Answer:
a. I = 30 A
b. E = 1080000 J = 1080 KJ
c. ΔT = 12.86°C
d. Cost = $ 4.32
Explanation:
a.
The current in the coil is given by Ohm's Law:
[tex]V = IR\\I = \frac{V}{R}[/tex]
where,
I = current = ?
V = Voltage = 120 V
R = Resistance = 4 Ω
Therefore,
[tex]I = \frac{120\ V}{4 \Omega}\\[/tex]
I = 30 A
b.
The energy can be calculated as:
[tex]E = VIt\\E = (120\ V)(30\ A)(5\ min)(\frac{60\ s}{1\ min})\\[/tex]
E = 1080000 J = 1080 KJ
c.
For the increase in the temperature of water:
[tex]E = mC\Delta T\\[/tex]
where,
m = mass of water = 20 kg
C = specific heat of water = 4.2 KJ/kg.°C
Therefore,
[tex]1080\ KJ = (20\ kg)(4.2\ KJ/kg.^oC)\Delta T[/tex]
ΔT = 12.86°C
d.
First, we will calculate the total energy consumed:
[tex]E=(Power)(Time)\\E=VI(Time)\\E = (120\ V)(30\ A)(0.5\ h/d)(30\ d)\\E = 54000\ Wh\\E = 54 KWh[/tex]
Now, for the cost:
[tex]Cost = (Unit\ Cost)(Energy)\\Cost = (\$ 0.08\KWh)(54\ KWh)[/tex]
Cost = $ 4.32
