Answer:
a) I = 13.04 A
b) R = 8.82 ohms
c) 1291.87 kilocalories are generated an hour.
Explanation:
let P be the power of the heater, V be the voltage of the heater, I be the current of the heater, R be the resistance.
a) we know that:
P = I×V
I = P/V
= (1500)/(115)
= 13.04 A
Therefore, the current of the heater is 13.04 A
b) we now have voltage and current, according to Ohm's law:
R = V/I
= (115)/(13.04)
= 8.82 ohms
Therefore, the resistance of the heating coil is 8.82 ohms.
c) the number of kilocalories generated in one hour by the heater is just the energy the heater produces in one hour which is given by:
E = P×t
= (1500)(1×60×60)
= 5400000 J
since 1 calorie = 4.81 J
1 kilocalorie = 0.001 calories
E = 5400000/4.18 ≈ 1291866.029 calories ≈1291.87 kilocalories
Therefore, 1291.87 kilocalories are produced/generated in one hour.
This question involves the concepts of power, energy, and Ohm's Law.
(a) The current in the heater will be "13.04 A".
(b) The resistance of the heating coil is "8.82 Ω".
(c) The energy generated by the heater in one hour will be "1291.87 Kilocalories".
The current in the heater can be found using the following formula:
P = IV
where,
Therefore,
[tex]1500\ W=I(115\ V)\\\\I=\frac{1500\ W}{115\ V}[/tex]
I = 13.04 A
The resistance of heater coil can be found using Ohm's Law, as follows:
V = IR
where,
Therefore,
[tex]115\ V = (13.04 A)(R)\\\\R=\frac{115\ V}{13.04\ A}\\\\[/tex]
R = 8.82 Ω
Energy can be found using the following formula:
E = Pt
where,
Therefore,
E = (1500 W)(3600 s)
E = 5400000 J[tex](\frac{1\ Calorie}{4.18\ J})(\frac{1\ kilocalorie}{1000\ calories})[/tex]
E = 1291.87 Kilocalories
Learn more about Ohm's Law here:
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