Mastering Physics Solutions: Exercise 17.44

Exercise 17.44

Part A = 82.62 dollars Click to use the calculator/solver for this part of the problem
Part B = 3.2 Ω Click to use the calculator/solver for this part of the problem

Solutions Below:

On average, an electric water heater operates for 1.8 h each day.

Part A

If the cost of electricity is $0.34 kWh, what is the cost of operating the heater during a 30-day month? The power of a typical electric water heater is 4500 W.
Express your answer using two decimal places.

Since the heater operates for 1.8 hours each day and is at 4500 W, 1.8 x 30 x 4500 = 243000 W. This equals 243kW. At 0.34 each (243 * 0.34), the total cost is:

82.62 dollars

Part B

What is the resistance of a typical water heater? The current requirements of a typical electric water heater is 37.5 A.
Express your answer using two significant figures.

The formula for power is W = I2 * R. Since I = 37.5 and W = 4500 (from Part A):

W = I2 * R
4500 = (37.5)2 * R
4500 = 1406.25R
R = 3.2 Ω

3.2 Ω

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