proportional to the difference of temperature, and the quantity of
heat in the gases is proportional to their temperatures. Draw a base
line representing -460° Fahr., the absolute zero of temperature. At
one end erect an ordinate, upon which set off T = 3,777°, the
temperature of the furnace. At 849° = _t_, on the scale of
temperature, draw a line parallel to the base, and mark on it a length
proportional to the heating surface of the boiler; join T by a
diagonal with the extremity of this line, and drop a perpendicular on
to the zero line. The temperature of the water in the boiler being
uniform, the ordinates bounded by the sloping line, and by the line,
_t_, will at any point be approximately proportional to the rate of
transmission of heat, and the shaded area above _t_ will be
proportional to the quantity of heat imparted to the water. Join T by
another diagonal with extremity of the heating surface on the zero
line, then the larger triangle, standing on the zero line, will
represent the whole of the heat of combustion, and the ratio of the
two triangles will be as the lengths of their respective bases, that
is, as (T - _t_) / T, which is the expression we have already used. The
heating surface was 220 square feet, and it was competent to transmit
the energy developed by 41 lb. of coal consumed per hour = 12,819 u. ×
41 u. = 525,572 units, equal to an average of 2,389 units per square
foot per hour; this value will correspond to the mean pressure in an
ordinary diagram, for it is a measure of the energy with which
molecular motion is transferred from the heated gases to the
boiler-plate, and so to the water. The mean rate of transmission,
multiplied by the area of heating surface, gives the area of the
shaded portion of the figure, which is the total work which should
have been done, that is to say, the work of evaporating 544 lb. of
water per hour. The actual work done, however, was only 485 lb. To
give the speculations we have indulged in a practical turn, it will be