Question

(a) A wire wound aluminum conductor (R = 50.00 Ω 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.

Answer 1

(a) Heat supplied = I²Rt = (V²/R)t
where I is supplied current , R is resistance and t is time
Heat supplied = (10²/50)*(3*60) = 360 J
Now this heat supplied = mcT
where m is mass of aluminium , c is specific heat and T is the change in temperature
360 = 15*(0.215)*(T-20)*(4.18)
T = 46.7 ^oC
(b) Heat received by the liquid nitrogen = mL
where m is the mass of liquid nitrogen evaporate
and L is the latent heat of vaporization.
Since temperature of LN2 is not mentioned that's why i am unable to give values
mL = m_Alc_AlT
(c) in this case some more heat will be added hence the equation will modifies
mL = m_Alc_AlT + (V²/R)t