(a) A particular household uses a 1.8-kW heater 2.0 h / day...

(a) A particular household uses a $1.8-\mathrm{kW}$ heater $2.0 \mathrm{~h} /$ day ("on" time), four 100 -W lightbulbs $6.0 \mathrm{~h} /$ day, a $3.0-\mathrm{kW}$ electric stove element for a total of $1.0 \mathrm{~h} /$ day, and miscellaneous power amounting to $2.0 \mathrm{kWh} /$ day. If electricity costs $\$ 0.105$ per $\mathrm{kWh}$, what will be their monthly bill $(30 \mathrm{~d}) ?(b)$ How much coal (which produces $7500 \mathrm{kcal} / \mathrm{kg}$ ) must be burned by a $35 \%$ -efficient power plant to provide the yearly needs of this household?

Key Concepts

Electrical Energy Consumption

This concept involves calculating the energy used by multiplying a device's power rating (in kilowatts) by the time it operates (in hours). The resulting energy is typically expressed in kilowatt-hours (kWh), which is the standard unit for measuring electrical energy consumption.

Electricity Cost Calculation

This refers to determining the monetary cost associated with energy usage by multiplying the total energy consumed (in kWh) by the cost per kWh. It is a straightforward application of unit pricing to compute bills over a specified period, such as monthly.

Unit Conversion for Energy

Many energy problems require converting between different units of power and energy, such as from watts to kilowatts, or from kilowatt-hours to other units like joules or kilocalories. Correct unit conversion is essential for ensuring consistency in calculations involving energy usage and fuel energy content.

Energy Efficiency in Power Generation

This concept deals with the fact that not all input energy (from fuels like coal) is converted into useful electrical energy due to losses in the conversion process. The efficiency factor indicates the percentage of the fuel's energy that is actually turned into electricity, affecting how much fuel is required to meet energy demands.

Calorific Value of Fuels

Calorific value is a measure of the energy released by burning a unit mass of a fuel, typically expressed in kilocalories (kcal) per kilogram. It is crucial for determining how much fuel is needed to produce a desired amount of energy, especially when considering the efficiency of conversion in power plants.

Transcript

00:01 The total amount of energy that an electric appliance uses today is going to be equal to the total number of kilowatt hours used.

00:13 If we say that our device is rated at 1 .8 kilowatts and it operates for two hours per day, then then we add to that our four light bulbs at 100 watts each, or four times 0 .1 kilowatts running at hours a day.

01:01 And then we add our electric stove at two kilowatt hours per day.

01:18 The total energy that we are using is 11 kilowatt hours per day.

01:29 If we charge about $1 .105 per kilowatt hours, then and the cost is that total energy, 11 kilowatt hours per day, multiplied by what we're being charged, $0 .105 per kilowatt hour.

02:07 That gets us how much we spend in a day if we would like to know the cost in, say, 30 days.

02:19 Multiply by 30 days.

02:23 Days cancel out with days, kilowatt hours cancel out with kilowatt hours, and we're left with about $35.

02:36 That's really 30 days per month.

02:42 So we're given $35 a month is what it costs to run these appliances.

02:48 If we say that this power is being generated by a coal plant with an efficiency of 35 % and which produces 7 ,500 kilo calories per kilogram of coal burned, how much coal do we need to provide one years, worth of energy.

03:25 Well, the total energy that we need to get to the house is going to be 0 .35 times the total mass of coal that we need to burn times the energy that we get per piece of coal.

03:55 The energy that we get by burning.

03:59 Solving this for the mass of the coal.

04:05 It's just right, m total, which is equal to the total energy we need to provide over that efficiency times the energy stored in the coal...

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