Mechanical Equivalent Of Heat Essay, Research Paper
Mechanical Equivalent of Heat
By: A.D. Haynes
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[Abstract / Introduction]
Long before physicists recognized that heat is a form of energy transfer resulting from the random microscopic motion of atoms, they defined heat in terms of the temperature changes it produces in a body. The traditional unit of heat is the calorie (cal), which is the amount of heat needed to raise the temperature of 1 g of water by 1˚C. The kilocalorie is 1000 cal: 1 kcal = 1000 cal. Incidentally, the calories marked on some packages of food in grocery stores are actually kilocalories, sometimes called large calories.
The heat necessary to raise the temperature of 1 kg of a material by 1˚C is called the specific heat capacity, or the specific heat, usually designated by the symbol c. Thus, by definition, water has a specific heat of: c = 1 kcal/kg·˚C. Specific heat varies from substance to substance (see Appendix A, Table A4), and varies with temperature. For example, the specific heat of water varies by about 1% between 0˚C and 100˚C, reaching a minimum of 35˚C. This variation must be taken into account for a precise definition of the calorie: a calorie is the heat needed to raise the temperature of 1 g of water from say, 14.5˚C to 15.5˚C. Finally, the specific heat depends on the pressure to which the material is subjected during the heating.
Since specific heat is defined as the amount of heat required to increase the temperature of 1kg of a given substance by 1˚C, the amount of heat Q required to increase the temperature of a mass m by ΔT is proportional to m and to ΔT and can be found by the equation: Q = m c ΔT. This merely says that a large mass or a large temperature change requires more heat, in proportion to the mass and to the temperature change. Incidentally, work is defined as a force applied through a distance. For rotational motion, work is equal to the torque applied through an angular displacement, which can be described as W = F r θ. Further still, the total work will be equal to the applied torque multiplied by the total angular displacement. So, for one full rotation, the angular displacement is 2π, which makes the total angular displacement 2πN, depending on how many times the device is rotated. Therefore, the total work done on a machine can be found by the equation: Wtotal = F r 2πN.
Since heat is a form of work, it can be transformed into macroscopic mechanical work, and vice versa. For example, the transformation of heat into macroscopic work is accomplished by the workings of a steam engine, a steam turbine, or a similar machine. However, transformation of macroscopic work into heat requires no special machinery any kind of friction will convert work into heat.
Since heat is a form of work, the calorie is the unit of energy and it must be possible to express it in joules. The conversion factor between these units is called the Mechanical Equivalent of Heat, and the traditional method for the measurement of this phenomenon is known as Joule s experiment: A set of falling weights drives a paddle wheel, which churns the water in a thermally insulated bucket. The churning raises the temperature of the water by a measurable amount, converting a known amount of gravitational potential energy into a known amount of heat. The best available experimental results give 1 cal = 4.186 J for the accepted value of the mechanical equivalent of heat.
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[Procedure]
For the scope of this experiment, work was accomplished by an apparatus, which transferred heat to a system consisting of two brass cones (one filled with water), a metal stirrer, and the submerged portion of a thermometer. A handle on the outside of the apparatus was turned, which caused the outer of the two brass cones to rotate, while the inner cone was held stationary by a string. This string was attached to a large metal disk, which was fastened to the top of the inner cone, while the other end of the string was attached to a spring scale that read the applied force in Newtons. The friction between the two cones caused the temperature of the system to rise.
To begin this procedure, the device was C-clamped to the work table and a piece of paper, cut in the shape of the provided template, was fitted between the cones to enable the cones to move smoothly against one another. Next, the device was assembled and a trial run was made. During this time, the black clamp on the top of the large disk was adjusted so that a steady force reading on the dial could be obtained while the machine was in use.
