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# Definitions - PowerPoint PPT Presentation

Definitions. energy -. the capacity to do work or produce heat. kinetic energy - energy of motion potential energy - energy of position. Exothermic. When heat ( q) flows away from object or reaction. Endothermic. When heat ( q) flows into an object or a reaction. Heat Transfer.

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## Definitions

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Definitions

energy-

the capacity to do work or produce heat

kinetic energy - energy of motion

potential energy - energy of position

Exothermic

When heat (q) flows away from object or reaction

Endothermic

When heat (q) flows into an object or a reaction

Heat Transfer

between two objects at different temperatures

Sensible heat

- heat associated with a temperature change

Specific heat capacity -

the amount of energy needed to change the temperature of 1 g of substance by 1 oC (or K)

Thermal Equilibrium -

when objects reach a constant temperature

substance specific heat J/g.K

Al ---------------------- 0.902

Cu ---------------------- 0.385

Fe ---------------------- 0.451

NH3(l) ----------------- 4.70

CCl4(l) ----------------- 0.861

H2O(l) ----------------- 4.18

H2O(s) ----------------- 2.06

CCl2F2(g) ------------- 0.598

Example Problem

heat transferred = (S.H.)(mass)(DT)

Example: A lake that is 1 square mile and 10 feet deep contains 7.9 x 109 L of water. How many joules of energy must be transferred to the lake to raise the temperature by 1 oC? (densityH2O = 1.0g/mL)

Mass of water =

1000 mL

1.0 g

7.9 x 109 L

x

x

= 7.9 x 1012 g

mL

1.0 L

4.184 J

Heat transferred =

x 7.9 x 1012 g

x (+1 oC)

g.oC

= 3.3 x 1013 J

Heat Transfer Between Substances

What is the final temperature of 323.4 g H2O initially at 21.7 oC, when 79.2 g of iron 71.9 oC are added?

Specific Heat Capacity

J/g.oC

H2O 4.184

Fe 0.449

The fundamental principle here is that the energy lost by the substance initially at higher temperature is equal to the energy gained by the substance initially at lower temperature.

-qhotobject = qcold object

- qFe = qH2O

-(S.H.Fe)(m)(Tf - Ti,Fe) = (S.H.H2O)(m)(Tf - Ti,H2O)

-(0.449 J/g.oC)(79.2 g)(Tf - 71.9 oC) =

(4.184 J/g.oC)(323.4 g)(Tf - 21.7 oC)

Tf = 23.0 oC