the value in terms of a unit of work of a thermal unit of heat. By his experiments Joule has found that 772.55 foot-pounds = pound of water by degree Fahr., the intensity of gravity being that of the sea-level at Greenwich. For ordinary calculations the value 772 is sufficiently exact. The value for any other unit of work and any other unit of heat can be found by the arithmetical process of conversion. Thus .'. i.e., 9 degs. Fahr. = 5 degs. Cent., 772 × 9 foot-pounds = 5 pound of water by deg. Cent., 1,390 foot-pounds = pound of water by deg. Cent. In the case of the French units, 424 kilogrammetres = kgm. of water by deg. Cent. It is customary to denote by J any one of these three factors. EXAMPLES. Ex. 1. How much boiling water and how much freezing water must be taken to make up a pound of water at blood-heat, that is, 97° Fahr.? Suppose a lb. of boiling water, then 1 x lb. of freezing water. The change of temperature in the first case is 212-97 deg. Fahr., therefore the heat given out is (21297) lb. of water by deg. Fahr. In the second case the change of temperature is 97-32 deg. Fahr., therefore the heat taken in is (1 − x)(97 – 32) lb. of water by deg. Fahr. Now, if there be no loss of heat to other bodies, hence 36 Answer-13 lb. of boiling, and 33 lb. of freezing water. Ex. 2. Reduce pound by deg. Cent. to ergs. 1,390 foot-pounds = lb. of water by deg. Cent., .. 1·356 × 1,390 × 107 ergs i.e., 1.9 × 1010 ergs = foot-pound; = lb. of water by deg. Cent., Ex. 3. In one of Rumford's experiments, a horse working for 2 hours 30 minutes raised the temperature of a mass, equivalent in capacity for heat to 26.6 lbs. of water, by 180 degs. Fahr. Taking the work of one horse at 30,000 foot-pounds per minute, calculate from this experiment the dynamical equivalent of heat. The heat produced was Now Hence 26.6 × 180 lb. of water by deg. Fahr. 30,000 foot-pounds per minute, the work expended was 150 x 30,000 foot pounds. 266 × 18 foot-pounds = lb. of water by deg. Fahr. i.e., 940 foot-pounds = lb. of water by deg. Fahr. EXERCISE XL. 1. Reduce pound of water by degree Fahrenheit to ergs. 2. Find the relation between the pound of water by degree Fahrenheit, and the kilogramme of water by degree Centigrade. 3. The temperature of a fluid is ascertained by the hand to be the same as that of a mixture of 3 pounds of water taken at 0° C. with 7 pounds of water taken at 100° C.; what is the temperature of the fluid? 4. What is the amount of available heat in a quart bottle, filled with water at 95o C., and capable of being cooled to 10° C. ? 5. A mass of 100 lbs. falls 100 feet, and after striking a fixed obstacle rebounds 30 feet; calculate the value in terms of the British thermal unit of the work done. 6. A leaden bullet of 2 oz. strikes a target at a speed of 1,000 feet per sec., and is stopped. Find in terms of the British unit the whole amount of heat generated in the impact? 7. The Niagara Falls are 165 feet in height. Find by how much the temperature of the water will be increased by the fall, supposing that the whole energy of the water due to the fall is transformed into heat. 8. Given that the frictional resistance to a passenger train is 17 pounds per ton of load; what is the amount of work done against friction by a train of 75 tons in going a journey of 10 miles? Also what is the amount of heat developed? 9. By the consumption of a gramme of carbon 8,000 units of heat are produced; if 40 per cent. of this be employed in raising a mass of one gramme, how high will it be raised, the intensity of gravity being 981 cm. per sec. per sec.? Р 10. The combustion of one pound of coal raises the temperature of 100 gallons of water through 4'4 degrees Cent. What is the mechanical equivalent of the absolute thermal effect of the coal? 11. An engine consumes 40 lbs. of coal of such calorific power that the heat developed by the combustion of one lb. is capable of converting 16 lbs. of water at 100° C. into steam at the same temperature, and during the process the engine performs 16 × 107 foot-pounds of work. What percentage of the heat produced is wasted? SECTION XLI.--THERMAL CAPACITY. ART. 180.-Thermal Capacity per Unit Mass. The thermal capacity per unit mass of a substance, commonly called the thermal capacity, is the number of units of heat required to raise unit of mass of the substance one degree in temperature. It is expressed in the form c H = M of substance by ✪ difference. The value of c varies from temperature to temperature, increasing as the temperature gets higher. When H is defined as 1 H M of water by = difference, the thermal capacity of water is evidently 1, unless the temperature of the water differs much from the temperature at which the unit is defined. When the mass is constant, say m M, then mc H = difference. This equivalence expresses the thermal capacity of a body. When, on the other hand, the interval of temperature is constant, t✪ to t2 ✪, we have a rate of the form k HM of substance. When t2-t is large, k is not equal to (t2-t1) c, unless c is the average value of the thermal capacity for the interval. ART. 181.-Thermal Capacity per Unit Volume. Given the thermal capacity per unit mass of a substance as c HM of substance by and the density of the substance as PM=V; then by eliminating M we get difference, pc HV of substance by → difference. This expresses what is called the thermal capacity per unit volume. When the unit of heat is defined in terms of water, and the unit of mass by the density of water, the thermal capacity per unit volume for water is 1, unless the temperature of the water differs greatly from its standard temperature. ART. 182.-Relative Heat and Specific Heat. Let the thermal capacities of two substances A and B be and then CH per M of B=✪ difference; c/c H per M of A = H per M of B. This means that c1 units of heat communicated to unit mass of the substance A are equivalent in respect of changing the temperature to c units of heat communicated to unit mass of the substance B. When the substance B is water, and H is a water-unit, the relative heat becomes the specific heat; c2 becomes 1, and c1 is equally the value of the thermal capacity per unit mass, and the specific heat referred to unit of mass. The specific heat is expressed by CM of water by ✪ difference per M of AM of water by ✪ difference per M of water; which is a ratio so far as dimensions of units are concerned. It is evident that when the thermal capacity per unit volume is taken the relative heat will be CP/CPH per V of A = H per V of B; and that the specific heat referred to volume will be Gs M of water by ✪ difference per V of A = M of water by → difference per V of water, where s denotes the value of the specific mass of A. ART. 183.-Resistance to Change of Temperature; Water Equivalent. Suppose that the thermal capacity of a substance A is This expresses equivalent masses, the equivalence being in respect of resistance to change of temperature. When B denotes water and H is defined in terms of water (Art. 177) c2 is 1, and C1 M of water = M of A. This is the water equivalent of unit of mass of the substance. When a body contains m, M of substance A, m2 M of B, and my M of C, then its water equivalent is M11 + M2C2 + MC3 M of water. In a similar manner we derive PC2/PCV of A = V of B, which expresses equivalent volumes with respect to the same property. When A is water about its standard temperature this sc V of water = V of substance; expresses water equivalent referred to volume. |