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---------- Forwarded message ---------- Date: Sat, 09 Oct 2004 08:59:08 -0400 From: Ludwik Kowalski <[EMAIL PROTECTED]> Reply-To: Forum for Physics Educators <[EMAIL PROTECTED]> To: [EMAIL PROTECTED] Subject: Conservation of energy. The experiment described several weeks ago was not good to measure L (latent heat of evaporation of water) but it can be used to demonstrate that energy is conserved. An extra step must be added to determine the value of L. It is a good experiment because the equipment needed (thermometer, stopwatch, voltmeter, ammeter, an immersion heater and a power supply) is usually available. A simple variac can be used instead of a dc power supply. PREPARATIONS AND A PRELIMINARY EXPERIMENT 1) Place a certain amount of water into a jar or beaker. 2) Insert the immersion heater and the thermometer into water. 3) Connect the heater to the power supply (through the ammeter to measure I). 4) Connect the voltmeter to measure the V at the heater's terminals. 5) Turn the current on to make sure that water can be brought to the boiling point, for example, in 10 minutes. 6) Replace hot water by the room temperature water and measure the heat capacity H of your setup. To accomplish this turn the current on for a time interval t1. During that time the water temperature should increase by dT equal to about 10 C. Calculate H = V*I*t1/dT; it might turn out to be 2000 J/deg. Verify that H is slightly larger that c1*m1, where m1 is the mass of water and c1 is its specific heat. MAIN EXPERIMENT 7) Bring water to the boiling point. Then turn the current off and start measuring temperature every two minutes for about 30 minutes. Plot the cooling curve (temperature versus time). This curve can be used to determine cooling rates, R, of the setup at desired temperatures, such as 80 C. A typical value of R is 3 degrees per minute. 8) Adjust the power supply to reach a desired temperature, say 80 C, and to keep it constant. 9) Measure V and I; these values should also become constant. 10) Turn the current off after time t2 (measured with a stopwatch). ANALYSIS OF DATA 11) How much energy was supplied during the time t2? E1=V*I*t2 12) How much energy was lost during the same time interval? That value, E2, can be determined from the cooling curve. Suppose the experiment is performed at 80 C and that the value of R, at that temperature, is 3 deg/min. Also suppose that H=2000 J/deg and that t2=5 minutes. In that case E2=R*t2*H=3*5*2000=30000 Joules 13) Are E1 and E2 equal? They should be equal at the level of your accuracy and precision. The most likely source of error has to do with R (unless your thermometer has divisions for fractions of degrees). OBSERVATIONS Note that R refers to heat losses due to all mechanisms, including evaporation. Therefore, the above data are not sufficient for finding L. To calculate L we plan to add another step. We will replace water with sand whose mass m2 is equal to c1*m1/c2 (where c1 and c2 are known specific heats of two substances and m1 is the mass of replaced water Bring the sand to the constant temperature of 80 C, turn the heater off and collect the cooling curve data. The rate of cooling with sand, R2, should by much smaller than the rate of cooling with water, R1, because there is no evaporation. We will assume that the rate of cooling by conduction, convection and radiation, at any given temperature, does not change when m1 of water is replaced by m2 of sand. Under that assumption the amount of energy lost by conduction, convection and radiation is E3=R2*t2*H. (Note that H does not change when c1*m1=c2*m2). We expect E3 to be a fraction of E2, for example 15%.. Suppose E3 turns out to be 4500 J. In that case we will be able to say that thermal energy lost through evaporation, in 20 minutes, was 30000-4500=25500 J. The value of L would then be calculated as 25500/M, where M is the mass of water evaporated in 20 minutes. The jar would be standing on a scale to measure M. Our scale is is very reliable but that does not mean that measuring of M is easy. The difficulty has to do with the vapor condensing on the walls of the beaker, on the thermometer, etc. Condensed droplets of water return back to the container and what one is measuring turns out to be a fraction on M. This difficulty is more pronounced when L is measured at 100 C. What can one do to reduce this source of error? Constant wiping of walls with tissue is not very practical, especially at 100 C.
