(a) A particular household uses a 1.8-k W heater 3.0 h / day ("on" time), four 100-W lightbulbs

(a) A particular household uses a 1.8-k W heater 3.0 h / day...

(a) A particular household uses a $1.8-\mathrm{k} \mathrm{W}$ heater 3.0 $\mathrm{h} / \mathrm{day}$ ("on" time), four $100-\mathrm{W}$ lightbulbs 6.0 $\mathrm{h} /$ day, a 3.0 $\mathrm{-kW}$ electric stove element for a total of 1.4 $\mathrm{h} / \mathrm{day}$ , 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 7000 $\mathrm{kcal} / \mathrm{kg} )$ must be burned by a 35$\%$ -efficient power plant to provide the yearly needs of this household?

Key Concepts

Fuel Energy Content

This concept involves understanding the energy density of a fuel, which is the amount of energy released when a given mass of fuel is burned. It is typically expressed in units such as kilocalories per kilogram or megajoules per kilogram and is fundamental for calculating the required fuel mass to produce a desired amount of electrical energy.

Thermal Efficiency

This concept refers to the efficiency with which a power plant or energy conversion device converts raw fuel energy into useful electrical energy. It is expressed as a percentage and plays a crucial role in determining the amount of fuel required to meet a specific energy demand.

Electricity Cost Calculation

This concept relates to finding the cost of electricity consumption by multiplying the energy used (in kWh) by the cost per unit of energy. It is an essential financial evaluation tool that helps in budgeting and monitoring household energy expenses.

Energy Consumption Calculation

This concept involves determining the total energy used by calculating the product of power (in kilowatts) and time (in hours). It is fundamental in assessing the energy requirements of electrical appliances over a given period and is typically expressed in kilowatt-hours (kWh).

Unit Conversion in Energy

This key concept involves converting energy between different units, such as between kilowatt-hours and kilocalories or joules. Understanding these conversions is crucial in comparing energy systems and fuel sources, as measurements may use different units in various contexts.

Transcript

00:01 Hello everyone in this problem it is given in a household different appliances are used which are given as there is a heater of power 1 .5 kilowatt heater used for 3 hour per day for 100 watt electric lamp that is 0 .4 kilowatt electric lamp used for 6 hours per day for 100 watt electric lamp that is 0 .4 kilowatt electric lamp used for 6 hours per day.

00:52 Per day 3 kilowatt electric stove used for 1 .4 hour per day miscellaneous power consumption in a house is 2 kilowatt for 1 hour daily.

01:43 If cost of electricity is 0 .105 per kilowatt hour hour, then in the first part we have to calculate cost of electricity will cost of electricity will of a month of a month of 30 days and in second part we have to find the mass of the coal used if it produced 7 ,000 kilo calorie of energy per k g and efficiency of power plant is 35 % now we start solving it first we solve first part the first part total energy consumed in a month in a month we will calculate in kilowatt hour and it is defined as power into time into number of days that is 30 p1 into t1 plus p2 into t2 3 3 plus p4 into t4 to 30 now we are going to substitute the value p1 is given 1 .8 kilowatt time is 3 hours p2 is 0 .4 kilowatt and time is 6 hours p3 is 3 kilowatt and time is 1 .4 p 4 is 2 kilowatt and time is 1 .4 p 4 is 2 kilowatt and time is 1 hour number of days that is 30s in a month so this will be in kilowatt hour when you solve it total consumption of energy will be for 20 kilowatt hour 420 kilowatt hour now we will estimate the cost of will cost of electricity will it will be number of kilowatt hour that is 420 and cost of one watt hour 105 so it will be in dollar 44 .1 dollar per month this is the answer of first part now we have to solve for second part in second part we have to calculate mass of the coal required in a month if its energy specification is energy produced by the coal is given 7 ,000 kilo calorie per k.

06:51 So we will convert it into jule...

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