Newton's law of Cooling and its applications Dr N K Srinivasan
Introduction Newton's law of cooling is one of the basic laws of physics with wide applications. The mathematics of this law helps you to understand many other phenomena---for instance radioactive decay of isotopes. The formulation is simple and you may review your knowledge of exponential functions. This article can be followed even without the familiarity with exponential functions which you may study in 'Precalculus". I am not using the formal notation of calculus--so a knowledge of calculus is not required. I shall explain this law with respect to the cooling of a hot pot of tea . Consider a tea pot with hot tea at a temperature of 105 deg F. Assume that the pot is tightly closed with a lid,but for a small hole through which some steam may escape. Assume also that the pot is kept on a wooden table. Wood is a bad conductor of heat or a thermal insulator.So heat energy is not conducted away from the pot through the base of the pot. Thus heat conduction is ruled out. Since the temperature of hot tea is only around 100 F, there is not much of heat radiation or infra-red waves from the surface
of the pot. Note that the amount of thermal radiation from a hot object is proportional to temp raised to the power of four,called Stefan-Boltzmann law. This loss is low for temperatures around 100 F and becomes significant only for red hot temperatures. So loss of heat due to radiation from the pot is neglected here. Heat transfer to the surroundings is by one or more of three modes-- conduction, convection and radiation. We have ruled out conduction and neglected radiation for cooling of the pot. We are left with convection as the only mode of heat transfer of cooling of the pot. Newton's law of cooling is an expression for convection only. In convection a thin layer of the surrounding medium---in our case the room air, gets heated by the pot surface and then this warm air moves away or swirls around and mixes with the cold air in the room. Then a fresh layer of cold air hugs the pot and gets heated ..and the process is repeated.This is the process of convection. Newton's Law Newton stated that the rate of loss of heat energy is proportional to the temperature difference between the hot surface and the surroundings. This can be stated in a simpler way: The rate of temperature drop of the pot is proportional to the temperature difference between the pot surface and the surroundings. The rate of temp drop can be measured as : temp drop over a
small time interval: or Δ T /Δ t where T is the temperature and t is the time. The triangle sign represents a small change. Let us take the small change in time as one minute : Δ t= 1 minute. Then Newton's law becomes: rate of temp drop = k (temp of pot -tmep of air) Δ T = - k [ T - T
air]
]
Where Δ T is the temperature drop in one minute.
Let us assume that the temp of air remains constant at 60 F. Let us also assume that the temp of pot's surface is close to the tea temperature and is only slightly less. Note that the rate of cooling or in our case temp drop on one minute is proportional to the temp diffference between the pot surface and the surroundings. Initially the temp drop is high. As the pot gets cooler ,the temp drop keeps decreasing every minute, as the temp difference is getting reduced. let us take a few instances and try to understand this. For our calculations, we take k = -0.05 Take the pot temp as 100 F. Then Δ T = k [100 - 60] = -0.05 x 40 = -2 deg F So, in one minute the temperature will cool from 100 F to 98 F. At a later time, when the pot temp is 80 F,
Δ T = k [80 - 60] = - 0.05 x 20 = -1 In one minute, the temp drops from 80F to 79 F. Thus the cooling rate is half of the cooling rate at 100 F. For the next case, consider the pot temp as 62 F.
Δ T = - 0.05 [62-60] = 0.1 The temp drops only
by 0.1 deg in one minute..
The rate of cooling gets drastically less as we approach the surrounding temperature. This is the nature of "exponential decay" of temperature from 100F to 60 F. Therefore if you make a plot of temperature vs time, you get an exponential curve. You can construct the curve using the following table for every minute: Time(min)
Temp[F]
0
100
1
98
drop in temp -2 -0.05x38=-1.98
new temp[F] 98 96.2
....................................................... This is indeed a tedious way of doing, but you can easily write a computer program for this and run in your PC or lap top computer. Half Time
We can calculate the time it takes to reduce the temp difference by half:
The half time = t1/2
=
0.69/k or approximately 0.7/0.05
=14 minutes. The initial difference in temp between the pot and the air is 40 F. This will get reduced to 20 F in about 14 minutes. That is the pot temp becomes 100 - 20= 80 F in the first 14 minutes. In the second 14 minutes ,the temp drop will be 10 F.The pot temp after 28 minutes would be 80-10 = 70F. This continues further as shown in this table: Time[min]
Temp [F]
0
100
14
80
28
70
42
65
56
62.5
70
61.25
84
60.63
98
60.32
112
60.16
----------------------------------------Thus it takes about two hours to cool to room temperature.! Make a graph of this table --- temp[y axis] vs time [x axis]--- to learn about the exponential decay resulting from Newton's law. The equation for the temp vs time We can write the equation for temp vs time using the exponential function: T (t) = T
air
+ (T
pot
+ T
] e
air]
- k t
Here 'k' is a positive quantity. What does "k" depend on? We have used "k" as a mysterious number and used it for calculations. The cooling rate depends on the surface area of the pot.So 'k" is directly proportional to the surface area A. K depends inversely on the amount of heat in the pot, Q ; "Q" in turn depends on the mass and specific heat of the liquid inside:
We also use a proportionality constant R for any insulation; in our case, the ceramic pot itself has sufficient insulation properties. So
K = A / (m C R)
where m is the mass, C specific heat and R, the insulation factor. mass m = density of liquid x volume of the pot Heat Conservation Principles 1 A major conservation method is to reduce the surface area A. For a given volume, the surface area will be minimum for a sphere. So spherically shaped vessels or pots can be used. Next would be cylinder-shaped vessels or tanks. Most chemical tanks with hot liquids are combinations of sphere and cylinders. 2 We cannot do much to reduce mass of the given liquid. But we can insulate the pot or vessel by a layer of insulating film or thick batting.In soalr field, fiber-glass batting is often used. In Chemical industry and in Solar heaters, the pipes are wound with insulating tapes or pastes. Applications of Newton's Law 1 Solar Water heaters: The solar water heaters get heated during day time ,till sun-set time. 'Let the maximum temp reached be 120F. Let the mass of 80 gallon water tank be 670 lbs . c= 1 BTU /lb/F Take R= 15 for a good insulating film over the tank.
