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Thermochemistry - PowerPoint PPT Presentation

Thermochemistry. Chapter 15 Part I –Physical Changes of Water. Factors that influence heat change (q). Heating curve: Ice  Water  Steam (H 2 O @ 1 atm). Temperature. Energy Added. Critical Information. 4.18 J = 1 cal D q = mc D T (same phase energy change)

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Thermochemistry

Chapter 15

Part I –Physical Changes of Water

Heating curve: Ice  Water  Steam (H2O @ 1 atm)

Temperature

4.18 J = 1 cal

Dq = mcDT (same phase energy change)

Dq = mDh (phase change)

cice = 2.1 J/goC

cwater = 4.18 J/goC = 1.00 cal/goC

csteam = 1.7 J/goC

Dhfus = 334 J/g

Dhvap = 2261 J/g

You should be able to solve for:

q total:

How much heat needs to be removed from 84 g of steam at 412 K to turn it into ice at 218 K?

You should be able to solve for:

Final temperature:

Tf problem

What is the final temperature if 78 kJ of heat are added to 22 g of ice at -14oC?

You should be able to solve for:

% sample in two different phases:

How many grams of each phase will be present if 50 kJ of heat are added to 22 g of ice at -14oC?

You should be able to solve for:

Mass:

What is the mass of a sample of H2O if the addition of 35 kJ of heat takes it from 250 K to 415 K?

You should be able to solve for:

q total: Add the energy for each step on the graph

Final temperature: Subtract the energy for each step from the total energy until the energy runs out during a step. Use the energy remaining for that step to solve for DT, TF.

% sample in two different phases: Subtract the energy for each step from the total energy until the energy runs out during a plateau. Divide the energy remaining for that step by the total energy needed to change phase completely. Multiply the mass by this fraction.

Mass –factor out m from each step.

A sample of ice at -25oC is heated to steam at 200oC by the addition of 300 kJ of energy. What is the mass of the sample?