2. Specific heat capacity (a.k.a. Specific heat) symbolized as c, units in J/g  C It’s the heat required to raise 1 gram of a substance by 1  C Heat capacity calculated by c x m, units in J/  C It’s the heat required to raise the temperature of an object by 1  C. Molar heat capacity Similar to specific heat capacity, but uses moles instead of grams, units in J/mol  C. It’s the heat required to raise the temperature of 1 mol of a substance by 1  C

3. Specific Heat Capacity Or the amount of energy needed to heat substances up

6. Specific Heat Capacity (C) of a substance is the amount of heat required to raise the temperature of 1g of the substance by 1 o C (or by 1 K). The units of specific heat capacity are J o C -1 g -1 or J K -1 g -1. Sometimes the mass is expressed in kg so the units could also be J o C -1 g -1 or J K -1 kg -1

7. The next table shows how much energy it takes to heat up some different substances. The small values show that not a lot of energy is needed to produce a temperature change, whereas the large values indicate a lot more energy is needed.

8. Approximate values in J / kg °K of the Specific Heat Capacities of some substances are: Air 1000Lead 125 Aluminum 900 Mercury 14 Asbestos 840 Nylon 1700 Brass 400 Paraffin 2100 Brick 750 Platinum 135 Concrete 3300 Polythene 2200 Cork 2000 Polystyrene 1300 Glass 600 Rubber 1600 Gold 130 Silver 235 Ice 2100 Steel 450 Iron 500 Water 4200

9. M

10. The amount of heat energy (q) gained or lost by a substance = mass of substance (m) X specific heat capacity (C) X change in temperature (ΔT) q = m x C x ΔT The equation:

11. An example of a calculation using the specific heat capacity equation: How much energy would be needed to heat 450 grams of copper metal from a temperature of 25.0ºC to a temperature of 75.0ºC? (The specific heat of copper at 25.0ºC is 0.385 J/g ºC.)

12. Explanation: (ΔT) is: 75ºC - 25ºC = 50ºC Mass is: 450g C is: 0.385 J/g ºC q = m x C x ΔT. q = (450 g) x (0.385 J/g ºC) x (50.0ºC) = 8700 J

13. BoundaryEndothermic Exothermic Thermometer identifies  T of water.  H is found using q=cm  T and law of conservation of energy (technically, pressure must be constant for q to equal  H) System

14. A basic calorimeter - see handout In our lab we tried to determine  H via q from “calorimetry”. Here are some terms assoc- iated with calorimetry (and thermochemistry) Endothermic = absorbing energy Law of conservation of energy = release and absorption of energy must be equal Surroundings everything else Exothermic = releasing energy System with can as boundary For more lessons, visit www.chalkbored.com www.chalkbored.com

15. How much heat is required to raise 15.0 g of water from 20.0 o C to 50.0 o C? Specific heat of water is 4.18 J/g o C

16. How much heat is required to raise 15.0 g of water from 20.0 o C to 50.0 o C? Specific heat of water is 4.18 J/g o C Step 1 extract all the information from the question\and assign variables Q ? M 15.0 g C 4.184 J/g/ o C ΔT 20.0 o C to 50.0 o C?

17. How much heat is required to raise 15.0 g of water from 20.0 o C to 50.0 o C? Specific heat of water is 4.18 J/g o C Step 2 make the necessary conversions to prepare values for substitution into the equation q = mC p ΔT:

18. Step 3 plug in values q = mC p ΔT: q = (15.0)(4.18)(50.0-20.0) = 1.88 x 10 3 J