1 / 15

Chapter 6.6 & 6.7 Entropy and Specific Heat

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. 1 st law- quantity of energy 2 nd law- quality of energy

zubin
Download Presentation

Chapter 6.6 & 6.7 Entropy and Specific Heat

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. 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?

  2. 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

  3. 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

  4. 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.

  5. 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.

  6. 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.

  7. 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.

  8. 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.

  9. 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

  10. 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

  11. 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)

  12. 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?

  13. 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

  14. 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?

  15. 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

More Related