Next, the device was disassembled and the mass of the two cones and the metal stirrer were measured. Then, the diameter of the large disk was measured. Next, the inner cone was filled with water to within 1 cm from the top, and the mass of both cones plus the water was measured. After that, the tip of the thermometer was moistened and placed in the slotted rubber stopper of the large disk, with the metal stirrer being placed in the other slot of the stopper. The apparatus was then reassembled and the number on the counter of the apparatus was recorded, as was the starting temperature.
Next, the handle of the apparatus was rotated so that a close to constant reading on the spring scale was maintained. During this step, the water was stirred several times, and the rotation was ultimately stopped when the temperature reached about 25˚C. After this, the water was stirred until it reached its highest temperature, and this value was recorded, as was the counter reading. Then, the thermometer was taken out of the water, and the volume of the submerged portion was determined by approximation the diameter and length of the submerged portion was measured and assumed to be a perfect cylinder. Lastly, the Mechanical Equivalent of Heat was computed by using the measurements carefully obtained during the procedure.
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[Data & Analysis]
Wtotal = F r 2πN
Average force from scale (F) 2.7 N
Radius of large disk (r) 0.075 m
Number of revolutions (N) 1181
Qtotal = m c ΔT
Mass of water (mw) 38 g
Specific Heat of water (cw) 1.00 cal/g-C˚-1
Mass of both brass cones (mc) 276 g
Specific Heat of brass (cc) 0.092 cal/g-C˚-1
Mass of the metal stirrer (ms) 6 g
Specific Heat of steel (cs) 0.106 cal/g-C˚-1
Vol. of submerged part of the therm. (Vt) 4.40 x 10-1 cm3
Volume specific heat of the therm. (Ct) 0.46 cal/(cm3 ˚C)
Change in temperature (ΔT) 5.1˚C
The Mechanical Equivalent of Heat as it relates to our experiment is given in the following equation:
MEH = Work / Heat (J/cal)
MEH = Wtotal/Qtotal
MEH = Fr2πN
(mwcw + mccc + mscs + VtCt) ΔT
MEH = (2.7 N) (0.075 m) 2π (1181)
[(38 g) (1.00 cal/g-C˚-1) + (276 g ) (0.092 cal/g-C˚-1) + (6 g) (0.106 cal/g-C˚-1) + (0.440 cm3) (0.46 cal/(cm3 ˚C))] 5.1˚C
MEH = 1502.639474 J
[(38 cal/C˚) + (25.392 cal/C˚) + (0.636 cal/C˚) + (0.2024 cal/C˚)] 5.1˚C
MEH = 1502.639474 J
(64.2304 cal/C˚) 5.1˚C
MEH = 1502.639474 J
327.57504 cal
MEH = 4.587161079 J/cal
MEH = 4.587 J/cal
As aforementioned, the accepted value for the Mechanical Equivalent of Heat is 4.186 J/cal. Thus, the percent error between the experimental and accepted MEH is given by the following:
Percent error = 4.587 J/cal 4.186 J/cal x 100 = 9.57%
4.186 J/cal
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[Results & Conclusions]
With the low percent error derived by our laboratory experiment (9.57%), it was demonstrated in class that by using the provided brass-cone-system-apparatus, we were able to closely approximate Joule s experiment for calculating the Mechanical Equivalent of Heat. One way that we were able to achieve such a close approximation was by taking painstakingly careful measurements of all variables involved in the system. Another way was by accurate calculations of all variables involved in the MEH formula. In sum, by using known physical laws and experimental data, this laboratory experiment properly demonstrated why heat is a form of energy transfer, hence the Mechanical Equivalent of Heat.
A.D. Haynes
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[Homework]
1. The iron cube below has width (a), mass density (ρ), and specific heat (c). The initial temperature is To. Calculate the heat energy (Q) if the final temperature is T1.
2. The solid iron ball below has radius (R). The other variables are the same as in question #1. Calculate the heat energy (Q) again.
3. In our experience, suppose the Force is 500 N, the Radius is 10 cm, and there are 80 loop rotations. Suppose all the work is absorbed by 1 kg of water with the initial temperature being 15˚C. Calculate the final temperature.