We left off with the mechanical basis for the kinetic energy of translation of a monatomic gas and now examine the mechanics -heat relationships further. All this is paving the way to enter the field known as

*thermodynamics*. Before getting to an experiment to determine what is called "the Mechanical equivalent of heat" we need to complete some more basics, which will lead to two differing statements of the First Law of Thermodynamics. We need to focus on the physical meaning of the internal energy, or U. At the end of the preceding section we worked out the translational contribution to U. But in fact U has three contributions:

U = U(transl.) + U(rotational) + U (vibrational)

On average then, the available energy is equally shared by each independent degreee of freedom, acknowledging at once that molecules can not only move from point a to point b in space (translation), but also rotate and vibrate. Each of these contributes (½ kT) of energy on average. For the different translational velocity directions (V_x, v_y and v_z) we than have a total contribution of 3((½ kT) = 3kT/2. To obtain the rotational energy contributions we can use the "dumbell" model shown in the graphic where the system can rotate either in the z-plane (e.g. about the x-axis) or in the x-plane (about the z-axis). Since there is (½ kT) of energy for each of these then the total contributed to internal energy U is: 2((½ kT)= kT. (Note that the moment of inertia of a dumbell about its own axis (y) is zero so rotations about this axis are irrelevant. The total contributions we have thus far, from translation and rotation add up to: 3kT/2 + kT = 5kT/2.

If we finally add in the possible vibrations (visualizing the dumbell atoms connected via 'springs') we get on more kT contribution to make: 5kT/2 + kT = 7kT/2, so U = 7kT/2. We will come back to this on considering the various "specific heat capacities". For now, we want to obtain the mechanical equivalent of heat.

The apparatus shown is very useful in this sense, as it's simple and can give a really decent result. The apparatus used consists basically of a copper drum that acts as a calorimeter (calorie measuring device) and around which several turns of a copper band are wound. Each winding of the band makes direct frictional contact with the drum so on any relative motion the friction will yield heat. The motion is generated by use of a hand crank and the drum is mounted to a base plate composed of non-conducting material. A thermometer can be inserted into the drum by making accommodation using a leakproof rubber washer with a locking nut. (The thermometer rotates with the drum and is read using the aid of a plane mirror).

In the West Indies we usually used a load of mass 5 kg. (suspended) but originally resting on the floor. The copper band is already attached to the load using a cord with a spring safety hook and adjusting device. The other end of the copper band is connected wo the apparatus itself using a spiral spring and pin. The mechanical work is then given by:

W = 2πr Mgn, where:

r = radius of the drum (cm), M = mass of the load (grams), g = 980 cm/sec^2, and n = revolutions turned. Then the heat in calories is proiduced by the work done (against friction) and several trial runs need to be performed to regularize the procedure and spot any errors - say in the set up of the apparatus. To get the heat (H) generated we use the experimental formula:

H = w(H2O) (t2 - t1) + [w(cal) + w(band)]c(Cu) [t2 - t1]

where:

w(H2O) denotes the weight of water in the copper drum,

w(cal) denotes the weight of the copper calorimater drum

c(Cu) = the specific heat of copper = 0.092 (cal/g-C)

and t2 is the final temperature of the water+ drum while t1 is the initial temp.

The mechanical equivalent of heat (J, in joules per calorie) can then be found from:

J = (2πr Mgn)/{w(H2O) (t2 - t1) + [w(cal) + w(band)]c(Cu) [t2 - t1)}

This is a ratio of W (mechanical work done) to heat (H) and is expressed by:

J = W/H = 4.18 J per calorie

Thus, we have the number of units of mechanical energy needed to generate 1 unit of heat. The experiment also provides an empirical basis to the first law of thermodynamics expressed as: Q = U + W, where Q denotes the quantity of heat added to a system, U is the internal energy, and W is the work done, and this is all for an iso-volumetric process. Thus, the mechanical equivalent of heat and spefically the generation of H in relation to W (mechanical work done) incorporated an increase in the internal energy, e.g. owing to the friction effect in the experiment. Thus, the first law is also the formal statement of equivalence given by: W = JH, and provided the mechanical and heat energy are measured in the same units, it may be stated as :

"

**In any transfer of other forms of energy into heat, or heat into other forms of energy, the heat energy is equal to the amount of transformed energy"**

This, of course, is simply a special case of the law of conservation of energy. Note that in the case of the statement Q = U + W that Q is associated with the energy transfer arising from a difference of temperature between the systems (taken into account in our experiment by the difference (t2 - t1), while U is associated with the energy possessed

*because of the increased molecular motion due to friction*. Microscopically then, temperature increase is associated with the energy of molecular motion.

Some problems:

1) Work is done on water by a rotating paddle wheel driven by two weights of 1 kg mass each, falling at constant speed. Water is then stirred by the paddles increasing its temperature. Assuming that the loss in potential (mechanical) energy = work done by the paddle wheel on water, where the latter is defined according to:

m(H2O) c(H2O) (T2 - T1), and T2 = 15.5 C, T1 = 14.5 C,

find the height needed for the weights to drop to obtain 4.18J. (take the specific heat of water, c(H2O) = 4200 J/Kg K, and g = 9.8 m/sec/sec and the mass of the water = 100 kg.

2) A 200 pound weight falls 40 feet. How much heat was developed? (Take g = 32 ft/sec/sec and 1 Btu = 778 ft-lbs.)

3) A lead bullet (mass m) travels at velocity v = 300 m/s, and is fired into a wood target. Assuming that half the heat developed goes into the target and the rest into the bullet, how much will the temperature of the bullet be raised?

4)A 2 kg object has a speed of 2.5 m/s. Find: (a) its kinetic energy in joules, and b) the equivalent energy in calories.

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