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Lecture 33. Chapter 11: Heat Specific Heat. Monday, November 16, 1998. Physics 111. Exam # 3. Friday, November 20, 1998 in class Chapters 8, 9 (S. 1-8), 10 (S. 1-2, 4-6), and 11 (S. 1-5, 7). Hint : Be able to do the homework (both the

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Physics 111

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Lecture 33

Chapter 11: Heat

Specific Heat

Monday, November 16, 1998

Physics 111


Exam # 3

Friday, November 20, 1998 in class

Chapters 8, 9 (S. 1-8),

10 (S. 1-2, 4-6), and 11 (S. 1-5, 7)

Hint: Be able to do the homework (both the

problems to turn in AND the recommended ones)

you’ll do fine on the exam!

You may bring one 3”X5” index card (hand-written

on both sides), a pencil or pen, and a scientific

calculator with you.

I will put any constants, math, and Ch. 1 - 7

formulas which you might need on a single

page attached to the back of the exam.


The slope of this graph

is given by the mass

of the substance times

the specific heat of the

substance:

Q

DT

Specific Heat

Where Q is the heat added (lost) and DT is

the temperature change, m the mass, and c

the specific heat.


Units!


Specific Heat

Specifically, the specific heat of a substance

is the amount of heat required to raise the

temperature of 1 kg of the substance by 1oC.

Therefore, the specific heat of water, which

requires 4.186 J to raise 1 g by 1oC, is

4186 J/kg/oC.

Water has a very HIGH specific heat compared

to most other substances.


The HEAT is on!

We can rewrite our last equation solving for heat

By convention, heat flows into a system

(noted by temperature increases) are usually

positive while heat flows out of a system

(noted by temperature decreases) are negative.

Note: we did NOT use this in lab last night.


If the summer solstice occurs on June

21st, and on that day, the sun is most

directly over head in the Northern

Hemisphere, why is it that July and

August are hotter months than June?

?

The heat supplied from the Sun goes into

warming up the atmosphere and the water

(67% of the planet’s surface). Recall, water

has a very high specific heat. Therefore, it

takes quite a while for the water to warm up.

Notice that in the summertime, it’s usually

a couple of degrees cooler down by Lake

Michigan than here in Valpo. Same physics!


Calorimetry

The study of the transfer of thermal energy

between two or more materials in a system.

We performed such an experiment in lab

last week, looking the temperature change

of the bath and melt water from the ice cube.

The basic technique to solving calorimetry

problems:

Heat lost by

substance 1

Heat gain by

substance 2

=


When 100 g of aluminum shot is heated

to 100oC and placed in a 500 g water

bath initially at 18.3oC, the final

equilibrium temperature of the mixture

is 21.7oC. What is the specific heat

of aluminum?

?

Let’s first calculate the heat gained by the water:


When 100 g of aluminum shot is heated

to 100oC and placed in a 500 g water

bath initially at 18.3oC, the final

equilibrium temperature of the mixture

is 21.7oC. What is the specific heat

of aluminum?

?

Now, let’s look at the heat lost by the aluminum

(which must equal the gain by the water:


d

solid

liquid

gas

Latent Heat

As we also saw in our lab experiment last week,

when a substance changes phase, extra energy

is either released or required.

We classify matter as existing in one of three

phases:


T

Latent Heat

If we start out with our substance in a gaseous

phase, then how do we get it to become a liquid?

Think of the result of a lower

temperature in terms of the

molecular argument about the

speed of the molecules...

Cool It!

Molecules are more

likely to stick together.

vrms

Energy is released as a gas becomes a liquid!


Latent Heat

So, in order to get a substance to change

phase from a liquid to a gas, we must...

Heat It!

We have to add energy to

vaporize a liquid.

The amount of energy we must add to get a

substance to change from a liquid to a gas

is the same as the amount of energy that is

released when the same substance changes

phase from a gas to a liquid…and is known as

Lv

the latent heat of vaporization


T

Latent Heat

SIMILARLY...

If we start out with our substance in a liquid

phase, then how do we get it to become a solid?

Again, think of the result of a

lower temperature in terms of

the molecular argument about

the speed of the molecules...

Cool It!

Molecules are more

likely to stick together.

vrms

Energy is released as a liquid becomes a solid!


Latent Heat

So, in order to get a substance to change

phase from a solid to a liquid, we must...

Heat It!

We have to add energy to

melt a solid.

The amount of energy we must add to get a

substance to change from a solid to a liquid

is the same as the amount of energy that is

released when the same substance changes

phase from a liquid to a solid…and is known as

Lf

the latent heat of fusion


Latent Heat

On what property of our substance could the

latent heats (that is, the amount of heat necessary

to melt our substance from a solid to a liquid

of to vaporize our liquid to a gas) possibly depend?

How much of the substance we’ve got.

Clearly, the more of it I’ve got, the more heat I’m

going to have to add to get a given temperature

change.

