Question

Physics question 28 plz help me

Answer 1

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:

$V = IR\\I = \frac{V}{R}$

where,

I = current = ?

V = Voltage = 120 V

R = Resistance = 4 Ω

Therefore,

$I = \frac{120\ V}{4 \Omega}$

I = 30 A

b.

The energy can be calculated as:

$E = VIt\\E = (120\ V)(30\ A)(5\ min)(\frac{60\ s}{1\ min})$

E = 1080000 J = 1080 KJ

c.

For the increase in the temperature of water:

$E = mC\Delta T$

where,

m = mass of water = 20 kg

C = specific heat of water = 4.2 KJ/kg.°C

Therefore,

$1080\ KJ = (20\ kg)(4.2\ KJ/kg.^oC)\Delta T$

ΔT = 12.86°C

d.

First, we will calculate the total energy consumed:

$E=(Power)(Time)\\E=VI(Time)\\E = (120\ V)(30\ A)(0.5\ h/d)(30\ d)\\E = 54000\ Wh\\E = 54 KWh$

Now, for the cost:

$Cost = (Unit\ Cost)(Energy)\\Cost = (\ 0.08\KWh)(54\ KWh)$

Cost = $ 4.32