0.8 Energetics of chemical reactions  (Page 2/7)

Our observations do reveal that we can relate the temperature rise produced in a substance to a fixed quantity ofheat, provided that we specify the type and amount of the substance. Therefore, we define a property for each substance,called the heat capacity , which relates the temperature rise to the quantity of heat absorbed. We define $q$ to be the quantity of heat, and $\Delta (T)$ to be the temperature rise produced by this heat. The heat capacity $C$ is defined by

This equation, however, is only a definition and does not help us calculate either $q$ or $C$ , since we know neither one.

Next, however, we observe that we can also elevate the temperature of a substance mechanically , that is, by doing work on it. As simple examples, we can warm water by stirring it, or warmmetal by rubbing or scraping it. (As an historical note, these observations were crucial in establishing that heat is equivalentto work in its effect on matter, demonstrating that heat is therefore a form of energy.) Although it is difficult to do, we canmeasure the amount of work required to elevate the temperature of 1g of water by 1°C. We find that the amount of work requiredis invariably equal to 4.184J. Consequently, adding 4.184J of energy to 1g of water must elevate the energy of the watermolecules by an amount measured by 1°C. By conservation of energy, the energy of the water molecules does not depend on howthat energy was acquired. Therefore, the increase in energy measured by a 1°C temperature increase is the sameregardless of whether the water was heated or stirred. As such, 4.184J must also be the amount of energy added to the watermolecules when they are heated by 1°C rather than stirred. We have therefore effectivelymeasured the heat $q$ required to elevate the temperature of 1g of water by 1°C. Referringback to , we now can calculate that the heat capacity of 1g of water must be $4.184\frac{J}{°C}$ . The heat capacity per gram of a substance is referred to as the specific heat of the substance, usually indicated by the symbol $c_s$ . The specific heat of water is $4.184\frac{J}{°C}$ .

Determining the heat capacity (or specific heat) of water is an extremely important measurement for tworeasons. First, from the heat capacity of water we can determine the heat capacity of any other substance very simply. Imaginetaking a hot 5.0g iron weight at 100°C and placing it in 10.0g of water at 25°C. We know from experience that theiron bar will be cooled and the water will be heated until both have achieved the same temperature. This is an easy experiment toperform, and we find that the final temperature of the iron and water is 28.8°C. Clearly, the temperature of the water hasbeen raised by 3.8°C. From and the specific heat of water, we can calculate that the water must have absorbed an amount of heat $q=10.0g4.184\frac{J}{g°C}3.8°C=159J$ . By conservation of energy, this must be the amount of heat lost by the 1g iron weight, whose temperature was lowered by 71.2°C. Again referring to , we can calculate the specific heat of the iron bar to be $c_s=\frac{-159J}{-71.2°C5.0g}=0.45\frac{J}{g°C}$ . Following this procedure, we can easily produce extensive tables ofheat capacities for many substances.