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The Weight Thermometer
The weight thermometer,^{(1)} consists of a glass tube closed at one end, drawn out to a fine neck, which is bent so that it can easily dip into a vessel of liquid. It is used (1) to determine the coefficient of expansion of a liquid relatively to glass; (2) to determine the coefficient of expansion of a solid, that of the liquid being known.
Our first operation will be to weigh the empty glass tube, which must be perfectly clean and dry. Let its weight be 5.621 grammes. We now require to fill it. For this purpose it is heated gently in a Bunsen burner or spirit lamp, being held during the operation in a testtube holder. Its neck is then dipped under the surface of the liquid whose coefficient of expansion is required  glycerine, suppose  and the tube allowed to cool. The pressure of the external air forces some of the glycerine into the tube. As soon as the liquid ceases to run in, the operation is repeated, and so on until the tube is nearly full. It is then held with its orifice under the glycerine, and heated until the fluid in the tube boils. The air which remained in is carried out with the glycerine vapour and the tube left filled with hot glycerine and its vapour. The flame is removed and the thermometer again cooled down, when the vapour inside condenses and more liquid is forced in by the external air pressure. If a bubble of air is still left inside, the operation of heating and cooling must be repeated until the bubble is sufficiently small to be got rid of by tilting the thermometer so that it floats up into the neck. There is another plan which may sometimes be adopted with advantage for partially filling the thermometer. Place it, with its beak dipping into the glycerine, under the receiver of an airpump and exhaust. The air is drawn both out of the thermometer and the receiver. Readmit the air into the receiver. Its pressure on the surface of the glycerine forces the liquid into the tube. It is difficult, however, by this method to get rid of the last trace of air. Suppose the thermometer is filled; it is now probably considerably hotter than the rest of the room. Hold it with its beak still below the surface of the glycerine and bring up to it a beaker of cold water, so as to surround with water the body of the tube and as much as possible of the neck. This of course must not be done too suddenly lest the glass should crack. Let the thermometer rest in the beaker of water  its orifice still being below the surface of the glycerine  and stir the water about, noting its temperature with an ordinary thermometer. At first the temperature of the water may rise a little; after a time it will become steady, and the tube may be removed. Let the observed temperature be 15° C. We have now got the weight thermometer filled with glycerine at a temperature of 15° C. Weigh the tube and glycerine; let the weight be 16.843 grammes. The weight of glycerine inside then is 16.8435.621, or 11.222 grammes. It is advisable to arrange some clamps and supports to hold the tube conveniently while it is cooling in the beaker of water. Instead of using water and cooling the thermometer to its temperature, we may use ice and cool it down to a temperature of 0°C. If we do this we must, as soon as the tube is taken out of the ice, place it inside a small beaker of which we know the weight, for the temperature will at once begin to rise and some of the glycerine will be driven out. Thus we should lose some of the liquid before we could complete the weighing. Our next operation is to find the weight of liquid which the tube will hold at 100° C. To do this we place it in a beaker of boiling water, setting at the same time a receptacle to catch the glycerine which is forced out. When the water has been boiling freely for some time take out the tube, let it cool, and then weigh it Subtracting the weight of the glass, let the weight of the glycerine be 10.765 grammes. Thus 10.765 grammes of glycerine at 100° C. apparently occupy the same volume  that of the thermometer  as 11.222 grammes did at 15° C. The apparent expansion for an increase of temperature of 85° (from 15°100°) is therefore 0.0425. The mean apparent expansion per 1°C. throughout that range is, therefore, 0.0425/85 = 0.00050. This is only the coefficient of expansion relatively to glass, for the glass bulb expands and occupies a greater volume at 100° C. than at 15° C. To find the true coefficient of expansion we must remember that the apparent coefficient is the true coefficient diminished