Energy Review

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# Energy Review - PowerPoint PPT Presentation

Energy Review. Things to know. How is heat transferred? Exo and endo Specific heat Heat Specific Heat of lead lab Hess’s law Stoich with enthalpy. How is heat transferred?. Remember heat always goes from hot to cold. Temperature – Thermal energy -. Motion of particles - friction.

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Presentation Transcript

### Energy Review

Things to know

How is heat transferred?

Exo and endo

Specific heat

Heat

Hess’s law

Stoich with enthalpy

How is heat transferred?

Remember heat always goes from hot to cold.

Temperature –

Thermal energy -

Motion of particles - friction

Size does affect the amount of thermal energy

Exo and endo

Exothermic –

Endothermic –

Signs of q and DH

Exo : -q and – DH (lost heat)

Endo: +q and + DH (gained heat)

Releasing of heat to surroundings

Feels warm to the touch (usually)

Taking in heat from the surroundings

Feels cold to the touch (usually)

Specific heat

The amount of heat needed to raise 1 gram of substance by 1 oC.

Large value –

Small value –

Which substance will have the greatest change in temp: Swater = 4.180 J/goC or Srock = 35 J/goC?

Water – smallest specific heat

Stores a lot of heat, slow temp changes

Stores very little heat, fast temp changes

Heat problems q = msDT

Calculate heat needed to raise 25.0 g of lead by 30.0 oC. S lead = 0.13 J/goC

q = ?

m = 25.0 g

s = 0.13 J/goC

DT = 30.0 oC

q = msDT

q = (25.0 g)(0.13 J/goC)(30.0 oC)

q = + 97.5 J

Endothermic: positive q and raised the temp

Calculate the specific heat(J/goC) of a 100.0 g object that loses 12.0 kJ of heat as it cools from 98.2 oC to 22.3 oC.

q = msDT

q =

m =

s =

DT = Tf – Ti

-12 kJ

algebra

100.0 g

S = q

mDT

? J/goC

22.3oC – 98.2oC = -75.9oC

S = 1.58 J/goC

Heat gained = - heat lost

The heat gained by the cold is equal to the heat lost by the hot lead.

Why the temp very close to the initial cold water?

Water has much, much greater specific heat

Hess’s law

Enthalpy change(heat change) of a series of steps in chemical process is equal the enthalpy change of the overall reaction.

(2 step process)

A + B  C DH = +40 kJ

C  D DH = - 25 kJ

A + B  D DH = +15 kJ

Overall (1 step process)

A + B  D DH = +15 kJ

Calculate DH for NO(g) + O(g)  NO2(g)

½

3/2 O2 (g)  O3(g) DH = - 213.5 kJ

Step 1) (2 O3(g)  3 O2(g) DH = -427 kJ)

Step 2) (O2(g)  2 O(g) DH = +495 kJ)

O (g)  ½ O2 (g) DH = - 247.5 kJ

Step 3) NO(g) + O3(g)  NO2(g) + O2(g) DH = -199 kJ

NO(g) + O(g)  NO2(g) DH = -660 kJ

Are all 3 chemicals from the overall equation on the correct side of the arrow?

Need to flip step 2 to get O on correct side and times by ½

Times step 1 by ½ and flip to get rid of O3 (g)

½

NO

Stoich with enthalpy

2H2(g) + O2(g)  2H2O(l) DH = -482 kJ/mol

• If 4.00 mols of H2 was used calculate DH.
• If 15.0 g of O2 was used calculate DH.

= -968 kJ

= -227 kJ