Calculating Heat During Change of Phase Heat Added

Calculating Heat During Change of Phase Heat Added (J)

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

As a substance heats up (or cools down) each phase & phase change needs to be calculated separately. p Review n 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. p. Specific heat of water = 4. 184 J/g°C p. Specific heat of ice =2. 09 J/g°C p. Specific heat of steam = 2. 03 J/g°C

Need to calculate heat for the WHOLE process of changing physical state p NEW CHANGING STATE requires “heat of vaporization” and “heat of fusion” n p qvap = m. Hvap Heat of vaporization = 2260 J/g n p 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 Heat of fusion = amount of heat needed to melt 1 g of a solid at its melting point qfus = m. Hfus Heat of fusion = 334 J/g

Values ***WRITE ON YOUR HEATING/COOLING CURVE Specific heat of water = 4. 184 J/g°C p Specific heat of ice =2. 09 J/g°C p Specific heat of steam = 2. 03 J/g°C p Heat of fusion = 334 J/g p Heat of vaporization = 2260 J/g p

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? Calculation is within the liquid phase of water (no phase changes) 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

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? p THINK!!! HOW MANY STEPS SHOULD THIS CALCULATION BE? Use you heating and cooling curve to help you understand the steps!

ANSWER (2 steps) p p Step 1 = raise temp of water from 55. 0 °C to 100. 0 °C Step 2 = vaporize water using heat of vaporization 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? q = mcΔT q = (135. 5 g)(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. 5 g)(2260 J/g) = = 306230 to (3 sig figs) = 306000 J ***ADD UP THE 2 NUMBERS TO GET THE OVERALL HEAT REQUIRED FOR THIS PROCESS

25500 J + 306000 J 331500 J = 332000 J

Example 3 p How much heat energy is required to convert 15. 0 g of ice at -12. 5 °C to steam at 123. 0 °C? p THINK!!! HOW MANY STEPS SHOULD THIS CALCULATION BE? ?

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 qice = (15. 0 g)(2. 09 J/g°C)(0. 0 --12. 5°C) = 392 J qfus = (15. 0 g)(334 J/g) = 5, 010 J qwater = (15. 0 g)(4. 184 J/g°C)(100. 0 -0. 0°C) = 6, 280 J qvap = (15. 0 g)(2260 J/g) = 33, 900 J qsteam = (15. 0 g)(2. 03 J/g°C)(123. 0 -100. 0°C) = 700. J 46, 282 J 46, 300 J

Pop Quiz - How much heat energy is released to convert 25. 0 g of steam at 135. 0°C to ice at -15. 0°C?