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Published byTrevon Lemmer Modified over 2 years ago

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State Changes Heat of Fusion Heat energy required to convert a solid at its melting point to a liquid Value for water: 333 J/g Heat of Vaporization Heat energy required to convert a liquid at its boiling point to a solid Value for water: 2260 J/g

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**endothermic process (H positive)**

Vaporization endothermic process (H positive) Condensation exothermic process (H negative) Melting endothermic process (H positive) Freezing exothermic process (H negative)

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Finding the Energy to Change the State of a Given Mass of a Substance - Pause the video and solve the problems below. J = (mass)(heat of vaporization or heat of fusion) Find the heat energy needed to change 50.0 g of water from liquid to gas. How much heat energy will be released when 50.0 g of water freeze (change from liquid to solid)?

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Finding the Energy to Change the State of a Given Mass of a Substance - Pause the video and solve the problems below. J = (mass)(heat of vaporization or heat of fusion) Find the heat energy needed to change 50.0 g of water from liquid to gas. (2260J/g)(50.0g) = 1.13 x 106 J How much heat energy will be released when 50.0 g of water freeze (change from liquid to solid)? (-333J/g)(50.0g) = x 105 J

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**Graphing Temperature v. Time During a State Change**

120 Temperature (C) -10 Time (min)

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**Why do the plateaus occur in this graph even though heat was steadily added over time?**

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Cpgas = 1.86 J/gK Heat of vaporization = 2260J/g Specific heat of water = 4.184J/gK heat of fusion = 333 J/g Specific heat of ice = 2.06 J/gK

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**Pause the video while you solve the problem.**

Find the energy needed to convert 500.g of ice at -50.0 C to steam at 200. C. Pause the video while you solve the problem. Solve this in 5 steps by finding energy needed to: Heat ice from to 0 using specific heat of ice Melt ice using heat of fusion of ice Heat water from 0 to 100 using specific heat of water Boil water using heat of vaporization of water Heat steam from 100 to 200. using specific heat of steam Then add all of these q values together to find the total amount of energy needed.

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**Pause the video while you solve the problem.**

Find the energy needed to convert 500.g of ice at -50.0C to steam at 200. C. Pause the video while you solve the problem. Solve this in 5 steps by finding energy needed to: Heat ice from to 0 using specific heat of ice q = (2.06J/gK)(50.0g)(0 C C) = 5.15 x 104J Melt ice using heat of fusion of ice q = (333J/g)(50.0g) = 1.67 x 105J Heat water from 0 to 100 using specific heat of water q = (4.184J/gK)(50.0g)(100 C - 0C) = 2.09 x 105J Boil water using heat of vaporization of water q = (2260J/g)(50.0g)= 1.13 x 106J Heat steam from 100 to 200. using specific heat of steam q = (1.86J/gK)(50.0g)(200. C – 100.0C) = 9.30 x 104J Then add all of these q values together to find the total amount of energy needed kJ

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**Let’s try something harder!!! **

Pause the video while you solve the problem. Suppose you pour 250. mL of tea (density 1g/mL) at 18.2C into a glass with 5 ice cubes (each weighing 15g). What quantity of ice will melt and what mass will remain? (Hint: the tea will cool until it is at thermal equilibrium with the ice at 0 C.) Assume the specific heat of tea is 4.2J/gK.

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