A cup of chai is poured at 90 °C on a table in a room at 30 °C. Five minutes later it is at 70 °C, still too hot to gulp. You wait for it to fall another 20 degrees, to 50 °C. Does that take another five minutes? Most people guess yes. The chai takes longer, and this lesson shows exactly how much longer.
The rate follows the excess
The last two lessons found how fast a body loses heat. By convection the rate is \(hA(T - T_0)\), where \(T_0\) is the temperature of the surrounding air. It is proportional to the excess temperature \(T - T_0\). By radiation the rate is \(e\sigma A(T^4 - T_0^4)\), which does not look proportional at all. This lesson uses \(\sigma = 5.67 \times 10^{-8}\,\text{W/(m}^2\,\text{K}^4)\) and \(c_\text{water} = 4200\,\text{J/(kg K)}\).
But when the excess is small, radiation behaves the same way. Write \(T^4 - T_0^4\) as a product and look at it when \(T\) is close to \(T_0\).
\[ T^4 - T_0^4 = (T^2 + T_0^2)(T + T_0)(T - T_0) \]
Why. a difference of squares, used twice: \(T^4 - T_0^4 = (T^2 - T_0^2)(T^2 + T_0^2)\).
\[ T^2 + T_0^2 \approx 2T_0^2, \quad T + T_0 \approx 2T_0 \]
Why. when \(T\) is close to \(T_0\), replace \(T\) by \(T_0\) in the two sums. The small difference \(T - T_0\) is kept.
\[ T^4 - T_0^4 \approx 4T_0^3\,(T - T_0) \]
Why. multiply: \(2T_0^2 \times 2T_0 = 4T_0^3\).
\[ e\sigma A(T^4 - T_0^4) \approx 4e\sigma A T_0^3\,(T - T_0) \]
Read it. the radiative loss is proportional to the excess, like convection, with an effective coefficient \(4e\sigma T_0^3\).
At room temperature, \(T_0 = 300\,\text{K}\), the effective coefficient is \(4\sigma T_0^3 = 6.1\,\text{W/(m}^2\,\text{K)}\) for a black surface. It is close to the \(h \approx 10\) of still air. So a warm body loses heat at a total rate proportional to its excess, as long as that excess is small. For chai 10 K above a 30 °C room the straight-line form is within 5 % of the exact loss. At 60 K it falls short by 25 %.
When a body is only a little hotter than its surroundings, it loses heat at a rate proportional to its excess temperature. So its temperature falls at the rate \(\dfrac{dT}{dt} = -k\,(T - T_0)\). The constant \(k\) depends on the body's surface, area and heat capacity.
A body of mass \(m\) and specific heat \(c\) loses heat \(mc\,dT\) when its temperature falls by \(dT\). If it loses heat at the rate \(hA(T - T_0)\), with \(h\) now covering convection and radiation together, then \(-mc\,dT/dt = hA(T - T_0)\). So \(dT/dt = -k(T - T_0)\) with \(k = hA/(mc)\). A large surface cools the body faster, and a large heat capacity slows it.
A cup holds \(0.15\,\text{kg}\) of chai, with \(c = 4200\,\text{J/(kg K)}\). Its surface open to the air, top and sides, is about \(0.05\,\text{m}^2\). Take \(h = 17\,\text{W/(m}^2\,\text{K)}\) for convection and radiation together. Find its cooling constant \(k\) in per minute. Ignore the cup's own heat capacity.
\(k = \dfrac{hA}{mc}\) \(= \dfrac{17 \times 0.05}{0.15 \times 4200}\) \(= \dfrac{0.85}{630}\) \(= 1.35 \times 10^{-3}\,\text{s}^{-1}\).
In per minute, \(k = 1.35 \times 10^{-3} \times 60\) \(= 0.081\,\text{min}^{-1}\).
At 90 °C, 60 K above the room, the chai cools at \(0.081 \times 60 = 4.9\,\text{K}\) per minute.
For the same shape and material, a small body cools faster than a large one. The area goes as \(r^2\) and the mass as \(r^3\), so \(k = hA/(mc)\) goes as \(1/r\). A sphere of half the radius cools twice as fast from the same start. This is why a spoonful of dal cools long before the pot, and why small animals must eat so much to keep warm.
Chai at \(80\,°\text{C}\) is cooling at \(4\,\text{K}\) per minute in a room at \(30\,°\text{C}\). At what rate is it cooling when it has reached \(55\,°\text{C}\), in K per minute?
