14
White Board Practice Problems

# Heating & Cooling Curves

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Heating & Cooling Curves. White Board Practice Problems. Review. q = m x C gas x D t. The heat quantity for each step is calculated separately from the rest. GAS. q = D H vap x grams. VAPORIZE. LIQUID. q = D H fus x grams. q = m x C liquid x D t. temperature. MELT. - PowerPoint PPT Presentation

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White Board Practice Problems

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Review

q = Hfus x grams

q = Hvap x grams

q = m x Cgas x t

q = m x Cliquid x t

q = m x Csolid x t

SOLID

MELT

LIQUID VAPORIZE

GAS

The heat quantity for each step is calculated separately from the rest.

The total energy amount is found by adding the steps together.

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How much heat is need to change 22.0 grams of water from -14.0 °C to 77.0 °C?

Start by planning how many steps are needed.

-14.0°C

0 °C 0 °C

77.0 °C

q solid = ?

q melt= ? q liquid= ?

Problem

1

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Next, calculate each step.

-14.0°C

0 °C 0 °C

77.0 °C

q = Hfus x gramsq= (333 J/g)(22.0 g)q melt= 7326 J

q = m x Csolid x tq = (22.0 g)(2.05 J/g∙°C)(14.0 C°)

q ice= 631 J

q = m x Csolid x tq = (22.0 g)(4.184 J/g∙°C)(77.0 C°)q liquid= 7088 J

How much heat is need to change 22.0 grams of water from -14.0 °C to 77.0 °C?

Problem

1

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q melt= 7326 J

q ice= 631 J

-14.0°C

0 °C 0 °C

77.0 °Cq liquid= 7088 J

q total = q ice + q melt + q liquid

q total = 631 J + 7326 J + 7088 J

q total = 15045 J = 15.0 kJ

How much heat is need to change 22.0 grams of water from -14.0 °C to 77.0 °C?

Problem

1

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How much heat must be removed from 9.00 grams of 34.0 °C water to freeze it?

Start by planning how many steps are needed.

0 °C 0 °C

34.0 °C

q freeze= ? q liquid= ?

Problem

2

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Next, calculate each step.

0 °C

0 °C

34.0 °C

q = Hfus x grams

q= (333 J/g)(9.00 g)

q melt= -2997 J

q = m x Cliquid x t

q = (9.00 g)(4.184 J/g∙°C)(34.0 C°)

q liquid= -1280 J

How much heat must be removed from 9.00 grams of 34.0 °C water to freeze it?

Problem

2

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q freeze= -2997 J

q liquid= -1280 J

q total = q liquid + q freeze

q total = -1280 J + -2997 J

q total = -4277 J = - 4.28 kJ

How much heat must be removed from 9.00 grams of 34.0 °C water to freeze it?

0 °C

0 °C

Problem

2

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Starting with a 13.0 g ice cube at 0.00 °C, how much heat is needed to completely boil it?

Start by planning how many steps are needed.

0 °C

0 °C

100. °C

100. °C

q melt = ?

q liquid= ? q boil= ?

Problem

3

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Next, calculate each step.

0 °C

100. °Cq = Hfus x grams

q= (333 J/g)(13.0 g)q melt= 4329 J

q = m x Cliquid x tq = (13.0 g)(4.184 J/g∙°C)(100.0 C°)

q ice= 5439 J

q = Hvap x gramsq = (2260 J/g)(13.0 g)q liquid= 29,380 J

0 °C

100. °C

Starting with a 13.0 g ice cube at 0.00 °C, how much heat is needed to completely boil it?

Problem

3

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q melt= 4329 J

q liquid = 54349 J

100.°C

0 °C 0 °C

100. °C

q boil = 29,380 J

q total = q melt + q liquid+ q boil

q total = 4329 J + 5439 J + 29,380 J

q total = 39148 J = 39.1 kJ

Starting with a 13.0 g ice cube at 0.00 °C, how much heat is needed to completely boil it?

Problem

3

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Starting with a 11.5 g liquid water at 0.00 °C, how much heat is needed to raise it’s temperature to 145°C?

Start by planning how many steps are needed.

145 °C

0 °C

100. °C

100. °C

q steam = ?

q liquid= ? q boil= ?

Problem

4

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Next, calculate each step.

0 °C

100. °C

q = m x Cgas x tq = (11.5 g)(2.02 J/g∙°C)(45.0 C°)

q gas= 1045 J

q = m x Cliquid x tq = (11.5 g)(4.184 J/g∙°C)(100.0 C°)

q liq= 4812 J

q = Hvap x gramsq = (2260 J/g)(11.5 g)q boil= 25,990 J

145 °C

100. °C

Problem

4Starting with a 11.5 g liquid water at 0.00 °C, how much heat is needed to raise it’s temperature to 145°C?

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q gas= 1045 J

q liquid = 4812 J

100.°C

145 °C

0 °C

100. °C

q boil = 25,990 J

q total = q liquid + q boil+ q gas

q total = 4812 J + 25,990 J + 1045 J

q total = 31847 J = 31.8 kJ

Problem

4Starting with a 11.5 g liquid water at 0.00 °C, how much heat is needed to raise it’s temperature to 145°C?