Temperature & Heat

Overview

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Temperature vs Heat

What is temperature (degrees)?• A measure of the average kinetic energy of

the particles in an object

What is heat (joules)?• That energy transferred between objects

because of a difference in their temperatures

How does heat flow?• Heat flows from a hotter to a colder body by

transfer of kinetic energy to adjacent molecules

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Units of Measurement

There are 3 units of temperature

measurement:

• Degrees Celsius (C) (Water: 0 – 100)

• Degrees Fahrenheit (F) (Water: 32 – 212)

• Kelvin (K), referencing absolute zero

Conversion - Celsius & Fahrenheit

• F – 32 = 9/5*C

• K = C + 273

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Measuring Devices

Thermometers

• Analog

• Digital

Thermocouples

• The temperature of the junction of 2 different

wires creates a small voltage, related to the

temperature of the junction

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Expansion

Each material expands, according to its natural property, as its temperature increases by ΔT degrees C.

Options• Linear: ΔL = Lo α ΔT, where

• α = coefficient of linear expansion

• Area: ΔA = Ao (2α) ΔT

• Volume ΔV = Vo β ΔT, where• Β = coefficient of volumetric expansion

Example

A metal bar is 1.6 m long at 21 C. If the

bar is heated to 84 C by how much does

it expand? ( = 1.7 x 10-5 / C)

L = Lo* *(T2 – T1)

• L = 1.6 * 1.7 x 10-5 * (84 – 21)

• L = 1.7 x 10-3 m

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Specific Heat – basic property

The amount of heat, in joules, required to raise the temperature of 1 kg of a substance by 1 degree Celsius.• Symbol of the specific heat of a substance is “c”

• Examples of c• Water = 4187 J/kg- C,

• ie 4187 joules of heat energy are required to raise the temperature of 1 kg of water by 1ºC.

• Iron = 448 J/kg- C

• Aluminum = 899 J/kg- C

• Ice = 2090 J/kg- C

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Implications of Specific Heat

The heat energy (ΔQ) transferred into or out of a mass (m) of material with a specific heat (c) is given by:• ΔQ = m*c*ΔT, where ΔT is the change in

temperature

This phenomenon can be used when bodies of different temperatures are brought into contact• Hot bodies lose, while cold bodies gain heat

Example of Heat Transfer

How much heat is required to raise the

temperature of a 2 kg block of copper

from 20C to 250C (cCU=386 J.kg/˚C)?

Solve Q = m*c*ΔT

• Q = 2 * 386 * (250 – 20)

• Q = 177,560 J

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Example

What would be the final temperature of a

5kg iron (c = 448) rod at 20C if it

received 110,000 J of heat?

Solve Q = m*c*ΔT

• 110,000 = 5 * 448 * ΔT

• 49.11 = ΔT

• But ΔT = Tf – To = Tf – 20

• Thus, Tf = ΔT + 20 = 69.11 C

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Mixture Examples

What happens when 2 bodies of different temperatures are brought into contact, such as a hot piece of metal being placed into a bucket of cold water?• The hot body (metal) cools while the cold body

(water) warms according to the conservation of energy

• Heat lost by hot body = Heat gained by cold body, or

• (m*c*ΔT)lost = (m*c*ΔT)gained

Heat flows from hot to cold!

Example Mixtures

See OHP…

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Phase Change – Change of State

All solids, liquids and gases are made of

particles, and the only difference between

them is how much energy the particles

have. If you give the particles energy, or

take energy away from them, then you can

change their state.

Energy in – Energy out

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Phase Change – Change of State

There are 2 different types of heat

• Sensible (or specific) – the addition of which

causes a rise in temperature of the material,

and

• Latent – the addition of which causes no

temperature change until the current state of

the material has changed completely, from solid

to liquid, liquid to gas or in reverse.

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Latent Heat

There are 2 different latent heats

• Solid > liquid = latent heat of fusion or melting• ( ex. turning ice to water)

• QF = mass * heat of fusion, or

• QF = m * hf (where hf = 3.35 x 105 J/kg for ice)

• Liquid > gas = latent heat of vaporization• (ex. turning water to vapor)

• QV = mass * heat of fusion, or

• QV = m * hv (where hv = 2.26 x 106 J/kg for water)

• In graphic form it looks like…

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Change of State

Heat can be applied to an object to change its state from solid, through liquid to gas, according to the 5 steps shown:

1. Heat the solid until it reaches melting point (specific heat of solid)

2. Heat the solid until it changes completely to liquid (latent heat fusion)

3. Heat the liquid until it reaches the boiling point (specific heat of liquid)

4. Heat the liquid until it changes completely to gas (latent heat of vaporization)

5. Heat the gas (specific heat of gas)

For example, H2O…

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Heat Phases – Water

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Steps in Solving Specific/Latent

Heat Problems

Identify the temperature range in the problem

Mark the temperature range on the heat graph

On the graph move from the lowest to the highest

temperature identifying the sloped/flat stages

Calculate the heat (Q) for each step as follows:

• Sloped > specific heat stage > Q = mcΔT

• Flat > latent heat stage > Q = mh (h for fusion/vaporiz’n)

Add the Q values for each step to determine the total

heat (Qtotal)

Example

How much heat, in joules, is required to raise the temp of 2 kg of ice from -10˚C to 20˚C?• Mark start and finish temp

• Count slopes and flats• 2 x slopes + 1 flat, meaning:

• 2 x specific heat segments

• 1 x latent heat segment

• Calculate Q for each segment, then total the Q values

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S

S

F

Calculations

Segment 1: sloped = specific heat for ice being heated from -10 to 0˚C• Q = mcΔT = 2 * 2090 * (0 – (-10)) = 41,800 J

Segment 2: flat = latent heat converting ice @ 0˚C to water @ 0˚C• Q = mhf = 2 * 3.35 x 105 = 670,000 J

Segment 3: sloped = specific heat for water from 0˚C to 20˚C• Q = mcΔT = 2 x 4187 * (20 – 0) = 167,480 J

Total heat (Q) = 879,280 J

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