Thermochemistry Ch 16 -the study of energy (in the form of heat) changes that accompany physical & chemical change.

Thermochemistry Part 1: Heat Intro to thermochem - Discuss HEAT v. TEMPERATURE

Both could have same temperature, but which has more heat???

HEAT v. TEMPERATURE  Heat energy transferred from one body to another by thermal interaction (conduction, convection or radiation)  Temperature The “hotness” of an object (i.e., a physical property that quantitatively expresses the common notions of hot and cold). This “hotness” is proportional to the average kinetic energy of the particles making up the substance. You can add heat to a system without causing the temperature to rise (phase changes)

Thermochemistry  Thermochemistry: the study of energy (in the form of heat) changes that accompany physical & chemical changes  Heat flows from high to low (hot → cool)

Endothermic v. Exothermic  endothermic reactions: absorb energy in the form of heat; show a positive value for quantity of heat (q > 0) where q = heat  Example: H 2 O(l)  H 2 O(g)q=+2870 kJ  exothermic reactions: release energy in the form of heat; show a negative value for quantity of heat (q < 0) please note: it is not a negative value for amount of heat, just a negative sign to show direction of heat flow (flowing OUT OF the system)  Example: H 2 O(g)  H 2 O(l)q=-2870 kJ

Magnitude of Heat Flow  Units of heat energy:  1 kcal = 1,000 cal = 1 Cal (nutritional)  1 kJ = 1,000 J  1 calorie = J  1 kcal = kJ

Magnitude of Heat Flow But wait! A can of coke in Italy says it has 140 kcal!!! Is that 140,000 Calories???  NO, 1 nutritional Calorie = 1 kcal = 1,000 calories

Magnitude of Heat Flow  For a pure substance of mass m (constant state), the expression of q can be written as: q = m  c  T  q = magnitude of heat  m = mass  c = specific heat of substance (J/gC)  T = change in temp (T f – T i )

Specific Heat  Specific heat = the amount of heat that must be added to raise the temp. of 1 g of a substance by 1C, with no change in state.  Specific heat values (in J/gC):  CO 2 (g) = J/gC  Cu(s) = J/gC  Fe(s)= J/gC  H 2 O (l)= J/gC

Example 1 How much heat is given off by a 50.0 g sample of copper when it cools from 80.0 to 50.0C? Cu(s) = J/gC q = mcΔT q = (50.0 g)(0.382 J/gC)(50.0 C C) q = -573 J q = (50.0 g)(0.382 J/gC)(-30.0 C) (heat is given off) T f - T i

Example 2 Iron has a specific heat of J/gC. When a 7.55 g piece of iron absorbs J of heat, what is the change in temperature? If it was originally at room temp. (22.0C), what is the final temperature? q = mcΔT J = (7.55 g)(0.446 J/gC)(T) T f = 25.1 C T = 3.07 C = T f – 22.0 C

Example 3 The specific heat of copper is J/gC. How much heat is absorbed by a copper plate with a mass of g to raise its temperature from 25.0C to oven temperature (420.F)? q = mcΔT q = (135.5 g)(0.382 J/gC)(190.6 C) q = 9863 J = 9.86 kJ F = (1.8)(C) = (1.8)(T f ) + 32 T f = C