Need Help? ausr. An office worker uses an 120-V power supply. Assume that the wi

Question

Need Help? ausr. An office worker uses an 120-V power supply. Assume that the wire is at 100 C throughout the 3.00-min time interval heater to warm 545 9 of water in a light, covered, insulated cup from 20*C to 1000C in 3.00 minutes. The heater is a Ni ) Calculate the average power required to warm the water to 100°C in 3.00 min. (The specific heat of water is 4186 l/kg·°C.) (b) Calculate the required resistance in the heating element at 100°C. c) Calculate the resistance of the heating element at 20.0C (d) Derive a relationship between the diameter of the wire, the resistivity at 20.0 Ao the resistance at 20.00 RO and the length L Do not substitute numerical values; use () L-3.00 m, what is the diameter of the wire? LG

Explanation / Answer

(A) Q = m C deltaT  

Q = 545 x 4.186 x (100-20) = 182509.6 J

P = Q / t = 182509.6 / (3 x 60)

P = 1014 Watt  

(B) P = V^2 / R

1014 = 120^2 / R

R = 14.2 ohm

(C) R = R0(1 + alpha deltaT )

14.2 = R0 (1 + (0.0004)(100-20))

R0 = 13.76 ohm

(D) R = rho L / A and A = pi d^2 / 4

R = 4 rho L / pi d^2

d = sqrt[ 4 rho L / pi R0]

(e) d = sqrt[ 4 x 100 x 10^-8 x 3 / pi x 13.76 ]

d = 0.53 x 10^-3 m Or 0.53 mm

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