| Physics help?

Physics help?

number1toolfool asked:


A particular household uses a 1.2-kW heater 3.0 h/day (“on” time), four 1.00E2-W lightbulbs 6.0 h/day, a 3.4-kW electric stove element for a total of 2.3 h/day, and miscellaneous power amounting to 1.6 kWh/day.

A.If electricity costs $0.120 per kWh, what will be their monthly bill (3.0E1 days)?
B.How much coal (which produces 7.0E3 kcal/kg) must be burned by a 35%-efficient power plant to provide the yearly needs of this household?

Figured out the first part to be 48 dollars, but not sure how to even start the second part. any help will be appreciated!

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One Response to “Physics help?”

  1. martian on April 10th, 2009 11:08 pm

    for the second question.. u need to find the daily energy usage. u can do that by multiplying the power requirement of each component with the time it remains active… like for example:
    for the heater:
    the energy consumed by heater each day is 1200*3*3600 JOULES.
    Likewise find the total energy consumed per day by the household and multiply by 365 to get the annual energy usage. Let this annual energy usage be E.
    We now have the energy required..now for the power plant, we know that 35% of the energy released from burning coal will be available as usable energy. i.e. 35% of x = E. Now x is the energy that needs to be directly released by burning coal. convert x from JOULES to CALORIES.. lets call this x1. the amount of coal to be burned is (x1)/(7.0E6) kg.