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Specific Heat and other things

GFR

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The specific heat is the amount of heat per unit mass required to raise the temperature by one degree Celsius. The relationship between heat and temperature change is usually expressed in the form shown below where c is the specific heat. The relationship does not apply if a phase change is encountered, because the heat added or removed during a phase change does not change the temperature.
shta.gif

The specific heat of water is 1 calorie/gram °C = 4.186 joule/gram °C which is higher than any other common substance. As a result, water plays a very important role in temperature regulation. The specific heat per gram for water is much higher than that for a metal, as described in the water-metal example. For most purposes, it is more meaningful to compare the molar specific heats of substances.
The molar specific heats of most solids at room temperature and above are nearly constant, in agreement with the Law of Dulong and Petit. At lower temperatures the specific heats drop as quantum processes become significant. The low temperature behavior is described by the Einstein-Debye model of specific heat.
 
The specific heat is the amount of heat per unit mass required to raise the temperature by one degree Celsius. The relationship between heat and temperature change is usually expressed in the form shown below where c is the specific heat. The relationship does not apply if a phase change is encountered, because the heat added or removed during a phase change does not change the temperature.
 
Sample Problem
How much heat is needed to raise the temperature of 5 grams of copper by 10° Celsius?

1. First we identify the information we are given in the problem:
* mass = 5 g
* change in temperature = 10°C
* specific heat of copper = 0.09
2. Next, we place this information into the formula:
* heat gained = mass x change in temperature x specific heat
* heat gained = 5 g x 10°C x 0.09 cal/g°C
3. Solving the equation gives a heat value of 4.5 cal
 
Sample Problem
A 50 g substance loses 23 calories of heat energy when the temperature falls from 35°C to 20°C. What is the specific heat of the substance?

1. We identify the information given in the problem:
* mass = 50 g
* heat lost = 23 cal
* change in temperature = 15°C
2. We now place the information into the formula:
* heat gained = mass x change in temperature x specific heat
* 23 cal = 50 g x 15°C x specific heat
* 23 cal/50 g x 15°C = 50 g x 15°C x specific heat/50 g x 15°C
3. Solving the equation gives a specific heat value of 0.03 cal/g°C
 
my cats in heat.
 
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