THERMOCHEMISTRY Specific Heat

Thermochemistry 17.1  Thermochemistry is the study of energy changes (HEAT) that occur during chemical reactions and changes in state.

Heat (q)  Heat is the energy that transfers from one object to another because of a temperature difference between them.  Heat ALWAYS flows from a warmer object to a cooler one.

Heat movement  Heat moves between the system (reaction) and the surroundings  *** must obey the law of conservation of energy (heat (energy) is never created nor destroyed, just transferred)  Thermochemical equations tell you the direction of heat flow by the “sign”, + or -

Endo vs. Exo-  Endothermic reactions: absorbs heat from surroundings (+). If you touch an endothermic reaction it feels COLD  Exothermic reactions: release heat to the surroundings (-) If you touch an exothermic reaction it feels HOT

HEAT energy  UNIT of energy = JOULE (J)  Heat capacity is how much heat (energy) is needed to increase the temperature of an object by 1 C.  *** heat capacity of an object depends on both its mass and its chemical composition The greater the mass, the greater its heat capacity

Both the iron & wooden bars have the same mass, BUT they are made of different materials. They heat at different rates BECAUSE they have different SPECIFIC HEAT CAPACITIES  Small SPECIFIC HEAT. Heats up fast, only a little bit of energy is needed to raise the temperature of the iron bar.  Large Specific Heat. Heats up slow. Lots of heat is needed to raise the temperature of the wooden bar. Iron BAR Wooden BAR

SPECIFIC HEAT (C) SPECIFIC Heat Capacity (specific heat, C)  specific to a substance  Amount of heat it takes to raise the temp. of 1 g of a substance by 1 C Units = (J/gC)  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

CALCULATING HEAT - You can calculate how much heat is needed to raise the temperature of a given amount of substance  q = mCT  Where: q = heat (Joules) m = mass (grams) C = specific heat (J/gC) T = change in temperature (C) T f -T i

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? 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)

Example 2 Iron has a specific heat of J/gC. When a 7.55 g piece of iron absorbs J of heat, (A) what is the change in temperature? (B) 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  A metal plate originally at 25.0 o C and a mass of g absorbs 9,870 J of heat when placed in a o C oven. Calculate the plate’s specific heat. q = mcΔt 9870 J = (135.5 g)(c)(190.6 C) c = J/gC

Based on the table of specific heats, what metal is this? c = J/gC  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