 |
 |
 |
 |
 |
 |
|
|||
|
|
|
|||||||||||||||||||||||||||||||||||||
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)
The answer is -5.6 kJ/mol.
|
|
 |
