1. Calculating Heat During Change of Phase Heat Added (J)

2. Heat in Changes of STATE  Heating and Cooling Curve  LABEL: *physical states (solid, liquid, gas) *heat of fusion *heat of vaporization *boiling point *melting point  (similar to pg. 523)

3. As a substance heats up (or cools down) each phase & phase change needs to be calculated separately.  Review Specific heat is the amount of heat that must be added to a stated mass of a substance to raise its temperature by 1°C, with NO change in state.  Specific heat of water = 4.184 J/g°C  Specific heat of ice =2.09 J/g°C  Specific heat of steam = 2.03 J/g°C

4. Need to calculate heat for the WHOLE process of changing physical state  NEW CHANGING STATE requires “heat of vaporization” and “heat of fusion” Heat of vaporization = amount of heat that must be added to 1 g of a liquid at its boiling point to convert it to vapor with NO temp. change  q vap = mH vap Heat of vaporization = 2260 J/g Heat of fusion = amount of heat needed to melt 1 g of a solid at its melting point  q fus = mH fus Heat of fusion = 334 J/g

5. Values ***WRITE ON YOUR HEATING/COOLING CURVE  Specific heat of water = 4.184 J/g°C  Specific heat of ice =2.09 J/g°C  Specific heat of steam = 2.03 J/g°C  Heat of fusion = 334 J/g  Heat of vaporization = 2260 J/g

6. USE YOUR HEATING/COOLING CURVE to help guide your calculations 1. How much heat is released by 250.0 g of water as it cools from 85.0 °C to 40.0 °C? q = mcΔT q = (250.0 g)(4.184 J/g°C)(40.0°C - 85.0°C) q = - 47070 J = (3 sig figs) - 47100 J Calculation is within the liquid phase of water (no phase changes)

7. Example 2 - How much heat energy is required to bring 135.5 g of water at 55.0 °C to its boiling point and then vaporize it?  THINK!!! HOW MANY STEPS SHOULD THIS CALCULATION BE? Use you heating and cooling curve to help you understand the steps!

8. ANSWER (2 steps)  Step 1 = raise temp of water from 55.0 °C to 100.0 °C  Step 2 = vaporize water using heat of vaporization q = mcΔT q = (135.5g)(4.184 J/g°C)(100.0°C-55.0°C) = 25511.94 to (3 sig figs) = 25500 J q = mass x heat of vaporization q = (135.5g)(2260 J/g) = = 306230 to (3 sig figs) = 306000 J ***ADD UP THE 2 NUMBERS TO GET THE OVERALL HEAT REQUIRED FOR THIS PROCESS  How much heat energy is required to bring 135.5 g of water at 55.0 °C to its boiling point and then vaporize it?

9. 25500 J + 306000 J 331500 J = 332000 J

10. Example 3  How much heat energy is required to convert 15.0 g of ice at -12.5 °C to steam at 123.0 °C?  THINK!!! HOW MANY STEPS SHOULD THIS CALCULATION BE??

11. How much heat energy is required to convert 15.0 g of ice at -12.5 °C to steam at 123.0 °C? = 5 steps q ice = (15.0 g)(2.09 J/g°C)(0.0- - 12.5°C) = 392 J q fus = (15.0 g)(334 J/g) = 5,010 J q water = (15.0 g)(4.184 J/g°C)(100.0-0.0°C) = 6,280 J q vap = (15.0 g)(2260 J/g) = 33,900 J q steam = (15.0 g)(2.03 J/g°C)(123.0-100.0°C) = 700. J 46,282 J 46,300 J