The Laws of Thermodynamics (01 of 38) Physics Lecture Notes The Laws of Thermodynamics.
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Laws of Thermodynamics
2014
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Laws of Thermodynamics1. The first law of thermodynamics is:
“Whenever heat flows into or out of asystem, the gain or loss of thermal energyequals the amount of heat transferred.”
The first law is also known by another name:
Conservation of energy.
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The basic rule is straightforward– However much heat is added or removed from an object must equal the change in the total energy content of that object.
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Second Law:Entropy Law Entropy = disorder, randomness or chaos
“In an isolated system, entropy tends to increase spontaneously”
Heat cannot by itself pass from a colder to a hotter body.
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Third law:
If an object reaches the absolute zero of temperature (0 K = -273.15 °C = −459.67 °F), its atoms will stop moving.
The statement is: At absolute zero, the entropy of a perfectly crystalline substance is zero.
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Specific Heat
The heat required to increase the temperature of a substance is given by the specific heat, Cp
Cp= Q/m∆T
Q =mCp∆T
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Specific HeatLarge specific heat means it takes large
quantities of heat with little change in temperature.
Example:Specific heat of 1kg water iscwater = 4184 J/(kg*K)
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Examples of Specific Heat
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Why do different materials have different specific heats?
There are two main reasons:1. One kilogram will have different numbers of molecules for different substances.2. The heat added can go into other forms of energy besides kinetic energy.
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Water’s High Specific Heat
Water has a very high specific heat for a few reasons…..
In part because H2O molecules are light and there are a lot of them in one kilogram
And because a lot of the added heat goes into vibrations and rotations of the molecules and not into kinetic energy.
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Conductors: Good vs. Bad
Low specific heat = good thermal conductors
High specific heat = bad thermal conductors
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Heat Transfer
The change of temperature depends on Heat transfer: Q in joules (j)Mass of the substance: m in kilograms (kg)The specific heat of the substance: Cp (J/kgK)
ΔT = Q/ mCp
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Q = mCpΔT
Energy - Q in J
Mass - m in kg
Specific heat - Cp in J/(kg∙K)
Temperature - T in K
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ExampleA 1.63-kg cast-iron skillet is heated on the stove
from 295 K to 373 K. How much heat had to be transferred to the iron?
m = 1.63 kg Q = CpΔTm
ΔT = 78 KCp = 450 J/kgK Q = 57,213 J