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Solving Problems: A Chemistry Handbook
Chemistry: Matter and Change
151
Energy and Chemical Change
Energy and Chemical Change
SOLVING PROBLEMS:
A CHEMISTRY HANDBOOK
CHAPTER
15
15.1 Energy
Energy is the ability to do work or produce heat. Heat is commonly
measured in joules or calories. One calorie is equivalent to
4.184 joules.
The amount of heat required to raise the temperature of one
gram of a substance by one degree Celsius is the specific heat of the
substance. Liquid water has a high specific heat of 4.184 J/(g и °C).
By contrast, the specific heat of iron is 0.449 J/(g и °C). When the
temperature of a substance changes, the amount of heat absorbed or
released is given by the following equation.
q ϭ c ϫ m ϫ ⌬T
In the equation, q ϭ the heat absorbed or released, c ϭ the specific
heat of the substance, m ϭ the mass of the sample in grams, and ⌬T
is the change in temperature in °C.
Example Problem 15-1
Calculating Heat
A silver bar with a mass of 250.0 g is heated from 22.0°C to 68.5°C.
How much heat does the silv1er bar absorb?
Use the equation for heat.
q ϭ c ϫ m ϫ ⌬T
The temperature change is the difference between the final temperature and the initial temperature.
⌬T ϭ 68.5°C Ϫ 22.0°C ϭ 46.5°C
From Table 16-2 in your textbook, the specific heat of silver is
0.235 J/(g и °C). Substitute the known values to solve for the amount
of heat absorbed.
q ϭ 0.235 J/(g и °C) ϫ 250.0 g ϫ 46.5°C ϭ 2730 J
Solving Problems: A Chemistry Handbook
Chemistry: Matter and Change
151
Energy and Chemical Change
Energy and Chemical Change
SOLVING PROBLEMS:
A CHEMISTRY HANDBOOK
CHAPTER
15
15.1 Energy
Energy is the ability to do work or produce heat. Heat is commonly
measured in joules or calories. One calorie is equivalent to
4.184 joules.
The amount of heat required to raise the temperature of one
gram of a substance by one degree Celsius is the specific heat of the
substance. Liquid water has a high specific heat of 4.184 J/(g и °C).
By contrast, the specific heat of iron is 0.449 J/(g и °C). When the
temperature of a substance changes, the amount of heat absorbed or
released is given by the following equation.
q ϭ c ϫ m ϫ ⌬T
In the equation, q ϭ the heat absorbed or released, c ϭ the specific
heat of the substance, m ϭ the mass of the sample in grams, and ⌬T
is the change in temperature in °C.
Example Problem 15-1
Calculating Heat
A silver bar with a mass of 250.0 g is heated from 22.0°C to 68.5°C.
How much heat does the silv1er bar absorb?
Use the equation for heat.
q ϭ c ϫ m ϫ ⌬T
The temperature change is the difference between the final temperature and the initial temperature.
⌬T ϭ 68.5°C Ϫ 22.0°C ϭ 46.5°C
From Table 16-2 in your textbook, the specific heat of silver is
0.235 J/(g и °C). Substitute the known values to solve for the amount
of heat absorbed.
q ϭ 0.235 J/(g и °C) ϫ 250.0 g ϫ 46.5°C ϭ 2730 J
