A wire wound aluminum conductor (R = 50.00 and m = 15.00g) is initially at 20.0

Question

A wire wound aluminum conductor (R = 50.00 and m = 15.00g) is initially at 20.0 C. The conductor, thermally insulated from its environment, is connected to a 10.00 V power supply for 3.00 minutes. What will be the final temperature of the aluminum? Assume the resistance remains constant. In this temperature range, cAl = 0.215 cal/g-C.

(b) With its thermal insulation removed and disconnected from the power supply, the resistor from part (a) is placed in LN2. Calculate the amount of LN2 that evaporates while the resistor is cooling from room temperature to LN2 temperature. Assume room temperature is 20.0 C. Do not use experimental values when doing this calculation. Use accepted values only.

(c) While the uninsulated resistor is in liquid nitrogen, it is supplied with 10.00 V for 3.00 minutes. How much additional LN2 will evaporate? Assume the resistance remains constant.

Please answer ALL three parts, especially part C, thank you!

Explanation / Answer

Power decipated in the resistor

P = IV = V2/R = 100/50 = 2 Watt

Energy  = Pt = 2*60*3 = 360 J

specific heat of Al = 0.215 cal/g-C

                             = 900 J/kg-K

mass of reisitor    m = 15.0 g

if T is change in temp of the resistor then

T*15.0e-3*900 = 360

   T = 26.67 deg C

Final temp = 20 +26.67 = 46.67 C

b) LN2 temp = -195 C = 78 K

latent heat of evaopration LN2 = 199 kJ/kg

when Al resistor (15g is put in LN2 it will loose heat temp brought down from 20 C to -195

heat lost = 215*15.0e-3*900 = 2.903 e+3 J

amount of LN2 to evaoparate = 2.903/199 = 0.0145 kg

                                                 = 14.5 gms

when the resistor is heated for 3 mts it will produce 360 J of additional energy

ampunt of LN2 to evaporate = 360/199 = 1.81 gms

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