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Thermal Properties

  1. Apr 16, 2005 #1
    HI im really stuck with this question, i dont know what to do with it??..can anyone help??..id like to learn how to solve this question...thanks.

    A man makes two cups of instant coffee four times a day. On every occasion he fills the kettle up to the 10-cup level. What is the cost of the energy he wastes in one year by heating more water than is neccessary??..make suitable assumptions to solve.

    Pleeeeeeeeeez help me!!
     
    Last edited: Apr 17, 2005
  2. jcsd
  3. Apr 16, 2005 #2

    Doc Al

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    Every time he makes his coffee the man "wastes" the energy required to heat the 8 cups of water he doesn't use. How much energy does it take to raise the temperature of 8 cups of water from room temperature to boiling?
     
  4. Apr 17, 2005 #3
    i dont know how to work it out??...what formula should i use??
     
  5. Apr 17, 2005 #4
    Is this the formula??....

    Energy = M x C x Change in temperature

    Room temp is roughly 15 C
    Boiling temp = 100 C

    So change in temp = 85 C

    C, specific heat capacity for water = 4190

    Not sure about "M" if this is the right formula???
     
  6. Apr 17, 2005 #5

    Doc Al

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    All of this is quite reasonable. (Note: 15 degrees C = 59 degrees F) Take care to use proper units. The specific heat of water is in J/kg. To find M: What is the mass of a cup of water?

    When you put it all together, you'll need to figure how many (unnecessary) cups of water the man heats in a year.
     
  7. Apr 17, 2005 #6
    Thanks...but i dont know how to work out the mass of a cup of water??
     
  8. Apr 17, 2005 #7

    Doc Al

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    Come on... don't be lazy! How many ounces in that coffee cup? (Make a reasonable assumption!) 1 fluid ounce is about 30 cm^3; the density of water is 1 gram per cm^3.
     
  9. Apr 18, 2005 #8
    so should i get a cup and measure its weight, then fill it with water and measure the weight again, subtract one from the other and il have the mass of water??..then i should convert the value accordingly??
     
  10. Apr 18, 2005 #9

    Doc Al

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    Why go to all that trouble? (Though you can if you have a scale handy!)

    Just guess as to how many ounces of water in each cup! (1 standard cup = 8 fluid oz.) Then use the data I provided to find the mass (in kg) of that cup of water.
     
  11. Apr 19, 2005 #10
    ok so does that mean that the mass of water per cup works out at 0.24kg so the amount of water being wasted evrytime he makes his 2cups is:
    E = M x C x CIT
    = 1.92 x 4190 x 78
    = 627494.4 j (This is for ONE boiling of the kettle)

    so that means that in a day the energy being wasted is 2,509,977.6 j

    so in a year would be 2,509,977.6 j x 365 = 916,141,824 j

    does that seem correct???
     
  12. Apr 19, 2005 #11

    Doc Al

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    Sounds good to me. It looks like you changed your assumption about the temperature difference from 85 degrees to 78 degrees. But no matter; the point is to give a reasonable estimation of the energy wasted. Since we're just estimating anyway, I'd round off that final answer. Given your assumptions, the energy wasted is approximately 900 million joules per year! (If you stuck with your first assumption -- 85 degree change -- the answer would be about 1 billion joules.)
     
  13. Apr 19, 2005 #12
    wow, it actually looks right??..thanks for your help..so for the final part how would i work out the cost of that energy???..??..i went to the npower web site to find out some figures but its all so confusing??
     
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