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Jaweria

  • one year ago

Please anyone help me!!

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  1. Jaweria
    • one year ago
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    When a cold drink is taken from a refrigerator, its temperature is 5°C. After 25 minutes in a 20°C room its temperature has increased to 10°C. (a) What is the temperature of the drink after 50C minutes? (b) When will its temperature be 15°C?

  2. Cha1234
    • one year ago
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    http://answers.yahoo.com/question/index?qid=20090326150117AA7jVrX :)

  3. chrismoon
    • one year ago
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    @Cha1234 That question is asking for different values.

  4. Jaweria
    • one year ago
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    but this site has really confused answer

  5. Jaweria
    • one year ago
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    can anyone explain this to me here?

  6. chrismoon
    • one year ago
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    I learned this in basic physics, I assume solving the same way. Have you been taught Newton's Law of Cooling?\[\frac{ dT }{ dt }=k(T-20)\]

  7. Jaweria
    • one year ago
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    can anyone explain this to me that how I m getting this answer ln[(20 - T)/(20 - 5)] = - 0.01622

  8. Jaweria
    • one year ago
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    when I m calculating it I m getting 1.10623

  9. satellite73
    • one year ago
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    don't make it so hard it is easier than you think

  10. satellite73
    • one year ago
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    When a cold drink is taken from a refrigerator, its temperature is 5°C. After 25 minutes in a 20°C room its temperature has increased to 10°C. (a) What is the temperature of the drink after 50C minutes? (b) When will its temperature be 15°C?

  11. satellite73
    • one year ago
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    work with the differences in the temperature, that is what decays to zero initial difference is \(20-5=15\) degrees, after 25 minutes the difference is \(10-20=10\) degrees so the difference has decreased by \(\frac{10}{15}=\frac{2}{3}\)

  12. satellite73
    • one year ago
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    i.e. every 25 minutes, the difference in the temperature decreases by \(\frac{2}{3}\) starting with an initial difference of \(15\) you can model this by \[15\left(\frac{2}{3}\right)^{\frac{t}{25}}\]

  13. satellite73
    • one year ago
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    after another 25 minutes, at 50 minutes, it will decrease by another \(\frac{2}{3}\) from \(10\) to \(10\times \frac{2}{3}=\frac{20}{3}=6\tfrac{2}{3}\)

  14. satellite73
    • one year ago
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    that of course means it is \(6\tfrac{2}{3}\) colder than the 20 degree room, so its temperature is \(20-6\tfrac{2}{3}=13\tfrac{1}{3}\)

  15. satellite73
    • one year ago
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    b) when will it be 15 degrees? that means b) when will it be 5 degrees colder than the room set \[15\left(\frac{2}{3}\right)^{\frac{t}{25}}=5\] and solve for \(t\)

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