Chapter 6.6 & 6.7 Entropy and Specific Heat

Why is it that the filling of a hot apple pie may be too hot to eat,

even though the crust is not?

Entropy

• 1st law- quantity of energy

• 2nd law- quality of energy– A further definition of 2nd law: In natural

processes, high-quality energy tends to transform into lower quality energy.

– A process in which disorder returns to order without external help does not occur

Entropy

• And as time increases…….disorder does too

• Time always points in the direction from order to disorder– Old movies with damsel in distress tied to a

train track.– Next time look for the telltale signs of the

smoke entering the smokestack

Entropy

• The idea of ordered energy tending to disordered energy is the concept of entropy

• Entropy is the measure of how energy spreads to disorder in a system

• When disorder increases- entropy increases– The heat from a camp fire cannot spontaneously

return to the fire and restore the firewood to its original state.

– The molecules of an automobiles exhaust cannot recombine to form highly organized gasoline molecules.

Entropy

• When ever a system is allowed to spread its energy freely, it always does so in a manner that increases entropy (disorder)

• But energy can be put into a system in the form of WORK. – Living organisms extract energy from their

surroundings and use this energy to increase their own organization

– This decreases the entropy of the living organism’s system

– Plants extract photons of light for energy and create glucose through photosynthesis.

Entropy

• Even though the entropy of a living organisms may decrease

• It does so by increasing the entropy of its surroundings.

• So life forms plus their waist products have a net increase in entropy.

• Energy must be transformed within a living system

• When it cannot, the organism soon dies and again tends toward disorder.

Specific Heat Capacity

• Opening question…discussion• Some foods remain hotter longer than others. • Different substances have different thermal

capacities for storing heat. – Water, iron, and silver– Equal masses of different substances require a

different quantity of heat to change their temperatures by a specific number of degrees.

– Recall calories…. 1g of water needs 1cal of energy to raise the temp 1 degree C.

Specific Heat Capacity

• We consider water to have higher specific heat capacity (often called specific heat for short) – The specific heat capacity of any

substance is defined as the quantity of heat required to change the temperature of a unit mass of the substance by 1 degree Celsius.

Specific Heat Capacity

• A great way to think of specific heat capacity is by thinking of it as thermal inertia

• Mathematically explaining specific heat– The specific heat of a substance can be calculated if

you know how much heat is put into the material and how much the material changes temperature.

– Q= cmΔT – Q is the quantity of heat– c is the specific heat– m is the mass of the object– ΔT is the change in temperature. Remember: Tf - Ti

Practice Problem A

• Aluminum has a specific heat of 0.902 J/g x 0C.   How much heat is lost when a piece of aluminum with a mass of 23.984 g cools from a temperature of 415.0 oC to a temperature of 22.0 oC?– Solve for Q.. How much heat?– Q= cmΔT

Practice Problem A

– Solve for Q.. How much heat?

– Q= cmΔT

– Q= (0.902 J/g x 0C)(23.984g)(22.0 oC - 415.0 oC)=

– Q= 8.50 x 103 J (use correct sigfigs and scientific notation)

Practice Problem B

• mass (m)  is the unknown:

• The temperature of a sample of water increases by 69.5 oC when 24 500 J are applied.  The specific heat of liquid water is 4.18 J/g x oC.  What is the mass of the sample of water?

Practice Problem B

– Solve for m: Q= cmΔT then, m=Q/(c ΔT)

– m = 24 500 J/69.5 oC x 4.18 J/g x oC

– m = 84.3 g

Practice Problem C

• Specific Heat (c) is the unknown:

• When 34 700 J of heat are applied to   a 350 g sample of an unknown material the temperature rises from 22.0 oC to 173.0 oC.  What must be the specific heat of this material?

Practice Problem C

• Solve for c: Q= cmΔT then, c=Q/(mΔT)• Cp = 34 700 J/350 g x 151.0 oC

• Cp = 0.66 J/g x oC