_t_ₙ₁ - _t_₁
------------ = _v_,
_n_₁ - 1
_tʹ_ₙ₁ - _tʹ_₁
------------- = _vʹ_,
_n_₂ - 1
_t_₁ + _t_₂ + _t_₃ ... _t_ₙ₁
--------------------------- = t,
_n_₁
_tʹ_₁ + _tʹ_₂ + _tʹ_₃ ... _tʹ_ₙ₂
and -------------------------------- = _tʹ_.
_n_₂
The correction to be made to the difference between Θₙ - Θ₁ for the
influence of the outside temperature is determined by the formula of
Regnault-Pfaundler, which is as follows:
_v_ - _vʹ_ ⁿ⁻¹ Θₙ + Θ₁
∑ Δ_t_ = -------------- ( ∑ Θ_r_ + -------- - _nt_) - (_n_ - 1)_v_,
_tʹ_ - _t_ ₁ 2
ⁿ⁻¹
in which ∑ Θ_r_
₁
is calculated from the observation of the thermometer Θ₁, Θ₂ etc.,
made immediately after the combustion. It is equal to the sum of
observations Θ₁, Θ₂ etc., increased by an arbitrary factor equivalent
to (Θ₂ - Θ₁)/9, which is made necessary by reason of the irregularity
of the temperature increase during the first minute after combustion,
the mean temperature during that minute being somewhat higher than the
mean of the temperatures at the commencement and end of that time.
The quantity of heat formed by the combustion of the iron wire used for
igniting the sample is to be deducted from the total heat produced.
This correction may be determined once for all, the weight of the
iron wire used being noted and that of any unburned portion being
ascertained after the combustion.
Ten milligrams of iron, on complete combustion, will give sixteen
calories.
In the combustion of substances containing nitrogen, or in case the
free nitrogen of the air be not wholly expelled from the apparatus
before the burning, nitric acid is formed which is dissolved by the
water produced.
The heat produced by the solution of nitric acid in water is 14.3
calories per gram molecule. The quantity of nitric acid formed is
determined by titration and a corresponding reduction made in the total
calculated calories.
In the titration of nitric acid it is advisable to make use of an
alkaline solution, of which one liter is equivalent to 4.406 grams of
nitric acid. One cubic centimeter of the reagent is equivalent to a
quantity of nitric acid represented by one calorie.
Since the materials of which the bomb is composed have a specific heat
different from that of water, it is necessary to compute the water
thermal value of each apparatus.
The hydrothermal equivalent of the whole apparatus is most simply
determined by immersing it at a given temperature in water of a
different temperature.[581] With small apparatus this method is quite
sufficient, but there are many difficulties attending its application
to large systems weighing several kilograms. In these cases the
hydrothermal equivalent may be calculated from the specific heats of
the various components of the apparatus.
In calculating these values the specific heats of the various
components of the apparatus are as follows:
Brass 0.093
Steel 0.1097
Platinum 0.0324
Copper 0.09245
Lead 0.0315
Oxygen 0.2389
Glass 0.190
Mercury 0.0332
Hard rubber 0.33125
_Example._—It is required to calculate the hydrothermal value of a
calorimeter composed of the following substances:
Hydrothermal
value.
Steel bomb and cover, 2850 grams × 0.1097 312.65 grams.
Platinum lining, capsule and wires, 120 grams × 0.0324 3.89 ”
Lead washer, 100 grams × 0.0315 3.15 ”
Brass outer cylinder, 500 grams × 0.093 46.50 ”
Mercury in thermometer, 10 grams × 0.0332 0.33 ”
Glass (part of thermometer in water), 10 grams × 0.19 1.90 ”
Brass stirring apparatus (part in water), 100 grams
× 0.093 9.30 ”
------
Total water value of system 377.72 ”
When a bomb of 300 cubic centimeters capacity is filled with oxygen at
a pressure of twenty-four atmospheres it will hold about ten grams of
the gas, equivalent to a water value of 2.40 grams. Hence the water
value of the above system when charged, assuming the bomb to be of the
capacity mentioned, is 380.12 grams.
If the cylinder holding the water be made of fiber or other
non-conducting substance, its specific heat is best determined by
filling it in a known temperature with water at a definite different
temperature.
It is advisable to have the water cylinder of such a size as to permit
the use of a quantity of water for the total immersion of the bomb
which will weigh, with the water value of the apparatus, an even number
of grams. In the case above, 2622.28 grams of water placed in the
cylinder will make a water value of 3,000 grams, which is one quite
convenient for calculation.
=565. Computing the Calories of Combustion.=—In the preceding paragraph
has been given a brief account of the construction of the calorimeter
and of the methods of standardizing it and securing the necessary
corrections in the data directly obtained in its use. An illustration
of the details of computing the calories of combustion taken from the
paper of Stohmann, Kleber and Langbein, will be a sufficient guide for
the analyst in conducting the combustion and in the use of the data
obtained.[582]
Weight of substance burned, 1.07 grams.
Water value of system (water + apparatus), 2,500 grams.