We specify this quantity with mass (m).


Qp

m

The slope of this line is the Latent heat (L) of

either fusion of vaporization, depending upon

the phase change we’re examining.

YES!!

Latent Heat

Should this graph pass

through the origin?

slope = L

That is, how much heat

do I have to add to get

the phase to change for

an amount m of the

substance?


A 100.0 g ice cube is initially at a

temperature of -30.0oC. The ice cube

is placed in a sealed container over

a heat source. Heat is added until all

of the ice has become steam at a final

temperature of 120.0oC. How much

total heat was added to the ice cube?

?

Use the following information:

cH2O(s) = 2090 J/kg/oCLf = 3.33 X 105 J/kg

cH2O(l) = 4190 J/kg/oCLv = 2.26X106 J/kg

cH2O(g) = 2010 J/kg/oC


A 100.0 g ice cube is initially at a

temperature of -30.0oC. The ice cube

is placed in a sealed container over

a heat source. Heat is added until all

of the ice has become steam at a final

temperature of 120.0oC. How much

total heat was added to the ice cube?

?

First, we must warm the ice to the temperature

at which it melts (0oC). This requires heat:


A 100.0 g ice cube is initially at a

temperature of -30.0oC. The ice cube

is placed in a sealed container over

a heat source. Heat is added until all

of the ice has become steam at a final

temperature of 120.0oC. How much

total heat was added to the ice cube?

?

Next, we must melt the ice. This requires heat:


A 100.0 g ice cube is initially at a

temperature of -30.0oC. The ice cube

is placed in a sealed container over

a heat source. Heat is added until all

of the ice has become steam at a final

temperature of 120.0oC. How much

total heat was added to the ice cube?

?

Then, we must warm the water to the temperature

at which it boils (100oC). This requires heat:


A 100.0 g ice cube is initially at a

temperature of -30.0oC. The ice cube

is placed in a sealed container over

a heat source. Heat is added until all

of the ice has become steam at a final

temperature of 120.0oC. How much

total heat was added to the ice cube?

?

Next, we must change the liquid water into

steam. This requires heat:


A 100.0 g ice cube is initially at a

temperature of -30.0oC. The ice cube

is placed in a sealed container over

a heat source. Heat is added until all

of the ice has become steam at a final

temperature of 120.0oC. How much

total heat was added to the ice cube?

?

Then, we must warm the steam to the temperature

of 120.0oC.This requires heat:


A 100.0 g ice cube is initially at a

temperature of -30.0oC. The ice cube

is placed in a sealed container over

a heat source. Heat is added until all

of the ice has become steam at a final

temperature of 120.0oC. How much

total heat was added to the ice cube?

?

The total amount of heat added is then just the

sum of the heats added at each step:


2) Heat gains (losses) can only be found

over intervals where no phase changes occur.

Calorimetry Problems

Tips for solving such problems:

1) Make sure your units are consistent!


3) Only use the latent heat formula

when phase changes occur!

Calorimetry Problems

Tips for solving such problems:

4) Set heat loss = heat gain and use ONLY

positive values of DT !!


Thermal Conductivity

One of three ways to transfer thermal energy

from one body to another, this method is the

most direct. It actually involves contact

between the two substances.

Recall during lab last week, after you poured

the hot water into the aluminum container,

how did that container feel to your hand?

HOT!!

Why?


kth

thermal conductivity

Thermal Conductivity

The heat from the water inside the aluminum

container was conducted through the

container to your hand!

The ability of a material to conduct heat is

yet another of the myriad of tabulated

properties of materials.

We’ll denote this property by the constant


Thermal Conductivity

How long after you poured the hot water into

the container did it take for you to notice that

the container was now hot?

Probably not very long...

How would things change if the aluminum

container were thicker?

Would have taken longer to notice.


Units!

Thermal Conductivity

We call the rate at which heat is conducted

across a given material the heat transfer rate(H)

and define this quantity as:


T2

T1

l

Thermal Conductivity

So, we’ve already deduced that H will go down

as the thickness of the material goes up.

What would H be in the following case:

T1 = T2

0!


T2

T1

l

Thermal Conductivity

Clearly, the bigger the temperature difference

across the material (known as the temperature gradient), the bigger H will be.

So H must be proportional to (T2 - T1).

Could H depend upon anything else?


T1

T2

T2

T1

l

l

Thermal Conductivity

Well, clearly the left plate will conduct heat

across itself faster than will the right plate,

given the same initial temperature gradients.

So H is proportional to the surface area, too.


Thermal Conductivity

Hence, our relationship for the heat transfer

rate is given by:

Athe cross-sectional area

T2 - T1 the temperature gradient

l the thickness of the material

kTH the thermal conductivity


Thermal Conductivity

It’s useful to note:

Builder’s refer to the R-value of a material

to describe the thermal insulation properties.

The R-value is related to the material’s

thermal conductivity by


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