11.3 Heat in Changes of State

Warm up Is it exo- or endo- thermic???? -negative ΔH -positive ΔH -Heat as a reactant -Heat as a product -Combustion of fossil fuels

Heat, Q or Δ H, has been discussed so far to describe an amount of heat absorbed or released for a substance in a single state. Q = mcΔT

Phase Changes

Solid  Liquid  Gas – Requires an addition of heat into the system – Endothermic Processes Gas  Liquid  Solid – Heat leaves the system – Exothermic Processes

Phase Changes Melting / Solidification – ΔH fusion Vaporization / Condensation – ΔH vaporization

Problem Solving for a Phase Change For Solid to Liquid or Liquid to Solid USE ΔH fusion For Liquid to Vapor or Vapor to Liquid USE ΔH vaporization

How many grams of ice at 0  C can be melted by the addition of 5 kJ of heat? For water Δh fus = 6.01 kJ/mol

How many grams of ice at 0C can be melted by the addition of 5 kJ of heat? For water Δh fus = 6.01 kJ/mol First we can solve for moles 5 kJ 1 mol = 0.83 mol 6.01 kJ Are we done??? 0.83 mol 18 g = g mol

How much heat is absorbed when 50 g of 100 °C liquid water is converted to steam at 100 °C ? ΔH vap = 40.7 kJ /mol How do you think you might start this problem???

How much heat is absorbed when 50 g of 100 °C liquid water is converted to steam at 100 °C ? ΔH vap = 40.7 kJ /mol 50 g 1 mol 40.7 kJ = kJ 18 g mol

Calculating heat change using Q = mΔH Instead of being given ΔH as a kJ/mol value, you may be given a J or cal per gram unit….. ΔH fusion for water is 334 J/g ΔH vaporization for water is 2260 J/g Then you can just plug and chug!!!!

How much heat is released when 2 grams of liquid water at 0°C turns to ice at 0°C? Q = mΔH ΔH fusion for water is 334 J/g

How much heat is released when 2 grams of liquid water at 0°C turns to ice at 0°C? Q = mΔH ΔH fusion for water is 334 J/g Q = ( 2 g )(334 J/g) = 668 J

Solving for heat when both a temperature change and phase change occur….. Q = mcΔT + mΔH

Let’s practice… How much heat is required to convert 20 g of water at 28°C into steam at 100°C ? Specific heat of water = 1 cal/g°C ΔH vaporization = 540 cal/g Q = mcΔT + mΔH

Let’s practice… How much heat is required to convert 20 g of water at 28°C into steam at 100°C ? Specific heat of water = 1 cal/g°C ΔH vaporization = 540 cal/g Q = mcΔT + mΔH Q = (20 g)(1 cal/g°C)(72 °C) + (20g)(540 cal/g) Q = cal

In class: Ch. 11 Practice Problems Ch Quiz on Friday!!!