1. Temperature(“Degree of hotness”) - is the distribution of energy through the molecules. - related to the motion (speed) of molecules. Heat ≠ Temperature Heat(“Quantity of hotness”) - is a form of energy.[units: Joules (J)] - related to both the motion and mass of molecules.

2. Heat = mass x ∆T x C p oCoCg J J/g o C Specific heat (C p ) = Specific heat for water = 4.18 J/g o C amount of heat energy required to raise 1 g of a substance 1 o C. 1 cal = 4.18 J Thermal caloriesFood Caloriesvs. 1 Cal = 1000 calories = 1 kcal q = m x ∆T x c

3. Practice Calculate the quantity of energy needed to heat 8.0 g of water from 42.0 o C to 45.0 o C. Heat = mass x  T x spec. heat (45.0 - 42.0) ? J = 8.0 g x 3.0  C x 4.18 J/g  C = 1.0 x 10 2 J(100.32)

4. Practice A 110.0-g sample of iron at 55.5 o C raises the temperature of 150.0 mL of water from 23.0 o C to 25.5 o C. Determine the specific heat of the iron in J/g o C. Heat = mass x  T x spec. heat Heat gain (H 2 O) = heat loss (metal) Heat = 150.0 g x 2.5  C x 4.18 J/g  C (gained by H 2 O) ( 25.5- 23.0) = 1567.5 J (2 sigs) 1600 J

5. = heat lost by metal 1600 J = 110.0 g x 30.0  C x spec.heat (55.5 – 25.5) Spec.heat (C p ) = 0.48 J/g o C Heat gained by H 2 O Heat lost by metal: Heat = mass x  T x spec. heat

6. Read Sect 15.1 pgs 516-522 Do Problems: 1 & 2, pg 519 4, 5, 6 pg 521 Read Sect 15.2 pgs 523-528 Do Problems: 12 - 15, pg 525