   Chapter 17, Problem 59AP

Chapter
Section
Textbook Problem

You are cooking breakfast for yourself and a friend using a 1.20-kW waffle iron and a 0.500-kW coffeepot. Usually, you operate these appliances from a 110.-V outlet for 0.500 h each day. (a) At 12.0 cents per kWh, how much do you spend to cook breakfast during a 30.0-day period? (b) You find yourself addicted to waffles and would like to upgrade to a 2.40-kW waffle iron that will enable you to cook twice as many waffles during a half-hour period, hut you know that the circuit breaker in your kitchen is a 20.-A breaker. Can you do the upgrade?

(a)

To determine
The cost spends to cook breakfast for 30 days at 12.0cents per kWh

Explanation

Given Info: The power utilized by waffle iron is 1.20kW and coffee pot is 0.500kW . Appliances are operated on 110V supply for 0.500hour each day. Cost per kWh is 12.0cents . Cost to cook breakfast is calculated for 30.0day period.

Explanation:

Formula to calculate the total cost to cook breakfast is,

C=P(Δt)r

• C is the total cost required to cook breakfast
• P is the total power utilized by appliances
• Δt is the time period for which cost is calculated
• r is cost per kWh

Appliance consists of waffle iron and coffeepot. The power of the appliance is the sum of both.

P=Pwaffle+Ppot

• Pwaffle is the power of the waffle iron
• Ppot is the power of the coffee pot

Use Pwaffle+Ppot for P in the above equation to rewrite C.

C=(Pwaffle+Ppot)(Δt)r

Substitute 1

(b)

To determine
Whether it is possible to upgrade the waffle iron into 2.40kW with a 20A circuit breaker.

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