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Heat Energy. Increasing the temperature of 1 kg of water from 10 o C to 15 o C takes _________ energy compared to the energy needed to increase its temperature from 15 o C to 20 o C. A. more B. less C. the same

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heat energy
Heat Energy

Increasing the temperature of 1 kg of water from 10oC to 15oC takes _________ energy compared to the energy needed to increase its temperature from 15oC to 20oC.

A. more B. less C. the same

Increasing the temperature of 1 kg of water from 10oC to 15oC takes _________ energy compared to the energy needed to increase the temperature of 2 kg of water by the same amount.

A. more B. less C. the same

heat energy1
Heat Energy

Increasing the temperature of 1 kg of water from 10oC to 15oC takes _________ energy compared to the energy needed to increase its temperature from 15oC to 20oC.

A. more B. less *C. the same

Increasing the temperature of 1 kg of water from 10oC to 15oC takes _________ energy compared to the energy needed to increase the temperature of 2 kg of water by the same amount.

A. more *B. less C. the same

latent heat

Latent Heat

3U Physics

a caveat
A Caveat

Note that the equation

Q = mcDT

can only be applied if the material does not change state while changing temperature.

the heating curve
The Heating Curve

Changes of state require additional energy:

latent heat of fusion
Latent Heat of Fusion

Additional heat is required to change a material’s state from solid to liquid (at the melting point):

Q = mLf

where Lf is the latent heat of fusion (different for different materials).

latent heat of fusion1
Latent Heat of Fusion

This heat is absorbed by the material when it melts and released when it freezes.

latent heat of fusion2
Latent Heat of Fusion

Latent heat explains why ice stays on the ground even when the temperature is above 0oC – it takes additional latent heat to do the melting.

latent heat of vapourization
Latent Heat of Vapourization

Similarly, additional heat is to change a material’s state from liquid to gas (at the boiling point):

Q = mLv

where Lv is the latent heat of vapourization (different for different materials).

latent heat of vapourization1
Latent Heat of Vapourization

This heat is absorbed by the material when it ____________ and released when it ____________.

latent heat of vapourization2
Latent Heat of Vapourization

This heat is absorbed by the material when it evaporates and released when it condenses.

latent heat of vapourization3
Latent Heat of Vapourization

Latent heat explains why sweating cools you down – the sweat absorbs heat energy from your skin when it evaporates.

latent heat example
Latent Heat: Example

How much heat must be removed from 0.10 kg of water at 80.0oC to cool it to ice at -30.0oC?

Note cwater = 4200 J/kgoC

cice = 2100 J/kgoC

Lf(water) = 3.3 x 105 J/kg

latent heat example1
Latent Heat: Example

How much heat must be removed from 0.10 kg of water at 80.0oC to cool it to ice at -30.0oC?

To cool the water to 0oC:

Q = mcDT

= (0.10 kg)(4200 J/kgoC)(80.0oC)

= 33 600 J

latent heat example2
Latent Heat: Example

How much heat must be removed from 0.10 kg of water at 80.0oC to cool it to ice at -30.0oC?

To cool the water to 0oC:

Q = mcDT

latent heat example3
Latent Heat: Example

How much heat must be removed from 0.10 kg of water at 80.0oC to cool it to ice at -30.0oC?

To freeze the water at 0oC:

Q = mLf

latent heat example4
Latent Heat: Example

How much heat must be removed from 0.10 kg of water at 80.0oC to cool it to ice at -30.0oC?

To freeze the water at 0oC:

Q = mLf

= (0.10 kg)(3.3 x 105 J/kg)

= 33 000 J

latent heat example5
Latent Heat: Example

How much heat must be removed from 0.10 kg of water at 80.0oC to cool it to ice at -30.0oC?

To cool the ice to -30.0oC:

Q = mcDT

latent heat example6
Latent Heat: Example

How much heat must be removed from 0.10 kg of water at 80.0oC to cool it to ice at -30.0oC?

To cool the ice to -30.0oC:

Q = mcDT

= (0.10 kg)(2100 J/kgoC)(30.0oC)

= 6 300 J

latent heat example7
Latent Heat: Example

How much heat must be removed from 0.10 kg of water at 80.0oC to cool it to ice at -30.0oC?

33 600 J + 33 000 J + 6 300 J

= 7.3 x 104 J

more practice
More Practice

Homework Set 9: Latent Heat

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