Solving the law
Newton's law says how fast the temperature falls at each instant. To find the temperature at any time, the rate must be added up over time. Write \(\theta = T - T_0\) for the excess. Then \(d\theta/dt = -k\theta\), because \(T_0\) is constant. One step below is left for you.
\[ \dfrac{d\theta}{dt} = -k\theta \]
Why. Newton's law written for the excess \(\theta = T - T_0\).
\[ \int_{\theta_i}^{\theta} \dfrac{d\theta}{\theta} = -k\int_0^t dt \]
Why. separate the variables and integrate from the start, when the excess is \(\theta_i\).
?
Your turn. the integral of \(d\theta/\theta\) is \(\ln\theta\). Which line is it?
\[ \theta = \theta_i\,e^{-kt} \]
Why. take the exponential of both sides.
\[ T - T_0 = (T_i - T_0)\,e^{-kt} \]
Read it. put back \(\theta = T - T_0\). The excess decays exponentially towards zero.
Before you play the figure below, predict. In a room at 30 °C the chai falls from 90 °C to 70 °C in 5 minutes. How long does it take to fall the next 20 degrees, to 50 °C?
Look at the purple lines in Fig 6.1. Every 5 minutes the excess is multiplied by the same factor, \(40/60 = 2/3\). That is the mark of an exponential: equal times give equal ratios, not equal falls. So the curve is steep at first and flatter later. The excess gets smaller and smaller but never reaches zero. After 30 minutes the chai is still at \(30 + 60 \times (2/3)^6 = 35.3\,°\text{C}\).
Chai cools from \(90\,°\text{C}\) to \(70\,°\text{C}\) in \(5\,\text{min}\) in a room at \(30\,°\text{C}\). Find \(k\), and the further time for it to reach \(50\,°\text{C}\).
Excesses: 60 K at the start, 40 K after 5 min, and 20 K at 50 °C.
\(40 = 60\,e^{-5k}\), so \(k = \dfrac{\ln(60/40)}{5}\). Then from 40 K to 20 K, \(20 = 40\,e^{-kt}\), so \(t = \dfrac{\ln 2}{k}\).
\(k = \ln 1.5/5\) \(= 0.405/5\) \(= 0.081\,\text{min}^{-1}\). Then \(t = 0.693/0.081 = 8.5\,\text{min}\).
The chai takes 8.5 minutes more, not 5, to fall from 70 °C to 50 °C. That is also the time for the excess to halve at any stage of the cooling.
Take the natural logarithm of the cooling law: \(\ln(T - T_0) = \ln(T_i - T_0) - kt\). That is a straight line in \(t\). Its slope is \(-k\) and its intercept is the logarithm of the starting excess. Plotting \(\ln(T - T_0)\) against time is the way to test Newton's law on real data, and the slope gives \(k\) directly.
The excess temperature of a body over a room at \(20\,°\text{C}\) halves every \(8\) minutes. The body starts at \(80\,°\text{C}\). What is its temperature after \(24\) minutes, in °C?
The average-temperature shortcut
Logarithms are slow to work by hand, so examinations often use a shortcut. Over an interval of time \(t\) the body falls from \(T_1\) to \(T_2\). Replace the changing excess by its average, \(\bar T - T_0\), where \(\bar T = (T_1 + T_2)/2\). Then the average rate of fall is \(k\) times the average excess.
Try it on the chai. From 90 °C to 70 °C in 5 minutes: \(20/5 = k(80 - 30)\), so \(k = 0.08\,\text{min}^{-1}\). For the next 5 minutes, \((70 - T_2)/5 = 0.08\,((70 + T_2)/2 - 30)\). Solving gives \(T_2 = 56.7\,°\text{C}\). The exact law gives \(30 + 40 \times 2/3 = 56.7\,°\text{C}\) too, exactly the same.
That is not luck. In the shortcut, a step of length \(t\) turns the excess \(\theta_1\) into \(\theta_2 = \theta_1(1 - kt/2)/(1 + kt/2)\). For a fixed step that is a fixed ratio, here \(0.8/1.2 = 2/3\). The first step fixed \(k\) so that this ratio matched the real one. Every later step of the same length repeats the same ratio, so it agrees with the exact law.
With a step of a different length the agreement fails. Use the shortcut for the fall from 70 °C to 50 °C: \(20/t = 0.08 \times (60 - 30)\), so \(t = 8.3\,\text{min}\). The exact answer was 8.5 minutes. The shortcut is 2.5 % short, because over a large fall the rate at the start is much larger than at the end. The smaller the fall compared with the excess, the better the shortcut.
