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Energy and cost savings from turning down the thermostat?

  1. Jan 19, 2014 #1
    I liked this problem because it's something you can use to talk about money savings w/ other people. Just making sure everything was done right though.

    1. The problem statement, all variables and given/known data

    the interior temperature of the room is typically 34.6°F (

    the thermostat is typically set for 70.7°F.

    you propose to lower the thermostat to 60.1°F.

    Currently the building measures 23.5 × 34.5 × 14.5 feet

    How much heat will be saved each morning by bringing the building up to the new operating temperature of 60.1°F instead of 70.7°F?

    Ignore heat and air losses to the outside and

    consider air an ideal diatomic gas.

    Assume that in the morning the pressure in the room is atmospheric.

    Express your answer as a positive quantity.

    ----

    part 2:

    electricity rate is $6.56 per kilowatt-hour. How much money (in dollars) is saved each morning by only heating the room to 60.1F?


    2. Relevant equations



    diatomic gas:

    C_v = (5/2) R

    R = 8.3144621 J/ mol*K



    3. The attempt at a solution




    edit: found out both of my answers were wrong.
     
    Last edited: Jan 19, 2014
  2. jcsd
  3. Jan 19, 2014 #2

    DrClaude

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    Staff: Mentor

    Advice: use units throughout to make sure everything is consistent.
     
  4. Jan 19, 2014 #3
    Ok.

    n=PV/RT

    n= [(101,325)(332.889)] / [(8.3144)(294.65)] = 13,768.25 mol



    both of these answers are wrong too. Not sure where I'm botching this
     
    Last edited: Jan 19, 2014
  5. Jan 19, 2014 #4

    DrClaude

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    Staff: Mentor

    You're not being consistent here. Number of moles should be calculated from the conditions in the morning.

    There might be other errors (I haven't checked thoroughly), but you should fix this first.
     
  6. Jan 19, 2014 #5

    gneill

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    Staff: Mentor

    Consider that as the temperature rises the air will want to expand. Presumably the building is not air tight and pressure should remain at 1 atm. This means the number of moles to be heated will decrease as temperature rises... I see an integration in your future :smile:
     
  7. Jan 19, 2014 #6

    DrClaude

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    Staff: Mentor

    Fortunately, no. The problem states: "Ignore heat and air losses to the outside."
     
  8. Jan 19, 2014 #7

    gneill

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    Staff: Mentor

    Aurgh! I must've skimmed too lightly over that line; I only registered the heat loss bit. Thanks for catching that for me!

    (Mind you, the integration would not be a difficult one)
     
  9. Jan 19, 2014 #8
    great. It worked out using moles calculated from the temp K during morning time.
     
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