{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

notes_Lecture_04_100810

# notes_Lecture_04_100810 - Lecture 4 Calorimetry Hess Law...

This preview shows pages 1–7. Sign up to view the full content.

Lecture 4: Calorimetry & Hess’ Law • Reading: Zumdahl 9.4, 9.5 • Outline – Calorimetry (9.4) Examples of calorimetry calculations – Hess’s Law (9.5) Motivation for and definition of Hess’s Law Examples of using Hess’s Law

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
What do we know so far? Ideal Monatomic Gas • C V = (3/2)R • C P = C V + R = (5/2)R Polyatomic Gas • C V > (3/2)R • C P > (5/2)R All Ideal Gases Δ E = nC V Δ T = q V Δ H = nC P Δ T = q P Δ E = q + w w = -P ext Δ V (for now) If Δ T = 0, then Δ E = 0 and q = -w (Summary of what we’ve covered in sections 9.1-9.3) mol K J 31451 . 8 mol K atm L 08206 . 0 R = =
Heat Capacity Heat Capacity (C) : the energy required to raise the temp. of a sample of a substance by 1 o C; the ability of a substance to absorb heat. C is an extensive property (depends on amount of substance) Recall: thermal energy is associated with the random motions of atoms and molecules Heat capacity gives you a measure of how much more random those motions can become. q C T = ⋅Δ f i T T T Δ = - Units of C: J/ o C Calorimetry (9.4)

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Specific Heat Capacity Specific heat (s) : the energy required to raise the temperature of 1 gram of a substance by 1 o C. You can also have molar heat capacities. Specific heat is an intensive version of heat capacity (like density is an intensive version of mass) (intensive – does NOT depend on amount of substance) If we know the specific heat of a substance, the mass, and the temperature change, we can determine the heat flow. q s m T = ⋅Δ { C Units of s: J/ o C·g
Specific Heat Example A 25.0-g block of iron is heated to 120.0 o C and placed in 225 mL of water at 30.0 o C. What is the final temperature of the water? (s Fe = 0.45 J/g· o C ; s H2O = 4.184 J/g· o C) T H2O = 30 o C T Fe = 120 o C 2 Fe H O q q - = + q The hot iron block will lose heat energy to the cooler water. Will the iron block lose all of its heat energy? No…it will retain enough so that its avg. atomic KE is the same as the avg. molecular KE of the water . Once the heat transfer is complete, will the iron block have a higher or lower temperature than the water? Neither…it will have the same temperature.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
A 25.0-g block of iron is heated to 120.0 o C and placed in 225 mL of water at 30.0 o C. What is the final temperature of the water? (s Fe = 0.45 J/g· o C ; s H2O = 4.184 J/g· o C) T H2O = 30 o C T Fe = 120 o C 2 Fe H O q q - = + q 2 2 2 Fe Fe Fe H O H O H O s m T s m T - ⋅Δ = + ⋅Δ
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}