Calorimetry Lab Abstract And Prelab Chemistry Assignment

895 words - 4 pages

NFHS – Chem I - HONORS Calorimetry Lab – Specific Heat of a Metal
Calorimetry Lab – Specific Heat of a Metal
Abstract
Chemists can identify substances on the basis of their chemical and physical properties. One unique physical property of a substance is the amount of energy it will absorb per unit of mass. This property can be measured quite accurately and is called specific heat (Cp). Specific heat is the amount of energy measured in joules, needed to raise the temperature of one gram of the substance one Celsius degree. Often applied to metallic elements, specific heat can be used as a basis for comparing energy absorption and transfer.
To measure specific heat in the laboratory, a calorimeter of some kind must be used. A calorimeter is a well-insulated container used in measuring energy changes. The calorimeter is insulated to reduce the loss or gain of energy to or from the surroundings. The Law of Conservation of Energy states that energy is always conserved during physical processes and chemical reactions. We have observed that energy always flows from an object at a higher temperature to an object at a lower temperature. The heat gained by the cooler substance equals the heat lost by the warmer substance, if we assume no loss of heat to the surrounding environment.
heat lost = heat gained
In this experiment, you will determine the specific heat of a metal sample. The metal sample will be heated to a high temperature and then placed into a calorimeter containing a known quantity of water at a lower temperature. Having measured the mass of the water in the calorimeter and the temperature change of the water (ΔT), and knowing the specific heat of water (4.184 J/g °C), the heat gained by the water (lost by the metal) can be calculated as follows:
qwater = mwater * Cp-water * ΔTwater (eqn 1)
heat gained by the water = mass of water * specific heat of water * change in temperature
(J) (g) (4 .184 J/g °C) (°C)
As described in the abstract, the amount of heat gained by the water = the amount of heat lost by the metal:
qwater = qmetal (eqn 2)
Once the amount of heat gained by the water (qwater ) is calculated, we know that the amount of heat lost by the metal is the same value (qmetal).
Substituting the values for the metal:
qmetal = mmetal * Cp-metal * ΔTmetal (eqn 3)
heat lost by metal = mass of metal * specific heat of metal * change in temperature of metal
(J) (g) (?? J/g °C) (°C)
Since all values are known except the specific heat of metal, simply rearrange the...

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