The heating coil of a hot-water heater has a resistance of 20 and operates at 21

Question

The heating coil of a hot-water heater has a resistance of 20 and operates at 210 V. If electrical energy costs $0.080/kWh, what does it cost to raise 280 kg of water in the tank from 15°C to 84°C?
=_$________

The current supplied by a battery in a portable device is typically 0.142 A. Find the number of electrons passing through the device in twenty-four hours.
=__________electrons

An aluminum wire having a cross-sectional area equal to 2.50 10-6 m2 carries a current of 4.50 A. The density of aluminum is 2.70 g/cm3. Assume each aluminum atom supplies one conduction electron per atom. Find the drift speed of the electrons in the wire.
=____________ mm/s

An electric heater carries a current of 12.5 A when operating at a voltage of 2.70 102 V. What is the resistance of the heater?
=_____________O

For a TeflonTM-filled parallel-plate capacitor having a plate area of 225 cm2 and insulation thickness of 0.0400 mm, determine the following:
(a) the capacitance
=____________nF

(b) the maximum voltage that can be applied to this capacitor
=______________V

Explanation / Answer

Part A

We know that Power can be defined as how fast work can be done, or work per unit of time

Since Work and Energy are equivalent, substituting allows us to say that P = energy / time

Or Energy is Power times time

Since we are given terms in kilowatt hours, that is Energy. (Power in kilowatts times time in hours)

We can determine the overall energy for a kilowatt hour

1 kW hr X 1000 W/kW X 3600s/hr leaves us with 3.6 X 106 J for each kW hr

To raise the the temp of water by adding heat, the formula Q = mcT will come in handy. We know that the specific heat for water is 4186 J/kg C

Substituting leaves

Q = (280)(4186)(84-15)

Q = 8.09 X 107 J

From our conversion from kWhr, we can ten determine how many kW hr is required to heat the water

(1 kWhr/3.6 X 106 J) x (8.09 X 107 J) = 22.5 kW hrs

Finally, since it cost $0.08 / kW hr it can be determined that the total cost is

$0.08 X 22.5 = $1.80

Part B

Current is measured in Amps, which is also Coulombs per sec

Since we have .142 C/s, we can do a few conversions

(.142 C/s) X (1 electron/1.6 X 10-19C) X (3600 s / hr) X 24 hrs

Total is 7.67 X 1022 electrons

Part C

The formula for drift velocity is

vd = I/nqA

Knowing the density of the Aluminum, allows us to find its volume

= m/V or V = m/

The atomic mass of aluminum is 26.98 g/mole

So V = 26.98/2.7 = 9.99 cm3

Since each aluminum atom is going to contribute one electron and there are 6.02 X 1023 electrons per mole

n = 6.02 X 1023 electrons/9.99 cm3

n = 6.03 X 1022 electrons / cm3 = 6.03 X 1028 electrons / m3

Now substituting

vd = (4.5)/[(6.03 X 1028)(1.6 X 10-19)(2.5 X 10-6)

vd = 1.87 X 10-4 m/s

Part D

Ohms law is used to solve this

V = IR

R = V/I

R = 2.7 X 102/12.5

R = 21.6

Part E

To find capacitance the formula is

C = kA/d (The dielectric constant (k) of Teflon is 2.1)

The area is 225 cm2 = 2.25 X 10-2 m2 and the d is .04 mm = 4 X 10-5 m

C = (2.1)(8.85 X 10-12)(2.25 X 10-2)/(4 X 10-5)

C = 1.05 X 10-8 F or 10.5 nF

Part F

The dielectric strength of Teflon is 60 X 106 V/m

So Max Voltage Vmax = Emaxd

Vmax = (60 X 106)(4 X 10-5)

Vmax = 2.4 X 103 V

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