Goal: To understand heat transfers

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Transcript Goal: To understand heat transfers

Goal: To understand heat

Objectives: 1) To explore internal energy 2) To learn about specific heat 3) To learn about latent heat 4) To learn about heat transfers 5) To learn about radiative heat 6) To learn about conductive heat transfer

Internal Energy

• • U = 1.5 k T k = Bolzmann constant = 1.38 * 10 -23 J/K

Specific Heat

• • • • • So, if you want to increase the temperature of an object you have to add energy.

The energy required is: Q = m * C * ΔT Q is heat energy and C is called specific heat C is different for different materials

Example

• • For water the specific heat of water is 4186 J/(kgC) If you have a gallon jug mostly full of water (3.7 kg) and you want to increase its temperature by 20 degrees C then how much energy do you need to add to the water?

Latent Heat

• • • • There are 4 phases of matter: solid, liquid, gas, and plasma If you go up in phase it takes energy.

This energy is called Latent Heat.

If you go down you gain energy.

• • • Solid to liquid is called Fusion Liquid to gas is called Vaporization Solid to gas is called sublimation

Energy for Latent Heat

• The energy required to go up a phase is: Q = m L Where L is the latent heat.

Example

• • • • • You take a block of ice with mass 2 kg initially at a temperature of 0 C. The latent heat of fusion for water is 334,000 J/(kg) and the latent heat vaporization is 2.260 * 10 6 J/(kg) A) If you melt the ice how much energy does it take to melt the ice?

B) You then heat the water to 100 degrees C. How much energy does it take to heat the water once you melt the ice?

C) How much energy does it take to turn the 100 degree C water into steam?

D) What is the total energy required to turn 2 kg of ice to 2 kg of steam?

Heat transfer

• • • • Energy is conserved.

Drool – zombie walk… That means if one object looses energy then another has to gain that energy.

This allows for heat transfer.

Example

• • • • • • A 10 kg sword is being forged. The heat of the steel is initially 300 C.

The steel is plunged into a 20 kg vat of 10 degree C water. Latent heat of water still 4186 J/(kgC) The water is heated to 25 degrees C.

A) How much energy did the water gain?

B) How much energy did the sword loose (this one should not require an equation)?

C) What is the specific heat of the steel (C)?

Slightly more complicated example

• • • • • • • A really cold batch of ice cubes (-15 C) is pulled out of the freezer and tossed into a pot of nearly boiling water in a hot pan.

The temperature of the water is 95 C. The temperature of the pan is 150 C.

The mass of the water is 1.2 kg. The mass of the pan is 0.8 kg. The mass of the ice is 0.4 kg.

The latent heat of fusion for water is 334,000 J/(kg). The specific heat of ice is 2090 J/(kgC). The specific heat of liquid water is 4186 J/(kgC). The specific heat of the metal in the pan is 390 J/(kgC) Find the final temperature of the system.

Hint: set up the equation for how much energy the ice will gain vs the energy the rest will loose and you will only have 1 unknown, Tfinal.

Note that this is similar to HW question 5.

Radiative heat

• • • • • • All “solid” objects radiate energy.

This is called blackbody radiation.

This includes the night side of Pluto (40 K).

The amount of energy radiated depends on ONLY 2 things: 1) the temperature of the object 2) the surface area of the object

For perfect blackbody

• • • • • L = 4π σ R 2 T 4 L is luminosity (which is the power) σ is the Stefan-Boltzmann constant and equals 5.67 * 10 -8 W/(m 2 K 4 ) NOTE: you have to use the Kelvin scale for T.

K = 272 + C

Example

• • • The sun has a surface temperature of 5600 K.

The sun has a radius of 6.4 * 10 8 m.

What is the luminosity of the sun?

However

• • • • Not everything emits with perfect efficiency. Some things for whatever reason don’t emit as a perfect blackbody.

You use a factor called emissivity (e) L = e * 4π σ R 2 T 4

Conductive Heat transfer

• • • • • • You have a window. In the winter heat will be conducted through the window from high heat (house) to low heat (outside).

Here is how you calculate how much energy is passing through your window (because energy is money. If you use gas heat then every 1 million J of energy that passes through the window is about a cents worth of energy).

Q = k A ΔT t / L k is the conductive coefficient for that material A is the area of the window and L is the thickness of the window.

ΔT is the temperature differential on the two sides of the window.

2 more versions of the same

• • • Usually in the real work conductivities are given as U or R.

U = k / L R = 1/ U • • So, Q = U A ΔT t Or Q = A ΔT t / R • • Double pained windows tend to have a U of 0.03 or so and a R of 3.5 or so.

Walls tend to have R values of 10.

Sample 1

• • • • • • • A window has an R value of 3.5 W C m 2 The window is 0.9 m by 1.6 m in size.

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On a cold winter day you try to keep your inside temperature at 20 C. The outside temperature is -5 C.

A) How much energy will go through the window over the course of a 6 month winter if this is the average temperature difference?

B) If 1 million J costs a cent then what is the yearly cost of the energy that passes through the window?

C) How much money would you save if you turned down the thermostat by 2 degrees F (1 degree C)?

Sample 2

• • • • • A metal fork of length of 0.3 m is used to poke some food in an oven. While in the oven the end of the fork is heated to 200 C.

The fork has a cross sectional area of 2 * 10 -4 m 2 .

The conductivity of steal (k) is 20 W/(mC) What will be the energy transfer rate (Q/t) of heat to the end of the fork not in the oven if the room temperature is 20 C?

Tougher Sample

• • • • • • • A metal bar has a length of 0.4 m.

However, half of the bar is iron and half is copper.

The conductivity (k) of iron is 40 W/(mC) and copper is 400 W/(mC).

The iron section is indoors and inside a furnace such that the end of the bar is 300 C.

The other end (copper) is outside at a temperature of 20 C.

What is the temperature in the center (call it Tc)?

Hint set up the energy transfer from iron to center and compare to transfer to the copper end…

Greenhouse Effect

• • • Often is published in the news incorrectly.

At best they will use the word “trapped” I think this is a dangerous oversimplification.

• The Greenhouse Effect is the atmosphere acting like a heat lamp powered by the Earth’s heat.

Step by Step

• • • • Sunlight passes through the atmosphere.

Some is reflected back into space – this cools the earth and is a separate issue.

The greenhouse gasses allow most to pass through The sunlight that reaches the earth is then absorbed, heating the earth.

Without the atmosphere

• • The earth would freeze!

The atmosphere through the greenhouse effect warms the earth by 60 degrees F.

How it works

• • • • • The earth emits heat as a blackbody.

Since the earth is 1/20 the sun this heat comes out in the infrared instead of optical.

th the temperature of Greenhouse gasses absorb some of this heat as it passes through the atmosphere (some gets past and goes into space).

The atmosphere then emits its own heat.

Half of this goes to space, and half to Earth.

Quick question

• What is the most abundant greenhouse gas in the Earth’s atmosphere?

Conclusion

• • • • We have learned about specific and latent heats.

We have learned about heat transfers.

We have learned about radiative heat and heat conduction.

We hopefully now understand the greenhouse effect.