Jump to a New ChapterIntroduction to the SAT IIIntroduction to the SAT II Chemistry TestStrategies for Taking the SAT II Chemistry TestThe Structure of MatterThe States of MatterReaction TypesStoichiometryEquilibrium and Reaction RatesThermodynamicsDescriptive ChemistryLaboratoryBasic Measurement and Calculation ReviewChemical Formulas Review: Nomenclature and Formula WritingPractice Tests Are Your Best Friends
 9.1 Enthalpy 9.2 Spontaneous Reactions 9.3 Heat Capacity and Specific Heat 9.4 Enthalpies of Reactions 9.5 Hess’s Law

 9.6 More About Entropy 9.7 Gibb’s Free Energy 9.8 Practice Questions 9.9 Explanations
Heat Capacity and Specific Heat
We can determine the value of DH for a reaction in the lab by using a calorimeter, which measures the heat flowing into and out of a system as the reaction proceeds. We’ll talk more about this in chapter 11, the Laboratory chapter. The heat capacity of a substance is the amount of heat energy it must consume in order to raise its temperature by 1K or 1ºC; it can be expressed using either the units joules or calories. A calorie is defined as the amount of heat needed to raise the temperature of 1.00 gram of water by 1.00ºC, and joules are the SI units for energy; 1 calorie = 4.184 joules.
Every pure substance involved in a chemical reaction has a unique heat capacity, and the heat capacity of 1 mol of a pure substance is known as its molar heat capacity (J/mol-K or J/mol-ºC). The heat capacity of 1 gram of a substance is known as its specific heat (J/g-K). The following equation relates the specific heat of a substance, the temperature change, the mass of the substance, and how much energy was put into the system:
q = mCpDT
where
q = quantity of heat (joules or calories)
m = mass in grams
DT = Tf - Ti (final – initial)
Cp = specific heat capacity (J/g ºC)
One thing about specific heats that you should probably remember for the SAT II Chemistry test is that the specific heat of liquid water is 4.184 J/g ºC (or 1.00 cal/g ºC), which is unusually high (this is due to hydrogen bonding).
Example
How much energy would be needed to heat 450 grams of copper metal from a temperature of 25.0ºC to a temperature of 75.0ºC? The specific heat of copper at 25.0ºC is 0.385 J/g ºC.
Explanation
Given mass, two temperatures, and a specific heat, you have enough values to plug into the specific heat equation
q = mCpDT
and plugging in your values you get
q = (450 g) (0.385 J/g ºC) (50.0ºC)
= 8700 J
Example
100.0 mL of 1.0 M NaOH and 100.0 mL of 1.0 M HCl are mixed in a calorimeter. Both solutions were originally at 24.6ºC. After the reaction, the final temperature is 31.3ºC. Assuming that the solution has a density of 1.0 g/cm3 and a specific heat capacity of 4.184 J/g ºC, calculate the enthalpy change for the neutralization of HCl by NaOH. Assume that no heat is lost to the surroundings or the calorimeter.
Explanation
We are looking for energy in this problem too. Remember that the solution has a density of 1.08/cm3, and we have a total of 200 ml of solution, so the total mass to be considered is 200 g. You have all of the information you need, so again plug it into the formula q = mCpDT:
q = (200 g) (4.184 J/g ºC) (6.7 ºC)