Lecture 5.3 Specific Heat

I. Specific Heat Specific heat capacity (C) is the amount of heat needed to raise the temp. of 1 g of a substance by 1°C. Units are J/g°C Molar heat capacity = J/mol°C

I. Specific Heat Substance Specific Heat (J/g°C) Water (l) 4.18 Water (s) 2.06 Water (g) 1.87 Ammonia (g) 2.09 Aluminum (s) 0.897 Calcium (s) 0.647 Carbon, graphite (s) 0.709 Copper (s) 0.385 Gold (s) 0.129 Iron (s) 0.449 Mercury (l) 0.140 Lead (s) I. Specific Heat

I. Specific Heat The heat equation is used to calculate heat gained or lost. q = mCΔT ΔT =Tf – Ti

I. Specific Heat Ex #1: You find a copper penny in the snow and pick it up. How much heat is absorbed by the penny as it warms from the temperature of the snow, –8°C, to that of your body, 37°C? The penny is 3.10 g. 53.7 J

I. Specific Heat Ex #2: You’re camping in cold weather and decide to heat things by the fire to put in your sleeping bag. You have a bag of charcoal (C) and a jug of water of equal mass. You heat them by the fire to a temperature of 38°C. Which should you put in your sleeping bag?

II. Calorimetry Constant-pressure calorimetry is a way to determine enthalpies of reactions.

II. Calorimetry Enthalpy of reaction = heat gained or lost by water. qP = –ΔHrxn Assume no heat is lost to surroundings.

II. Calorimetry Ex #3: 0.158 g of Mg is combined with excess HCl to make 100.0 mL of solution. The temperature rises from 25.6°C to 32.8°C. Find ΔHrxn in kJ/mol. Mg + 2 HCl  MgCl2 + H2 Use 1.00 g/mL as the density of the solution and 4.18 J/g°C as C. -463 kJ/mol

II. Calorimetry Temperature equilibrium of a mixture. Use modified heat equation. m1C1ΔT 1 = −m2C2ΔT2

II. Calorimetry Ex #4: What is the final temperature of a mixture of 20 g of lead pellets at a temperature of 100°C with 300 g of water at 25.0°C?

PS 5.3 Read sections 6.2 Ch. 6: #29, 41, 42, 43, 45, 47, 49, 51, 53