You are cooking breakfast for yourself and a friend using a 1220 W waffle iron and a 530 W coffeepot. Usually, you operate the appliances from a 110 V outlet for 0.500 h each day. (a) At 12 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 2440 W waffle iron that will enable you to cook twice as many waffles during a half-hour period,but you know that the circuit breaker in your kitchen is a 20 A breaker. Can you do the upgrade?

Question

You are cooking breakfast for yourself and a friend using a 1220 W waffle iron and a 530 W coffeepot. Usually, you operate the appliances from a 110 V outlet for 0.500 h each day.  (a) At 12 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 2440 W waffle iron that will enable you to cook twice as many waffles during a half-hour period,but you know that the circuit breaker in your kitchen is a 20 A breaker. Can you do the upgrade?

pierretteA · Accepted Answer

(a)  The total power is given by  P=(1120+460) W  = 1580 W  = 1.58 kW  the coast to use this power for 0.500 h each day for30 days will be coast = [P (?t)] (rate)  = [(1.58 kW) (0.500 h /day) (30.0 days)] ($0.120 / kWh)  = .......  (b)  if you upgrade the new power requirement will be  P=(2240 + 460) W  = 2600 W  the required current will be  I = P / V  = 2600 W / 110 V  = 23.6 amps  this is greater than 20 amps  so your circuit breaker cannot handle the upgrade

You are cooking breakfast for yourself and a friend using a 1220 W waffle iron and a 530 W coffeepot. Usually, you operate the appliances from a 110 V

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