and the table of specific heats. and the table of phase changes

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Presentation transcript:

and the table of specific heats. and the table of phase changes THINGS TO REMEMBER ΔH for temperature changes uses the formula ΔQ = c m ΔT and the table of specific heats. ΔH for phase changes uses the formula ΔQ = m l and the table of phase changes

STUDY OF THE EFFECTS OF WORK, HEAT AND ENERGY ON A SYSTEM THERMODYNAMICS STUDY OF THE EFFECTS OF WORK, HEAT AND ENERGY ON A SYSTEM

ZEROTH LAW – Law of Thermodynamic Equilibrium Heat energy moves from hot to cold objects until a state of thermodynamic equilibrium (same temperature = same KE) exists between the objects

FIRST LAW – Law of Conservation of Heat Energy

SECOND LAW – Law of Entropy

THIRD LAW – Absolute Zero As the absolute temperature approaches zero (0 K), the entropy of the system approaches zero.

Example: An 8.0 g ice cube at -10oC was placed in a Thermos flask containing 100 ml of water at 20oC. The specific heat of ice is 2.05 J/g-K, the specific heat of water is 4.186 J/g-K and the heat of fusion of ice-water is 334 kJ/g-K. What is the final equilibrium temperature?

= 164 J + 2672 J + 33.49 J/oC Tf + 418.6 J/oC Tf – 8372 J ΔQ = Hgained + Hlost = 0 = cicemice(0 – (-10)oC) + mice (Hf) + cwmice(Tf-0oC ) + cwmw(Tf-20oC) = (2.05 J/g-oC)(8.0 g)(10oC) + (8.0 g)(334 J/g-oC) + 4.186 J/g-oC)(8.0 g) (Tf) + (4.186 J/g-oC)(100 g)(Tf-20oC) = 164 J + 2672 J + 33.49 J/oC Tf + 418.6 J/oC Tf – 8372 J = -5536 J + 452 J/oC Tf Tf = -5536 J/(-452 J/oC) = 12.2 oC