Download Calorimetry In all physical and chemical reactions energy is either

Calorimetry
In all physical and chemical reactions energy is either released or absorbed. If
energy is released in a reaction, it is said to be exothermic and the potential
energy of the reactants will be higher than the potential energy of the products.
If energy is absorbed in a reaction, it is said to be endothermic and the
potential energy of the reactants will be lower than the potential energy of the
products. The amount of energy being released or absorbed in a reaction can
be measured using a device called a calorimeter. A calorimeter is a device
made out of insulating materials and contains a reaction chamber within that is
surrounded by water, or a calorimeter could be something as simple as a
styrofoam cup. When the reaction takes place within the reaction chamber,
energy is either released to the water surrounding the reaction chamber or it is
absorbed from the water surrounding the chamber. The amount of energy
released or absorbed in the reaction can be calculated if you know the
following three things:



Temperature change of the water in the calorimeter
Mass of the water in the calorimeter
Specific heat of water
The specific heat of water is the amount of heat needed to raise the temperature
of 1 gram of water by 1 °C. The accepted value for the specific heat of water is
4.184 Joules per gram per °C. The equation that relates the 3 variables above is:
q = mcΔT
Where q is equal to the amount of heat released or absorbed in joules.
Where m is the mass of water in the calorimeter in grams.
Where c is the specific heat of water in J/g.°C.
Where *T is the change in temperature of the water in °C.
Example 1:
How many joules of heat are released in a reaction where 350.0 grams of water
in a calorimeter increase in temperature from 20.0 °C to 35.0 °C?
q=?
m = 350.0 g
c = 4.184 J/g.°C
*T = 35.0 °C - 20.0 °C = 15.0 °C
q = (350.0 g) (4.184 J/g.°C) (15.0 °C)
q = 21966 joules = 2.20 x 104 joules
Example 2:
What is the specific heat of a piece of aluminum if a 25.0 gram sample at
85.0°C is placed in a styrofoam cup calorimeter containing 150.0 mL of water
at 25.0°C and the temperature of the water in the calorimeter increases to
27.1°C?
When the hot piece of aluminum is placed into the cooler water, the aluminum
will cool down and the water will warm up until they are at the same
temperature (27.1°C). The heat lost by the aluminum (qAl) should equal the
heat gained by the water (qH2O). If we use the equation q= mc*T, we can
calculate the heat gained by the water and use that number to calculate the
specific heat of the aluminum.
q = mcΔT
qH2O = ?
mH2O = 150.0 grams (the density of water is 1 g/mL, so the milliliters of water
equal the grams)
cH2O = 4.184 J/g.°C
*TH2O = 2.1°C (27.1°C - 25.0°C)
qH2O = 150.0 g x 4.184 J/g.°C x 2.1°C
qH2O = 1.3 x 103 Joules
qH2O = qAl
qAl = 1.3 x 103 Joules
mAl = 25.0 grams
cAl = ?
*TAl = 57.9°C (85.0°C - 27.1°C)
q = mcΔT
1.3 x 103J = 25.0 g x c x 57.9°C
cAl =
1.3 x 103J
(25.0 g)(57.9°C)
cAl = 0.90 J/g.°C