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MECHANICAL EQUIVALENT OF HEAT
In this lab, you will demonstrate the equivalence of mechanical energy/work and heat (increase of internal energy of an aluminum cylinder, as indicated by its increase in temperature). Whenever an amount of mechanical work W (expressed in Joules) is transformed completely into an equivalent amount of heat Q (expressed in calories), the ratio W/Q is given as 4.186 Joules/calorie. The equality 1 cal ≡ 4.186 J is known, for purely historical reasons, as the Mechanical Equivalent of Heat. These days we consider calories and joules simply as different units for energy. The apparatus consists of an aluminum cylinder that can be rotated by a crank, with a thermistor inside the cylinder to measure its temperature. A weight Mg is suspended from a nylon rope, the other end of which is wrapped around the cylinder. A counter measures the number of turns of the crank and cylinder. You will work hard turning the crank yet the weight stays in the same position. Your work does not change the mechanical energy of the system; instead, it produces an increase in the internal energy of the cylinder.
This happens because of the friction between the nylon rope and the cylinder as you turn the crank. In fact, the rope is wrapped around the cylinder enough so the weight is supported by the friction between the cylinder and the rope: f=tension in rope=Mg. The work done by you is equal and opposite to the work done by friction, which is given by the force of friction times the distance moved. The distance moved is the circumference of the cylinder 2πr times the number of turns N. The work done by you is thus:
W = 2πMgrN.
The energy gained by the cylinder in this process is given by Q = mc(Tf ­ Ti), where m is the mass of the cylinder, c=0.220 cal/goC is the specific heat of the cylinder, and Tf ­ Ti is the change in temperature of the cylinder. The temperature of the cylinder is measured using the thermistor (a resistor whose resistance changes rapidly with temperature) and a digital ohmmeter. A table showing the relationship between resistance and temperature is printed on the apparatus. The mass and radius of the cylinder are also written on it. Record them here: m = ___________________ r = __________________
Procedure
1. First make sure the weight (mass 2 kg plus 1 kg mass of holder) is off the floor at the start, and that the resistance reading is stable. Note the scale. Try a few turns to see that the counter records turns reliably. Note how much the resistance changes. DON’T TURN THE CRANK BACKWARDS. STOP IMMEDIATELY IF THE STRINGS BECOME TANGLED.
Mass of the suspended weight: M = _________
2. Reset the counter to zero by turning the black knob on the left side.
3. Record the initial resistance: Ri = _______ . The initial temperature of the cylinder should be close to room temperature. Determine it by approximate extrapolation from the table on the apparatus, to three significant figures: Ti = _________.
4. Look up in the table the resistance of the thermistor corresponding to a temperature about 5 oC above Ti. R+5 = ___________. 5. Now you have work to do. Turn the crank until the thermistor reading reaches the value R+5, i.e., the temperature has risen by about 5 oC. Turn the crank smoothly at moderate speed. (This will take several hundred turns.)
6. After you stop cranking, the temperature measured by the thermistor will rise a little higher than the value at which you stopped cranking. This is due to the finite speed at which heat is conducted from the perimeter of the cylinder to the thermistor. Watch this temperature rise carefully and record the lowest resistance reached and thus the highest temperature (again, three significant figures). Lowest resistance = _________ Tf = _________
7. Record the number of turns: N = ________
8. Calculate the work done: W = 2πMgrN = ___________ Joules
9. Calculate the energy transferred by heat: Q = mc(Tf ­ Ti) = _________ calories
10. Calculate the ratio W/Q = _________ J/cal
Compare this to the accepted ratio W/Q = 4.186 J/ cal. What is your percentage error? What factors could cause the error?
Questions
1. If you turned the crank much more slowly, how would your result be affected?
2. Note that when you are not turning the crank, the end of the nylon rope not attached to the mass is taut where it is tied to the hook. When you are turning the crank, the rope is slack there. Why? If it were not slack, how would this affect your result?