by that of the glass  had the glass at 100° been of the same volume as at 15° more glycerine would have been expelled. The coefficient of expansion of glass may betaken as 0.000026. Thus the true coefficient of expansion of the glycerine is 0.000526. To obtain the temperature when we take the tube from the bath of boiling water, we may use a thermometer, or, remembering that water boils at 100° C. for a barometric pressure of 760 mm. of mercury, while an increasing pressure of 26.8 mm. of mercury raises the boiling point by 1°C., we may deduce the temperature of the boiling water from a knowledge of the barometric pressure. It is better, if possible, to raise the temperature of the weight thermometer to the boiling point by immersing it in the steam rising from boiling water, as in the hypsometer." A suitable arrangement is not difficult to make if the laboratory can furnish a hypsometer somewhat wider than the usual ones, with a good wide opening in the top of the cover. (2) To obtain the coefficient of expansion of a piece of metal  iron, for example  relatively to glycerine, we take a bar of the metal whose volume is obtained from a knowledge of its weight and specific gravity, and place it in the tube before the neck is drawn out. The bar should be bent so as only to touch the tube at a few points, otherwise it will be impossible to fill the tube with the glycerine. The tube is filled after having been weighed when empty, and the weight of glycerine in it at a known temperature is determined. Let the temperature be 0°C. It is then raised to say 100° C. and the weight of the glycerine within again determined. The difference between these two gives the weight of glycerine expelled. Let us suppose we know the specific gravity of glycerine; we can obtain the volume of the glycerine originally in the tube by dividing its weight by its density. Let us call this v_{1}. We can also find the volume of the glycerine expelled; let this be v, and let v_{2} be the volume of the iron, at the lower temperature, v, the volume of the thermometer, t, the change in temperature, α, the coefficient of expansion of the glycerine, β, the coefficient of expansion of the metal, γ, the coefficient of expansion of the glass. Then v = v_{1}+v_{2}. When the temperature has risen t° the volume of glycerine is v_{1}(1+αt) and that of the metal is v_{2}(1+βt); thus the whole volume of glycerine and iron will be v_{1}(1+αt) + v_{2}(1+βt).The volume of the glass is v(1+γt). The difference between these must clearly give the volume of glycerine which has escaped, or v. Thus v_{1}(1+αt) + v_{2}(1+βt)  v(1+γt) = v But v = v_{1}+v_{2}. Thus v_{1}(αγ)t + v_{2}(βγ)t = v. v_{1}(αγ)t is the volume of glycerine which would have been expelled if the volume of the tube had been v_{1}; that is to say, if the tube had been such as to be filled entirely with the glycerine which was contained in it at the first weighing. This can be calculated from the knowledge of the weight and specific gravity of the glycerine and of the value of the coefficient of expansion of the glycerine relatively to the glass. Subtract this from the volume actually expelled. The difference is the increase in volume of the metal relatively to glass for the rise in temperature in question. Divide the result by the volume of the metal and the rise in temperature; we get the coefficient of relative expansion of the metal. Thus, let the original weight of glycerine be 11.222 gms., then the amount which would be expelled, due to the rise of temperature of the glycerine only, will be 0.457 gramme, since the coefficient of expansion of glycerine relative to glass is 0.0005. Suppose that we find that 0.513 gramme is expelled. The difference, 0.056 gramme, is due to the expansion of the metal. Taking the specific gravity of glycerine as 1.30, the volume of this would be 0.043 c.c. Suppose that the original volume of the metal was 5 c.c. and the rise of temperature 100° C, the coefficient of expansion is given by dividing 0.043 by 500, and is, therefore, 0.000086. Experiments.  Determine the coefficient of expansion of the given liquid and of cubical expansion of the given solid. Enter results thus: Weight of empty tube: 5.06 g Weight of tube full at 15.5°: 11.58 g Weight of tube full at 100.6°: 11.32 g Weight of liquid at 15.5°: 6.52 g Weight expelled: 0.26 g Coefficient of expansion relative to glass: 0.000488 Coefficient of expansion of glass: 0.000026 True coefficient of expansion: 0.000514Similarly for the second experiment.


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