1. Laws of Thermodynamics
2014

2. Laws of Thermodynamics
1. The first law of thermodynamics is:
“Whenever heat flows into or out of a
system, the gain or loss of thermal energy
equals the amount of heat transferred.”
The first law is also known by another name:
Conservation of energy.

3. 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.

4. 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.

6. Third law:
If an object reaches the absolute zero of temperature (0 K = °C = − °F), its atoms will stop moving.
The statement is: At absolute zero, the entropy of a perfectly crystalline substance is zero.

9. Cp= Q/m∆T Q =mCp∆T 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

10. Specific Heat
Large specific heat means it takes large quantities of heat with little change in temperature.
Example:
Specific heat of
1kg water is
cwater = 4184 J/(kg*K)

11. Examples of Specific Heat

12. 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.

13. 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.

14. Conductors: Good vs. Bad
Low specific heat = good thermal conductors
High specific heat = bad thermal conductors

15. ΔT = Q/ mCp 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

16. Specific heat - Cp in J/(kg∙K)
Q = mCpΔT
Energy - Q in J
Mass - m in kg
Specific heat - Cp in J/(kg∙K)
Temperature - T in K

18. Example
A 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 K Cp = 450 J/kgK
Q = 57,213 J