Answer:
R=1923Ω
Explanation:
Resistivity(R) of copper wire at 20 degrees Celsius is 1.72x10^-8Ωm.
Coil length(L) of the wire=37.0m
Cross-sectional area of the conductor or wire (A) = πr^2
A= π * (2.053/1000)/2=3.31*10^-6
To calculate for the resistance (R):
R=ρ*L/A
R=(1.72*10^8)*(37.0)/(3.31*10^-6)
R=1922.65Ω
Approximately, R=1923Ω
Answer:
I think it is process or technology
Answer:
t = 6179.1 s = 102.9 min = 1.7 h
Explanation:
The energy provided by the resistance heater must be equal to the energy required to boil the water:
E = ΔQ
ηPt = mH
where.
η = efficiency = 84.5 % = 0.845
P = Power = 2.61 KW = 2610 W
t = time = ?
m = mass of water = 6.03 kg
H = Latent heat of vaporization of water = 2.26 x 10⁶ J/kg
Therefore,
(0.845)(2610 W)t = (6.03 kg)(2.26 x 10⁶ J/kg)

<u>t = 6179.1 s = 102.9 min = 1.7 h</u>