Download Understanding Calorimetry: Heat Measurements and Specific Heats of Common Substances and more Exercises English in PDF only on Docsity!
Calorimetry
Calorimetry is the study of heat measurements of chemical and physical systems using the calorimeter, a device consisting of closed container. The name calorimeter, however, is based on the calorie , a basic unit of energy.
Units of Energy
The calorie originally defined as the amount of heat required to raise the temperature of one gram of water by one degree centigrade. The calorie is the old unit but still it is in usage. Now there is a new SI unit. This unit is joul e (abbreviated as J ) (named after the English scientist James Joule), which is defined as the amount of energy or work required to accelerate 1 kilogram (kg) of object through 1 meter (m) per second(s). Therefore,
1 J = kg m^2 s -
One joule is a rather small amount of energy and in most cases larger unit kilojoules (kJ) is used.
1 kJ = 1000 J =1 x 10 3 J
The calorie and joule are related through the following relationship:
1 cal = 4.184 J (exactly)
The larger unit of calorie is the kilocalorie (kcal), which is 1000 calories that can also be related to kilojoules.
1 kcal = 1000 cal = 1 x 10 3 cal
1 kcal = 4.184 kJ
The Dietary (Nutritional) Unit
The energy content of food items are expressed in Calorie (note the capital C), which is actually 1 kilocalorie.
1 Cal = 1000 cal = 1 kcal = 4.184 kJ
The measurement of heat changes requires the understanding of two quantities, heat capacity and specific heat.
Heat Capacity and Specific Heat
The heat capacity (C) is defined as the amount of heat necessary to raise the temperature of a given amount of substance by one degree Celsius. The unit of heat capacity is cal/ 0 C or J/ 0 C. The specific heat (sH) , on the other hand, is defined as the amount of heat required to raise the temperature of one gram of the substance by one degree Celsius. The unit of specific heat is cal/g x 0 C or J / g x 0 C. The following table lists few specific heats of some common substances.
Substance
Specific Heat ( J/gx^0 C)
Aluminum 0. Carbon (graphite) 0. Carbon(diamond) 0. Copper 0. Ethanol 2. Glass(Pyrex) 0. Gold 0. Granite 0. Iron 0. Lead 0. Mercury 0. Sand 0. Silver 0. Water 4.
Note the difference between the heat capacity and the specific heat; the heat capacity refers to the given amount of substance whereas the specific heat refers to one gram of substance. The heat capacity is the extensive property that means it depends on the volume of the object. The specific heat is an intensive property that means it does not depend on the volume of an object. The relationship between these two is,
C = mass x specific heat = m x s (^) H
where m is the mass of the substance in grams. The amount of heat (q) is nothing but the heat capacity times the change in temperature, that is,
q = C x Δt = m x s (^) H x Δt
where Δt is the change in temperature, which is
Δt = tfinal - t (^) initial
The heat gained by the water = (mass x specific heat x change in temperature) (^) water = (m x s (^) H x Δt) (^) water
At the equilibrium,
Heat gained by the water = - Heat lost the metal
(q x s (^) H x Δt ) (^) metal = - (q x s (^) H xΔt) (^) water
Solving this equation for the specific heat of metal assumes the following equation.
( ) ( )
H water H metal
m x s x t s
m x Δ t
Knowing the mass of water, the specific heat of water, the temperature change of water, the mass of metal, and the temperature change of metal, the specific heat of metal is calculated using the above equation.
Once the specific heat of the metal is determined in this way, the approximate atomic mass of the metal can be calculated using the Dulong-Petit’s law:
Atomic mass x Specific heat (J / g x 0 C) ≈ 25
or Atomic mass = 25 / Specific heat (J / g x 0 C)
Example
A 35.8 gram of metal is heated to 100 0 C and then poured into 40.0 ml (40 g ) of water (^) , which was originally at 25.0 0 C. If the equilibrium temperature was 30.7 0 C, what is the specific heat of the metal? What is its approximate atomic mass according to Dulong- Petit’s laws?
Answer
The heat gained by the water = (m x sH x Δt) (^) water = 40 g x 4.184 J/gx 0 C x (30.7 – 25) 0 C = 953.952 J
The heat lost by the metal = -953.952 J
Therefore, the specific heat of the metal is,
s (^) H = - 953.952 J / ( m x Δt) (^) metal = - 953.952 J / ( 35.8 g x ( 30.7 – 100) 0 C = 0.385 J/g x 0 C
According to Dulong-Peti’s law,
Approximate atomic mass = 25 / 0.385 = 64.
Click on
Hands on Practice to Calculate Specific Heat
