Heat Energy Transfer

Introduction

Outline

• Heat Transfer
  • Definitions of terms
  • Specific Heat
• Practice with the Concepts
• Discussion Points

Heat Transfer

Energy can be transfered from one discrete piece of matter to another. We've already seen how this happens when a few particles collide, and the kinetic energy and momentum of the particles change.

Heat is a way of keeping track of the general movement of energy when lots of particles are involved. Because we don't observe individual particles any more, we need to use statistical methods, which describe tendencies for the overall behavior of the group of molecules understudy, but which cannot precisely describe the movement of individual particles. This powerful technique has some interesting consequences, as we'll see in the next chapter.

Definitions of Terms

We need to be very clear about our definitions of the terms heat, energy, and temperature.

Heat is a quantity of energy which flows from one substance to another. It is the result of energy released during collisions between particles. The unit of heat is the calorie, which usually defined as the amount of heat required to raise 1 gram of pure water 1 degree Centigrade (from 14.5 C to 15.5 C) at 1 atmosphere of pressure.

Another unit of heat which is often used is the British Thermal Unit, or BTU. This is the amount of heat required to raise the temperature of one pound of water by 1 degree Fahrenheit, from 63°F to 64°F. A BTU is about 250 calories, or 1/4 of a kilocalorie. When a heat device (like a butane burner) is rated at 3000 BTUs, the number is really a rate--3000 BTUs output per hour.

Energy we have already defined in two kinds as kinetic and potential energy. Since heat energy is based on collision energy, it is directly related to kinetic energy and can be converted to this form of mechanical energy. A calorie of heat energy can do 4.186 Joules of mechanical work (as determined by James Joule).

Temperature is a measurement of the average kinetic energy of the particles (atoms and/or molecules) in a substance. In determining this energy, we must consider all kinds of kinetic energy. A gas molecule, for example, can have translational motion, moving from one side to another of a container; it can also have rotational motion as it flips end-over-end, and it can also have vibrational motion. Molecules in solids do not generally have translational or rotational motion, but they do possess vibrational motion, so even solids generate some heat.

We can equate the internal energy of a gas to its thermal energy by adding the average kinetic energy of all particles. You recall that the expression for the kinetic energy of a single particle is 1/2 m * v². We need v, the average velocity of all the particles: $$\overline{v}=\frac{\sum _1^n{v}_1+v_2+v_3....+v_n}{n}$$ This allows us to determine the total kinetic energy by multiplying the average KE by the number of particles in the gas:

U = N * (½ m *v²)

But for an ideal gas, $$U=\mathrm{nkT}=\frac{3}{2}\mathrm{nRT}$$ (depending on whether we are measuring particles in molecules or moles).

Example: what is the average speed of a helium atom in a helum balloon at room temperature? Assume the mass of a helium atom is 6.65* 10^-27kg.

What information are you given?	the mass of the particle, 6.65* 10-27kg room temperature, usually assumed to be 20° C
What do you need to find?	v, the root mean velocity
What relationships do we know?	$$\mathrm{nkT}=n(\frac{3}{2})\mathrm{nRT}=N(\frac{1}{2}m{\overline{v}}^2)$$ We already know m and T, so the remaining missing information is the constant k (from the front cover) = 1.38 * 10-23 J/K
Check UNITS!	Our information on temperature is in Centigrade; the formula requires Kelvin, so convert: 20°C = 273K + 20K = 293K
Now we are ready to put it all together	Solve for v in our relationship: $$\overline{v}=\sqrt{\frac{3kT}{m}}$$ Substitute the values and do the math: $$\overline{v}=\mathrm{sqrt}\left(\frac{3\ast 1.38\ast {10}^{-23}J/K\ast 293K}{6.65\ast {10}^{-27}\mathrm{kg}}\right)$$ $$\overline{v}={\mathrm{1.35\; *\; 10}}^{-23}\sqrt{\mathrm{J/K}}$$ $$\mathrm{Now\; a\; Joule}=\mathrm{1\; N-m}=\frac{\mathrm{kg}}{{\mathrm{m}}^2}$$ so J/kg = m2/sec2, and √ (J/kg) = m/sec...the units of velocity which we need.

Specific Heat

The amount of heat an object must absorb or lose in order to change temperature depends on the composition of the object. Just as objects resist change in direction and speed (Newton's first law), and we can measure that resistance as a function of their mass and velocity, so objects resist changes in temperature. A change of 1 degree Centigrade in temperature requires a different amount of heat for 1 gram of water, 1 gram of copper, and 1 gram of wood. The ability of an object to absorb and give off heat is called its specific heat. It is related to the heat change, mass of the object, and temperature change by the formula

Q = m * c * δT

Since m = mass in kg and T is temperature in degrees C, the specific heat c must have units which make Q come out in calories or Joules. In the chart below, c is in J/kg*C:

Aluminum	900 J/kg*C
Lead	130
Wood	1700

Practice with the Concepts

Discussion Points

• A pot of water is boiling on a gas stove, and then you turn up the heat. What happens?
• What happens to the work done on a jar of orange juice when it is vigorously shaken?

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