Question

Your manager wants you to design a new superinsulated hot-water heater for the residential market. A coil of Nichrome wire is to be used as the heating element. Calculate the length of wire required. (The water heater is cylindrical in shape, and is completely filled with water. Let the diameter of the water heater be 50 cm and its height 0.8 m. The coil of Nichrome wire is located inside the water heater, in contact with the water. Assume that the diameter of the Nichrome wire is 2.0 mm and that the water, initially at 20°C, is to be heated to 77°C in 1.0 h. Let the voltage used by the water heater be 120 V.)

Answer 1

Volume of the water heater = Base area x height
= r² x h
= x (25)² x 80
= 157079.63 cm³
1000 cm³ of water weighs 1 kg
628318.5 cm³ of water weighs 157079.63^/1000 kg = 157.08 kg
Mass of water, m = 157.08 kg

Heat energy required to increase the temperature of water from 20 ^oC to 77 ^oC is
Q = mCdT
Where C is the specific heat of water, C = 4185.5 J/(kg?K)
Q = 157.08 x 4185.5 x (77 - 20)
= 3.75 x 10⁷ J

Heat energy released by the Nichrome wire, Q = V²/R x t
Where R is the resistance of the wire and t is the time
Q = [(120)²/R] x 60 x 60
= 5.184 x 10⁷ / R

Equating the two energies,
3.75 x 10⁷ J = 5.184 x 10⁷ / R
R = 1.383 Ohm

Resistance, R = L/A           ...(1)
Where is the resistivity, L is the length and A is the area.
From the given diameter of Nichrome wire, A = x (1 x 10^-3)²
= 3.14 x 10^-6 m².
Resistivity of Nichrome wire, = 1 x 10^-6 Ohm. m

From (1),
L = RA/
= [(1.383) x (3.14 x 10^-6)] / (1 x 10^-6)
= 4.35 m