma = 5.67 \times 10^{-8}\,\text{W/(m}^2\,\text{K}^4)\).
The power radiated by each square metre of a black body is proportional to the fourth power of its absolute temperature: \(E_b = \sigma T^4\). A black body of surface area \(A\) radiates the power \(P = \sigma A T^4\).
The unit of \(\sigma\) follows from the law itself. Write \(\sigma = P/(AT^4)\). Power is in watts, area in square metres and \(T^4\) in \(\text{K}^4\), so \(\sigma\) is in \(\text{W/(m}^2\,\text{K}^4)\), written also as \(\text{W m}^{-2}\,\text{K}^{-4}\).
A steel plate painted dull black has a total surface area of \(0.2\,\text{m}^2\). Treat it as a black body. How much power does it radiate at \(400\,\text{K}\)? And at \(800\,\text{K}\)?
At 400 K: \(P = \sigma AT^4\) \(= 5.67 \times 10^{-8} \times 0.2 \times 400^4\). Here \(400^4 = 2.56 \times 10^{10}\), so \(P = 290\,\text{W}\).
At 800 K the temperature is doubled, so \(T^4\) is multiplied by \(2^4 = 16\). \(P = 16 \times 290 = 4640\,\text{W}\), about 4.6 kW.
A plate that is twice as hot radiates sixteen times as much.
The fourth power
A fourth power grows very fast. Double the absolute temperature, and the radiated power becomes \(2^4 = 16\) times as large. Triple it, and the power grows \(81\) times. Move the slider in the figure below from 300 K upward, and watch how long the curve stays near the axis before it rises steeply.
Turn the question round. For the power to become four times as large, \(T^4\) must become four times as large. So \(T\) must grow by the factor \(4^{1/4} = \sqrt{2} \approx 1.41\). A small rise in temperature gives a large rise in radiation. This is why a glowing iron rod loses heat so quickly when it leaves the forge.
Two black bodies of equal area are at \(27\,°\text{C}\) and \(327\,°\text{C}\). What is the ratio of the power radiated by the hotter body to that of the cooler one?
Two black spheres are radiating. Sphere B has twice the radius of sphere A, and sphere A is at twice the absolute temperature of sphere B. What is the ratio of the power radiated by A to that radiated by B?
Real surfaces radiate less
A real surface radiates less than a black body at the same temperature. The ratio of its emissive power to that of a black body is its emissivity \(e = E/E_b\). It lies between 0 and 1. A surface whose emissivity is the same at every wavelength is called a grey body. It radiates \(P = e\sigma AT^4\).
Kirchhoff's law of the last lesson gives \(E = aE_b\). Dividing by \(E_b\) gives \(e = a\): a surface emits the same fraction it absorbs. Polished silver has \(e \approx 0.03\). Dull black paint and soot have \(e\) about 0.95 or more. Human skin, of any colour, has \(e \approx 0.97\) in the infrared that it emits. Skin is almost black at those long wavelengths.
The tungsten filament of a lamp works at \(2800\,\text{K}\), has emissivity \(0.35\) and radiates \(60\,\text{W}\). What is its surface area, in cm²? Take \(\sigma = 5.67 \times 10^{-8}\,\text{W/(m}^2\,\text{K}^4)\).
What a body actually loses
A body never radiates into nothing. Its surroundings radiate back at it, as Prevost said. Suppose a body of area \(A\) and emissivity \(e\) is at temperature \(T\), and its surroundings are at \(T_0\). The derivation below finds its net rate of loss.
\[ P_\text{emit} = e\sigma A T^4 \]
Why. the body radiates at its own temperature \(T\), whatever surrounds it.
\[ P_\text{fall} = \sigma A T_0^4 \]
Why. surroundings at \(T_0\) that enclose the body send it the radiation of a black body at \(T_0\).
\[ P_\text{abs} = a\,\sigma A T_0^4 = e\sigma A T_0^4 \]
Why. the body absorbs the fraction \(a\) of what falls on it, and by Kirchhoff's law \(a = e\).
\[ P_\text{net} = P_\text{emit} - P_\text{abs} = e\sigma A\,(T^4 - T_0^4) \]
Read it. the net loss is what goes out minus what comes in. It is zero when \(T = T_0\), as Prevost requires.
