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cub. cm. in a tube over mercury, the mercury standing 73 cm. higher inside the tube than outside. (1 gramme of air at o° C. under a pressure of 76 cm. of mercury measures 773.4 cub. cm.)

Int. Sc. 1885.

Let h denote the height of the barometer, then the pressure of the air in the tube over mercury is h-73. At this pressure, and at 27°, a milligramme of air measures 20 c.c., whereas at o° and 76 cm. pressure it measures o.7734 c.c. By equation (9), p. 102, we have

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...h-73=76× 0.7734 × 300/20 x 273=3.23, and the required barometric height is 76.23 cm.

116. How may the relation between the pressure and temperature of a given mass of air at constant volume be determined?

A quantity of air occupies 10 cubic feet at o° C. and ́under a pressure of 20 inches of mercury. What will be its volume at 30° C. and under a pressure of 1200 inches of mercury? Matric. 1884.

117. Explain accurately what is meant by the statement that the coefficient of expansion of air is 1/273. The volume of a certain quantity of air at 50° C. is 500 cubic inches. Assuming no change of pressure to take place, determine its volumes at 50° C. and at 100° C. respectively.

Matric. 1887.

118. What is meant by saying that the absolute temperature of a gas is 300° C.? If the absolute temperature be 260° C., what is the temperature in the Centigrade scale?

Two condensers contain equal quantities of air. One of them at temperature 47° C. is 30 in. long, 20 in. broad, and 10 in. high; and the other at 57° C. is 30 in. long, 25 in. broad, and 81 in. high. pressure of the air is the same in both.

Show that the Camb. B.A. 1883.

119. A vessel contains air at o° C. and at atmospheric

pressure.

It is heated to 100° C., and during the process one ounce of air escapes. How many ounces of air were there originally in the vessel, the expansion of the vessel itself being neglected? The coefficient of expansion of air at constant volume is

1/273.

Matric. 1886.

120. A given quantity of a gas is made continually to occupy the same space. Explain what changes will take place in its pressure when changes take place in its temperature.

A straight vertical tube, the section of whose bore is one inch, is closed at its lower end, and contains a quantity of air which supports an air-tight piston whose weight is 1 lb. The position of the piston is observed when the temperature of the air is 31° C., and the weight of the piston is then increased by 1 lb. Find what increase of temperature will be required to bring back the piston to its former position, the atmospheric pressure being 15 lbs. per square inch, and the absolute zero of the air thermometer being - 273° C. Camb. B.A. 1884.

CHAPTER IV

SPECIFIC AND LATENT HEAT

1. Specific Heat.

1. A COIL of copper wire weighing 45.1 gm. was dropped into a calorimeter containing 52.5 gm. of water at 10°. The copper before immersion was at 99°.6, and the common temperature of copper and water after immersion was 16°.8. Find the specific heat of the copper wire. The quantity of heat (Q) given out by a body of mass m and specific heat s in cooling through an interval of temperature 0 is Q=mse. Thus if s denote the specific heat of the copper, the amount of heat evolved by it in cooling from 99.6 to 16°.8 is 45.1 × s × (99.6 – 16.8).

Since the specific heat of the water is unity, the amount of heat required to raise its temperature from 10° to 16°.8 is 52.5 × 1 × (16.8 - 10), and as no heat is supposed to be gained or lost these two quantities are equal.

and

... 45.1 × sx 82.8 = 52.5 × 6·8,
S=357/3734-3=0.0956.

2. What is the temperature of an iron ball weighing 5 lbs., which, when immersed in 8 lbs. of water at 13°, raises the temperature to 48°? The specific heat of iron is 0.112.

If the temperature of the ball before immersion was t°, the number of heat-units 1 which it gives out, in cooling to the final temperature of 48°, is 5 × 0.112 × (t − 48).

1 Since specific heat is merely a number, or a numerical ratio between quantities of heat, it is independent of the unit of

I

Assuming that no heat is gained or lost during the experiment, this must be equal to the number of heat-units absorbed by the cold water, i.e. to 8 × (48 - 13). Equating these two quantities, we have

and

5 × 0.112 × (t-48) = 8 × (48 - 13),

... 0.56t=8× 35 + (0·56 × 48),
=280+26.88=306.88,

t=548°.

3. In order to determine the specific heat of silver, a piece of the metal weighing 10.205 gm. was heated to 101.9 and dropped into a calorimeter containing 81.34 gm. of water, the temperature of which was raised from 11°.09 to 11°.71. The water equivalent of the calorimeter, agitator, and thermometer employed was 2.91 gm.: find the specific heat of the silver.

The heat evolved by the hot body is 10.205 × 5 × (101.9 – 11.71), where s is the sp. heat of the silver. The heat is partly absorbed by the water and partly by the calorimeter, etc., and these are together equivalent to (81.34 +2.91) gm. 84.25 gm. Since these are raised from 11°.09 to 11.71, the heat absorbed is 84.25 (11.71 - 11.09). Equating these quantities, we have

and

=

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mass (or weight) employed; and this is also true for latent heat. In the statement or solution of a problem it is a matter of indifference whether we take as our unit of mass the pound, the kilogramme, or the gramme, provided that we use this unit consistently, the corresponding units of heat in the three cases being the pound-degree (or amount of heat required to raise one pound of water through 1o C.), the kilogramme-degree, and the gramme-degree. The last, which is the C. G.S. unit, is sometimes called a "calorie," and Berthelot distinguishes the kilogramme-degree from this by calling it a "Calorie" (1 Calorie = 1000 calories).

4. The same piece of silver was heated to 102°.2 and immersed in 75.3 gm. of turpentine at 10°.98: the experiment was performed with the same apparatus as in Ex. 3, and the final temperature was 12°.47. Calculate

the specific heat of the turpentine.

Taking the sp. heat of the silver as 0.05677, the amount of heat which it gives out is

10.205 × 0.05677 × (102.2 – 12.47) = 51.97.

Of this, 75.3 × × (12.47 – 10.98) is absorbed by the turpentine, s being its sp. heat; and the calorimeter absorbs

2.91 × (12.47 – 10.98)=4.336.

Equating the amounts of heat absorbed and emitted, 51.97 = (75.3 × 5 × 1·49) + 4·336,

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5. A certain vessel holds 800 c.c. of water at its temperature of maximum density (4°). How much heat must be imparted to the water before it begins to boil? 6. Define specific heat. A body of mass M and specific heat S at a temperature T° is dropped into a mass m of a liquid of specific heat s at t°: prove that the final temperature is

0=

MST+mst
MS+ms

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7. How many units of heat are required to raise the temperature of 150 grammes of copper (of specific heat 0.095) from 10° to 150°?

8. What amount of heat must be given to an iron armour-plate 2 metres long, I metre broad, and 20 cm. in thickness, in order to heat it from 10° to 140°? [Sp. gr. of iron, 7.7; sp. heat, o.112.]

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