Suppose you are provided with the following apparatus: An insulating container o

Question

Suppose you are provided with the following apparatus: An insulating container of negligible heat capacity containing100 grams of water at 25(degree Celsius) a. Draw a diagram, using the symbols described above, to showhow you should connect these components to heat water as rapidly aspossible. The meters should be connected so that from the two meterreadings alone you could determine at what rate the water is beingheated. b. Suppose the emf of the battery is 50 volts andthe current through the battery is 5 amperes(A). Assume thespecific heat of the water(c) 4 joules per gram(J/g) perCelsius degree, and the heat of vaporization is 2200 joulesper gram(J/g). Calculate the number of seconds requiredfor all the water to boil away.   

Explanation / Answer

There is no resistance anywhere in the circuit except that whichpasses through R1 & R2, which you are asked to combine in oneof 2 ways, in series or in parallel, in such a way that the circuitdissipates heat as quickly. The rate of dissipation is thewattage of the circuit : W = */R =2/R, where R is the total resistance of whicheverarrangement of resistors you chose. Since does notdepend on R, W is proportional to the inverse of the resistance,therefore you want to pick the circuit with the lesser totalresistance. R(series) = R1 + R2. R(parallel) = 1/R1 + 1/R2. You can show that the ratio of R(series) to R(parallel) = 2+ R2/R1 +R1/R2, which is easily > 1. So, you want to pickthe parallel arrangement: (I can't get the diagram thing to work, so you'll have to draw itfrom my description. Ask me if anything's iunclear).    The resistors in parallel are immersed in the tankwith the water. The A meter is in series with the battery andthe resistors. The V meter is connected anywhere across thebattery or the resistors (preferably out of the water, eh?) That way, the ammeter(A) measures the current through the circuit,I = /R, and the voltmeter (V) measures . Theenergy dissipated into the water can also be calculated from W =*I = 50*5 = 250 J/s. To boil the water, you need to heat it from 25C to 100C =75C. Heat added to reach boiling = 75 C *100g*4 J/(C g ) =3*104 J. Then you must add enough heat toevaporate 100g water at 100 C: 100g * 2.2*103 J/g= 2.2*105 J. Total heat needed = 3*104+ 2.2* 105 = 2.5* 105 J. Time needed = 2.5*105 J /250 J/s = 1000 s.

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