Preliminary thermometric readings, _t_₁ = 26.8; _t_₂ = 27.2; _t_₃ =
27.7; _t_₄ = 28.1; _t_₅ = 28.5; _t_ₙ₁ = 28.9.
Thermometric reading after combustion, Θ₁ = 28.9; Θ₂ = 202; Θ₃ = 213;
Θ₄ = 214.2; Θₙ = 214.0.
Final thermometric readings, _tʹ_₁ = 214.0; _tʹ_₂ = 213.8; _tʹ_₃ =
213.6; _tʹ_₄ = 213.5; _tʹ_₅ = 213.3; _tʹ_₆ = 213.1; _tʹ_₇ = 212.9;
_tʹ_₈ = 212.7; _tʹ_₉ = 212.6; _tʹ_₁₀ = 212.4; _tʹ_ₙ₂ = 212.2.
From the formulas given above the following numerical values are
computed:
_v_ = 0.42.
_vʹ_ = -0.18.
_t_ = 27.9.
_tʹ_ = 213.1.
_n_ = 5.
ⁿ⁻¹ Θ₂ - Θ₁
∑ Θ_r_ = Θ₁ + Θ₂ + Θ₃ + Θ₄ + ------- = 667.
₁ 9
Substituting these values in the formula of Regnault-Pfaundler, the
value of the correction for the influence of the external air is
0.42 - (-0.18) 214 + 29
∑ Δt = [--------------- (677 + --------- - (5 × 27.9))
213.1 - 27.9 2
- (4 × 0.42)] = 0.45,
which is to be added to the end temperature (Θₙ = 214.0).
The computation is then made from the following data:
Corrected end temperature (Θₙ + 0.45) 214.45 = 15°.3699
Beginning temperature (Θ₁) 28.90 = 12°.8406
Increase in temperature 185.55 = 2°.5293
Total calories 2.5293 × 25000 = 6323.3
Of which there were due to iron burned 9.1
” ” ” ” nitric acid dissolved 8.2
Total calories due to one gram of substance 5893.5
The thermometric readings are given in the divisions of the thermometer
which in this case are so adjusted as to have the number 28.90
correspond to 12°.8406, and each division is nearly equivalent to
0°.014 thermometric degree.
The number of calories above given is the proper one when the
computation is made to refer to constant volume. By reason of the
consumption of oxygen and the change of temperature, although mutually
compensatory, the pressure may be changed at the end of the operation.
The conversion of the data obtained at constant volume referred to
constant pressure may be made by the following formula, in which [_Q_]
represents the calories from constant volume and _Q_ the desired data
for constant pressure, _O_ the number of oxygen atoms, _H_ the number
of hydrogen atoms in a molecule of the substance, and 0.291 a constant
for a temperature of about 18°, at which the observations should be
made.
_H_
_Q_ = [_Q_] + (--- - _O_) 0.291.
2
=566. Calorimetric Equivalents.=—By the term calorie is understood the
quantity of heat required to raise one gram of water, at an initial
temperature of about 18°, one degree. The term ‘Calorie’ denotes the
quantity of heat, in like conditions, required to raise one kilogram of
water one degree.
For purposes of comparison and for assisting the analyst in adjusting
his apparatus so as to give reliable results, the following data,
giving the calories of some common food materials, are given:
Substance. Chemical composition.
Proteids. Calories. C. H. N. S. O.
Per Per Per Per Per
cent. cent. cent. cent. cent.
Serum albumin 5917.8 53.93 7.65 15.15 1.18 22.09
Casein 5867.0 54.02 7.33 15.52 0.75 22.38
Egg albumin 5735.0 52.95 7.50 15.19 1.51 22.85
Meat free of
fat and
extracted
with water 5720.0 52.11 6.76 18.14 0.96 22.66
Peptone 5298.8 50.10 6.45 16.42 1.24 25.79
Proteids (mean) 5730.8 52.71 7.09 16.02 1.03 23.15
Glycerids.
Butterfat 9231.3
Linseed oil 9488.0
Olive oil 9467.0
Carbohydrates. Formula.
Arabinose 3722.0 C₅H₁₀O₅
Xylose 3746.0 C₅H₁₀O₅
Dextrose 3742.6 C₆H₁₂O₆
Levulose 3755.0 C₆H₁₂O₆
Sucrose 3955.2 C₁₂H₂₂O₁₁
Lactose 3736.8 C₁₂H₂₂O₁₁ + H₂O
Maltose 3949.3 C₁₂H₂₂O₁₁
=567. Distinction between Butter and Oleomargarin.=—Theoretically
the heats of combustion of butter fat and oleomargarin are different
and de Schweinitz and Emery propose to utilize this difference for
analytical purposes.[583] The samples of pure butter fat examined by
them afforded 9320, 9327 and 9362 calories, respectively. The calories
obtained for various samples of oleomargarin varied from 9574 to
9795. On mixing butter fat and oleomargarin, a progressive increase
in calorimetric power is found, corresponding to the percentage of the
latter constituent. Lards examined at the same time gave from 9503 to
9654 calories.