Then k= 0.0035 Taking the surrounding temp as 60F, we get for t= 24 hours, T (24) = 60 + [120-60] e
-k.24
= 116 F The temp drop is only 4 deg F over 24 hours. 2 Coroner's calculations John was found murdered in his apartment . A coroner visted the apartment at 2 PM and checked the temp of the body of John; it was 71 F. The ambient temp was 60 F--- maintained by the thermostat in the air-conditioner in the room.Now can you calculate his time of death assuming that his normal body temperature was 98.6 F at the time of his death? [k= 0.0345]? Temp difference = 98.6 - 60 = 38.6 F The half time = 0.69/k = 0.69/0.0345 = 20 minutes Therefore the temp drop is as follows: Time
Temp
0
98.6
20
79.3
40
69.7
----------------So, John must have died approximately 40 minutes before the arrival of the coroner at 2PM. The approximate time of death is 1.20 PM.
-------------------------------------------------------------------------------------------Some mathematical notes 1 Since the cooling rate steadily decreases with time, the pot reaches the surrounding temperature after a long ,long time. We say that the temperature of the pot 'asymptotically' reaches the surrounding temperature. Mathematically it means that the pot reaches the air temperature at infinite time! or as the time tends towards infinity.! 2 The heat loss equation is actually a differential equation: d Q / dt = - k' {T - T'} where T is the temperature at a given time and T' is the surrounding temperature. This is also called " Fourier's first law" . This form is used for both conduction and conduction. k' is called the " Convection coefficient". This is called "natural convection coefficient". Forced convection occurs when you use a blower or fan for steady stream of air passing over the pot. When you use an electric hair drier, you are using forced convection. 3
Since Q = m c ΔT, we can modify
the above equation and
write in terms of temperature only: dT/dt = -k[T-T'] where k is another constant. 'k' was explained earlier. 4 Solar thermal devices like water heaters and cookers employ Newton's law and are designed with proper insulation to reduce
convection losses. You can read up on that. Interestingly small air gap [5mm width ] or air bubbles can serve as effective insulation. Expanded polystryrene or styrofoam is one such product for thermal insulation. 5 Geologists and meteorologists use Newton's law of cooling to study the cooling of the Earth's surface at night. 6 The half -time equation How do we get the half-time equation: half-time = 0.69/k Let us rewrite the temp-time equation: T (t) = T air + (T pot - T air) e-kt [ T(t)-T air] / [ T pot-Tair] = e-kt We wish to find the time t when T(t) - T air = 1/2(T pot -Tair) let us take the same example. The initial temp of pot = T pot =100F T air = 60F T pot - Tair = 40 We wish to find the time when the temp is the middle value between 100 and 60 or T(t)= 80 F Then in the above equation, we get: [80-60]/[100 - 60] = 1/2 = e-kt Or
2 = e
kt
To solve for t, we use logarithmic function. Here is a simple peek into logarithmic functions: If then Logarithmic
y= ex x = ln y
function is the inverse of exponential function! Let us take y =2 ;
Using calculator, you find ln y = ln2 =0.69 = x So, in our equation: 2 = e
kt,
x = k t = ln 2 =0.69 So
t = 0.69/k
This formula is useful to calculate half time or other similar things like half-life of a radioactive isotope.
-----------------------------------------------------------------------------------------Exercises 1 Draw the exponential curve for cooling of a hot kettle with initial temperature of 160F and room temperature of 60F , with k= 0.03. Find its half-time of cooling. 2 Take a polished steel container with 200 ml water and insert a thermometer. use a wooden or plywood or plastic lid with a hole for inserting the thermameter. Keep the container over a
wooden block. Avoid wind draft. You can enclose the container with a circular cardboard piece to isolate from wind currents. Use a watch
and measure the temperature every minute and plot
the graph of temperature versus
time.
Find the half time from the graph and use that number to calculate the 'k' value. 3 Repeat the experiment after painting the container with mat [dull] black paint. What is the value of 'k' now? can you explain why the cooling rate is greater with black surface.? 4 To reduce losses due to radiation, use the polished container and surround it with aluminum kitchen foil as a cylinder with an air gap of 10 mm.This foil will reflect back the radiation and thus correct for any radiation loss. Again make the measurements and plot the temp-vs time curve. Compare the half time and 'k'value with the previous two cases. 5 Plot the three temp-time curve on the same graph
paper and
make a presentation to the class. curve1 : polished container without aluminum foil reflector curve 2: Polished container with aluminum reflector curve 3: Black container --------------------------------------------------------------------------------Ref: Any physics text book & Internet sources. I suggest : Physics by Giancoli (any edition).! You can send feed back or enquiries to :
[email protected] [No fees whatever.]
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