88 Chemical and Physical Notes 



to record the measurements in terms of the centimetre, and 

 the volume is then given in cubic centimetres. The external 

 cooling surface of the bulb is found by adding the hemi- 

 spherical surface of the exposed end to the cylindrical surface, 

 which is the product of the length of the cylinder into its 

 circumference. The area of the external surface is given in 

 square centimetres (cm. 2 ). It is important to observe that, 

 when cooling, a thermometer loses heat through the whole 

 area of its external surface ; when receiving heat from a 

 particular direction, as, for instance, from the sun, the re- 

 ceiving area is, for the cylindrical part of the bulb, its length 

 multiplied by its diameter, and for the hemispherical ends 

 the area of a great circle of the sphere. 



All these measurements have been carefully made on the 

 bulb of the thermometer by Chabaud, col. 4, Table XIII. 

 For the determination of the circumference, a fine thread 

 was wound forty times round the cylindrical part. When 

 unwound it measured 1 1 1 cm., whence we have : 



Circumference 2*775 cm - 



Diameter 0*883 cm- 

 Circular area of cylinder ... ... ... 0-6124 cm. 2 



Length of cylinder 4-08 cm. 



Volume of cylinder 4*08x0*6124 = 2*498 c.c. 



Volume of sphere 0-883 cm - m diameter ... 0-377 c.c. 



Whence total volume of bulb 2-875 c - c - 



Water value of bulb 0*475 x 2-875 = 1-366 grms. 



It will be seen that by mensuration we arrive at 2-875 c - c - 

 as the volume of the bulb in place of 2*835 c - c - as derived from 

 the weights supplied by the maker. The difference, 0*04 c.c., 

 is under \\ per cent, of the whole volume. Applying 0*475 

 as the specific heat per unit volume of the bulb, we obtain for 

 the water value 1-366 in place of 1-338 grms., a difference of 

 0-028, which is just 2 per cent. The agreement is quite 

 satisfactory, and we can therefore use this method with 

 perfect confidence in determining the water value of the 

 bulbs of all our thermometers. 

 The area of the external cylindrical surface is 4-08 x 2*775 = J r 3 22 cm - 2 



Surface of one hemisphere = ^225 cm. 2 



Whence the total external surface =12*547 cm. 2 



