1. Temperature and Specific Heat Capacity 11/5/10

2. I.Temperature & Heat  temperature and heat are not the same thing!  temperature = a measure of the average kinetic energy (KE) of the particles in a sample of matter. High KE = high temp. °C + 273 = K  Scale = Kelvin  °C + 273 = K  heat = the energy transferred between samples of matter because of a difference in their temps. Always moves from high temp. matter to lower temp. matter. Unit = joule (J) II.Calorimetry & Specific Heat Capacity  calorimetry = the process of measuring the amount of energy absorbed or released as heat in a chemical or physical change.

3.  a calorimeter consists of known quantities of reactants in a sealed reaction chamber, immersed in a known quantity of water in an insulated vessel.  the reactants are ignited, and the amount of energy (heat) given off by the reaction is measured by the amount of heat absorbed by the surrounding water.  the amount of energy transferred (as heat) during a temp. change depends on the mass and the nature of the material changing temp., as well as the size of the temp. change.  specific heat capacity (SHC) = physical property of a substance used to compare heat absorption capabilities for different materials. Measured in J/(gK).  specific heat (c p ) = energy required (q) to raise the temp of 1 gram (m) of a substance by 1 Kelvin (  T). c p = q m   T c p = q m   T

4.  specific heat (c p ) = energy required (q) to raise the temp of 1 gram (m) of a substance by 1 Kelvin (  T).  notice the unit for SHC is c p. The subscript “p” stands for pressure. This reminds you that SHC is always calculated at a constant pressure.  Remember, pressure = force per unit area on a surface.  specific heat capacities for some common substances [in J/(g K)]: III.Enthalpy & Entropy  another way to state “q” from the above equation (energy gained/lost during the rxtn) is ΔH, which stands for enthalpy, or more specifically, enthalpy change (another physical property).  ΔH (enthalpy) is the diff. between the stored energy of the reactants and products in the rxtn. c p = q m   T c p = q m   T water4.18aluminum0.897iron0.449 ammonia2.09carbon0.709copper0.385 ethanol1.42calcium0.647lead0.129

5.  if ΔH is negative, the rxtn is exothermic; if ΔH is positive, then the rxtn is endothermic.  most rxtns are exothermic, occuring spontaneously.  endothermic rxtns can also do this, due to:  entropy = degree of randomness of particles in a system.  +  S in a rxtn = normal; –  S = rare  the KMT helps us understand entropy.  gases = high movement = high S. solids = movement limited to vibration only = low S.  as temp decreases, S decreases. Random motion ceases at absolute zero.  absolute zero = theoretical temp (0K) at which a pure crystalline solid has zero entropy. +  S = high degree of randomness –  S = low degree of randomness

6.  the tendency throughout nature is for a reaction to proceed in a direction that leads to a higher entropy state. Ex:  a naturally occurring endothermic process is melting.  an ice cube melts spontaneously at room temperature as energy is transferred from the warm air to the ice.  the well-ordered arrangement of water molecules in the ice crystal is lost, and the less-orderly liquid phase of higher energy content is formed.  a system that can go from one state to another without an enthalpy change does so by becoming more disordered

7. IV. Specific Heat Calculations  Let’s go back to Specific Heat to do some calculations.  Steps to follow to solve Specific Heat word problems: 1.Write out a column of information down the left-hand side of your work space. 2.Make sure all of your variable’s units match the units required. If not, convert it to match the needed unit, using a conversion table if necessary. Put a question mark in the space for the variable you are trying to solve for (what you DON’T have). 3.Use the appropriate rearrangement to solve for the variable you need. 4.Plug in the values and units you have in to the equation. 5.Calculate, write the appropriate unit, then box your answer! c p = q m  T m = q c p  T  T= q c p  m q = c p  m  T

8. Ex1: The specific heat capacity of gold is 0.129 J/[g  K]. How much heat energy is needed to raise the temperature of 5.0 g of gold from 352 K to 377 K? c p = __________ q = __________ m = __________  T = __________ Ex2: What mass of aluminum must be used to produce 2870 J of heat energy if the sample was heated from 276 K to 321 K? q = c p  m  ΔT q = (0.129 J/[g  K]) (5.0g) (25K ) = 16.13 J 0.129 J/[g  K] ? 5.0 g 25 K 377 K – 352 K = 25 K

9. Ex2: What mass of aluminum must be used to produce 2870 J of heat energy if the sample was heated from 276 K to 321 K? c p = __________ q = __________ m = __________  T = __________ m = q c p   T m = 2870 J_____ = (0.897J/[g  K]) (45K) 71.1 g 0.897 J/[g  K] 2870 J ? 45 K 321 K – 276 K = 45 K