**Newton’s Law of Cooling:** Newton was the first person to investigate the heat lost by a body in air. He found that the rate of loss of heat is proportional to the excess temperature over the surroundings. This result which is called Newton’s law of cooling is approximately true in still air only for a temperature excess of about 20^{o}C or 30^{o}C; but it is true for all excess temperatures in conditions of forced convection of the air, i.e. in a draught. With natural convection Dulong and Petit found that the rate of loss of heat was proportional to *0*^{5/4}, where *0* is the excess temperature, and this appears to be true for higher excess temperatures, such as from 50^{o}C to 300^{o}C. At low excess temperatures, however, less than 1^{o}C, G. T. P. Tarrant has pointed out that radiation, not convection, is the major contributing factor to the rate of cooling of an object.

To demonstrate Newton’s Law of cooling, we plot a temperature (*0*) time (*t)* cooling curve for hot water in a calorimeter placed in a draught. If *0 _{R}* is the room temperature, then the excess temperature of the water is (

*0 – 0*). At various temperatures, such as

_{R}*0*in a graphical illustration, we can draw tangents such as APC to the curve. The slope of the tangent, in degrees per second, gives us the rate of a fall of temperature, when the water is at the temperature

*0*:

We then plot these rates against the excess temperature, *0 – 0 _{R}*, and find a straight line passing through the origin. Since the heat lost per second by the water and calorimeter is proportional to the rate of fall of the temperature, Newton’s law is thus verified.

**Heat loss and Temperature Fall**

Besides the excess temperature, the rate of heat loss depends on the exposed area of the calorimeter, and on the nature of its surface: a dull surface loses heat a little faster than a shiny one, because it is a better radiator. This can be shown by doing a cooling experiment twice, with equal masses of water, but once with the calorimeter polished, and once after it has been blackened in a candle-flame. In general, for anybody with a uniform surface at a uniform temperature *0*, we may write, if Newton’s law is true,

Where S is the area of the body’s surface, *0 _{R}* is the temperature of its surroundings, k is a constant depending on the nature of the surface, and Q denotes the heat lost from the body.

When a body loses heat Q, its temperature *0* falls; if m is its mass, and c its specific heat capacity, then its rate of fall of temperature, d*0* / *dt*, is given by

Now the mass of a body is proportional to its volume. The rate of heat loss, however, is proportional to the surface area of the body. The rate of fall of temperature is therefore proportional to the ratio of surface to volume of the body. For bodies of similar shape, the ratio of surface to volume of the body. For bodies of similar shape, the ratio of surface to volume is inversely proportional to any linear dimension. If the bodies have surfaces of similar nature, therefore, the rate of fall of temperature is inversely proportional to the linear dimension: a small body cools faster than a large one. This is a fact of daily experience. A small coal which falls out of the fire can be picked up sooner than a large one. A tiny baby should be more thoroughly wrapped up than a grown man. In calorimetry by the method of mixtures, the fact that a small body cools faster than a large one means that the larger the scale of the whole apparatus, the less serious are the errors due to loss of heat from the calorimeter.

**Correction for Heat Losses in Calorimetry**

Newton’s law of cooling enables us to estimate the lost in an experiment on the methods of mixtures. In doing the experiment, we take the temperature of the mixture at half-minute intervals, and plot it against time. The broken line shows how we would expect the temperature to rise if no heat were lost; we have therefore to estimate the difference, p, between the plateau of its imaginary curve, and the crest of the experimental curve, C. p is known as the ‘cooling correction’.

We start by drawing an ordinate CN through the crest, and another LM through any convenient point L further along the curve; OM should be not less than twice ON – the greater it is, the more accurate the correction. We next draw an abscissa O’PQ through the room temperature, *0 _{R}*; and by counting the squares of the graph paper; we measure the areas O’CP (A1), PCLQ (

*A*2). Then, if q is the fall in temperature from C to L:

Before establishing this equation let us see how to use it. Suppose m1, c, are the mass and specific heat capacity of the specimen; m, c w are the mass and specific heat capacity of water; and C the heat capacity of the calorimeter. Then the heat which these lose to their surroundings is the heat which would have raised their temperature by p. thus,

To correct for the heat losses we must therefore add the correction p to the crest temperature *0* on each side of the heat balance equation. In equation (1), p must be added to *0 _{2 }*in both numerator and denominator.

**Theory of the Correction**

To establish equation (3), we write down the expression for the heat lost per second from the calorimeter, assuming Newton’s law of cooling:

**Measurement of Specific Heat Capacity of Liquid**

Specific heat capacities of liquids which react with water are often measured by the so-called method of Cooling. The cooling curve of a calorimeter is plotted, first when it contained a known volume of hot water, and then when it contains an equal volume of hot liquid. The volumes are made equal so as to make the temperature distribution, over the surface of the calorimeter, the same in each experiment. From the curves, the respective times t1 and tw are found the calorimeter and contents take to cool from *0 _{1}* to

*0*. Whatever the contents of the calorimeter, it gives off heat at a rate which depends only on its excess temperature, since the area and nature of its surface are constant. Therefore, at each temperature between

_{2}*0*and

_{1}*0*the calorimeter gives off heat at the same rate whatever its contents. Thus the average rate at which it loses heat, over the whole rage, is the same with water and with liquid. Consequently,

_{2}