| M3 | Lesson #2 |
Topics;
Laws of Thermodynamics
Heat vs Temperature
Calculating Energy Changes ( with temperature change
)
Enthalpy
Representing enthalpy
Energy in the lab is most easily measured as heat. As you can see from the demonstration relating to heat and temperature , the amount of heat transferred can be readily determined by changes in temperature of the surroundings. When a substance undergoes a temperature change without a phase, chemical or nuclear change occurring , three factors determine the amount of heat (Q) a substance will absorb or release.The three factors are:
1) mass of the substance (m).
2) specific heat capacity of the substance (a measure of how much energy the substance absorbs to change the temperature 1oC).
3) temperature change the substance undergoes (T).
Graphing relationships between
a) heat and temperature change
b) heat and mass.
Click on the pencil and complete the assignment
Heat Calculation formula
The factors indicated above all have a direct relationship with the amount of heat transferred. Therefore the following formula can be developed for the calculation of heat when a temperature change occurs (without a phase, chemical or nuclear change happens).
Heat (Q): is measured in Joules (note that the units below multiply to give you Joules) Joules is a metric unit of measurement.
Mass (m): measured in grams
Specific Heat Capacity (c): measured in J/ g * oC
Temperature change (Examples of Calculation of Heat (with no phase, chemical or nuclear change)
Example : How much heat, in joules is absorbed by 300 grams of water, if the temperature change the water undergoes is 15 oC ? (The specific heat capacity of water is 4.19 J/g* oC.)
Answer: 1) Write down the given information:
mass of water = 300 grams
temperature change in water = 15 oC
specific heat capacity = 4.19 J/ g * oC2) Determine the information required:
amount of heat in joules
3) Analyze the problem , make a plan
To solve for heat we multiply mass in grams with by the temperature change in oC and specific heat capacity.
4) Apply the formula and solve question.
Q = (300 g) * ( 4.19 J/g * oC ) * ( 15 oC )
Q = 18,855 J or 19 kJ
5) State the answer
The heat absorbed was 19 kJ
Check your understanding
Question: Calculate the specific heat capacity of concrete if 3,620 J of heat causes the temperature of a 250 gram sample to change from 10.15 oC to 26.50 oC?
click on the eye to view answer
Assignment: Heat Calculation #1