The person in Fig 2.1 radiates \(\sigma AT^4 = 5.67 \times 10^{-8} \times 1.5 \times 306^4\) \(= 746\,\text{W}\). That is more than seven 100 W bulbs. But the walls send back \(\sigma AT_0^4 = 5.67 \times 10^{-8} \times 1.5 \times 298^4\) \(= 671\,\text{W}\), and the skin absorbs it. The net loss is only \(746 - 671 = 75\,\text{W}\). A resting body makes about 100 W of heat, so radiation removes most of it.
Desert sand at \(300\,\text{K}\) faces a clear night sky that radiates like a black body at \(250\,\text{K}\). Treat the sand as black. How much heat does each square metre lose per second? How much under a cloud base at \(280\,\text{K}\)?
Area \(A = 1\,\text{m}^2\), \(e = 1\), \(T = 300\,\text{K}\). For the clear sky \(T_0 = 250\,\text{K}\); for the cloud \(T_0 = 280\,\text{K}\).
\(P_\text{net} = \sigma A\,(T^4 - T_0^4)\).
Clear sky: \(250^4 = 3.91 \times 10^9\), so \(P = 5.67 \times 10^{-8} \times 4.19 \times 10^9\) \(= 238\,\text{W}\). Cloud: \(280^4 = 6.15 \times 10^9\), so \(P = 5.67 \times 10^{-8} \times 1.95 \times 10^9\) \(= 111\,\text{W}\).
Under a clear sky each square metre loses 238 W; under the cloud only 111 W, less than half. This is Fig 1.2 in numbers.
The net loss is \(e\sigma A(T^4 - T_0^4)\), never \(e\sigma A(T - T_0)^4\). For the person, \((306 - 298)^4 = 4096\) would give a net loss of a third of a milliwatt, which is absurd. Take the fourth power of each absolute temperature first, then subtract.
A sphere of surface area \(0.2\,\text{m}^2\) and emissivity \(0.5\) is at \(500\,\text{K}\) in a room at \(300\,\text{K}\). Find its net rate of loss of heat by radiation, in watts. Take \(\sigma = 5.67 \times 10^{-8}\,\text{W/(m}^2\,\text{K}^4)\).
A blacksmith pulls an iron bar out of the forge. Its surface area is about \(0.01\,\text{m}^2\) and it glows orange at about \(1300\,\text{K}\). Treat it as black and ignore what the room sends back. Roughly how much power does it radiate?
In 1879 Josef Stefan in Vienna studied measurements of how fast hot bodies cool. He found that the radiated power grew as the fourth power of the absolute temperature. He used the law to make the first sound estimate of the Sun's surface temperature, about 5400 °C. Five years later his former student Ludwig Boltzmann derived the same law from the laws of thermodynamics. The law carries both their names.
At railway stations and hospitals, a nurse points an infrared thermometer at your forehead and reads your temperature in a second. The instrument measures the infrared power arriving from a small patch of skin. It is set for skin's emissivity, about 0.98, and it turns that power into a temperature by the Stefan-Boltzmann law. Because the power goes as \(T^4\), a change of one degree changes the power by more than 1 %, which is easy to detect.
- Putting Celsius temperatures into \(T^4\).Always convert to kelvin first: \(T = t + 273\).
- Writing the net loss as \(e\sigma A(T - T_0)^4\).Take each fourth power first: \(e\sigma A(T^4 - T_0^4)\).
- Forgetting that the surroundings radiate back.\(\sigma AT^4\) is what the body emits. Its loss is the emission minus what it absorbs.
- Thinking a larger emissivity means a hotter body.\(e\) is a property of the surface. At the same \(T\), a larger \(e\) means more radiation, not a higher temperature.