You are working in a laboratory and for an experiment you want to perform, you n

Question

You are working in a laboratory and for an experiment you want to perform, you need to heat 30 ml of water from a temperature of 20 C to 40 C in 1 minute. Looking in an equipment catalogue, you find you can order a device to do this, but because of the paperwork required to order equipment in your lab, it would take about 6 weeks for you to get it.

You don't want to wait that long. You remember from your physics class that current generates heat and low resistance produces high current that generates a lot of heat. You have some Nickel-Chromium alloy Nichrome wire that has a resistivity of about 10-6 O-m. Suppose that you want to build a small heater out of a coil of Nichrome wire and a 6 V battery. Assume the battery has negligible internal resistance. (This is probably a bad assumption, given the low resistance of the wire, but let's use it as a starting point!)


a) How much heat energy (in Joules) do you need to achieve your heating goal?
b) How much power (in Watts) do you need to do it in the time indicated?
c) What resistance should your Nichrome coil have in order to produce this much power in heat?
d) Can you create a coil from the Nichrome wire having these properties? (Hint: Can you find a plausible length and cross sectional area for your wire that will give you the resistance you need?)
e) If the internal resistance of the battery were 1/3 O, how would it effect your calculation? (Only explain what you would have to do; don't recalculate the size of your coil.)

Explanation / Answer

a) Heat energy required
   = m*c*(change in temp)
   = 30 * 4.186 * 20
   = 2511.6 J

b) Power
   = Heat/time
   = 2511.6/60
   =41.86 J/s or Watts

c) Power = V^2/R
    R = V^2/P = 36/41.86 = 0.86 ohm

d) R = p*L/A
    Thus, L/A = R/p = 0.86/10^-6 = 0.86 * 10^6 m^-1

e) If internal resistance = 0.33 ohm
    Then , required coil resistance = 0.86 + 0.33 = 1.16 ohm

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