A body cools from \(70\,°\text{C}\) to \(60\,°\text{C}\) in \(4\,\text{min}\) in a room at \(20\,°\text{C}\). Using the average-temperature form, find the time it takes to cool from \(60\,°\text{C}\) to \(50\,°\text{C}\).
First interval: \(T_1 = 70\), \(T_2 = 60\), \(t = 4\,\text{min}\), average 65 °C. Second: \(T_1 = 60\), \(T_2 = 50\), average 55 °C.
\(\dfrac{10}{4} = k(65 - 20)\), then \(\dfrac{10}{t} = k(55 - 20)\).
\(k = 2.5/45 = 0.0556\,\text{min}^{-1}\). Then \(t = 10/(0.0556 \times 35)\) \(= 5.14\,\text{min}\).
About 5.1 minutes. The exact law gives 5.16 minutes, so the shortcut is within half a per cent here.
A body cools from \(75\,°\text{C}\) to \(65\,°\text{C}\) in \(6\,\text{min}\) in a room at \(25\,°\text{C}\). Using the average-temperature form, how long does it take to cool from \(65\,°\text{C}\) to \(55\,°\text{C}\), in minutes?
Comparing two liquids by their cooling
Pour equal masses of water and of oil into two identical thin cups and let them cool in the same room. At the same temperature the two cups lose heat at the same rate. They have the same surface, the same \(h\) and the same excess. So over the same fall of temperature they lose the same heat. The one with the smaller heat capacity loses that heat in less time.
For the same fall, \(t \propto mc\). In Fig 6.3 the masses are equal and the oil takes half as long, so its specific heat is half that of water: \(c_\text{oil} = 4200 \times 2.5/5\) \(= 2100\,\text{J/(kg K)}\). If the cup itself has a heat capacity \(C\), include it on both sides: \(t_1/t_2 = (m_1c_1 + C)/(m_2c_2 + C)\). The same reasoning compares two bodies by their cooling constants: a body whose excess halves in half the time has twice the \(k\).
A copper cup has a water equivalent of \(10\,\text{g}\). With \(100\,\text{g}\) of water in it, it takes \(6\,\text{min}\) to cool from \(60\,°\text{C}\) to \(50\,°\text{C}\). With \(100\,\text{g}\) of a liquid instead, the same fall takes \(3\,\text{min}\). Find the liquid's specific heat, in J/(kg K). Take \(c_\text{water} = 4200\,\text{J/(kg K)}\).
Two bodies P and Q start at \(80\,°\text{C}\) in a room at \(20\,°\text{C}\). P reaches \(50\,°\text{C}\) in 10 minutes and Q in 20 minutes. What is \(k_P : k_Q\), and what could explain it?
Newton's law is an approximation. It fails for a large excess, because radiation then grows faster than the excess. It also fails when evaporation carries much of the heat, as from an open cup at 90 °C, and when the surroundings grow warmer. In exam problems, trust it for falls of a few tens of degrees, and use the exact form when the answer must be precise.
In 1701 an anonymous paper in the Philosophical Transactions described a scale of temperatures up to the heat of a coal fire. Its author was Isaac Newton. He heated a block of iron red-hot and let it cool in a steady wind. He noted the moments at which small pieces of different metals laid on it became solid. He assumed that the heat lost in a given time was proportional to the iron's excess heat, which is his law of cooling. The steady wind gave him forced convection, where the law works best.
To set curd, milk is boiled and left to cool to about 45 °C before the starter is stirred in. Hotter milk kills the bacteria. In a kitchen at 30 °C, the excess falls from 70 K to 15 K. By Newton's law the excess halves in equal times. So the first 40 degrees, down to 60 °C, take only a little longer than the last 15 degrees, down to 45 °C. Pouring the milk into a wide steel plate raises \(A\) and \(k\) and saves the wait.
- Working with temperatures instead of excesses.Newton's law is about \(T - T_0\). Subtract the room's temperature first.
- Expecting equal falls in equal times.Equal times give equal ratios of the excess, so each fall is smaller than the one before.
- Using the average-temperature form over a large fall.It is good for small falls. Over a large fall use \(T - T_0 = (T_i - T_0)e^{-kt}\).
- Forgetting the vessel's heat capacity when comparing two liquids.Include the cup's heat capacity \(C\) with each liquid: \(t \propto mc + C\).