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Jaweria
 2 years ago
Please anyone help me!!
Jaweria
 2 years ago
Please anyone help me!!

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Jaweria
 2 years ago
Best ResponseYou've already chosen the best response.0When 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?

Cha1234
 2 years ago
Best ResponseYou've already chosen the best response.0http://answers.yahoo.com/question/index?qid=20090326150117AA7jVrX :)

chrismoon
 2 years ago
Best ResponseYou've already chosen the best response.0@Cha1234 That question is asking for different values.

Jaweria
 2 years ago
Best ResponseYou've already chosen the best response.0but this site has really confused answer

Jaweria
 2 years ago
Best ResponseYou've already chosen the best response.0can anyone explain this to me here?

chrismoon
 2 years ago
Best ResponseYou've already chosen the best response.0I learned this in basic physics, I assume solving the same way. Have you been taught Newton's Law of Cooling?\[\frac{ dT }{ dt }=k(T20)\]

Jaweria
 2 years ago
Best ResponseYou've already chosen the best response.0can anyone explain this to me that how I m getting this answer ln[(20  T)/(20  5)] =  0.01622

Jaweria
 2 years ago
Best ResponseYou've already chosen the best response.0when I m calculating it I m getting 1.10623

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1don't make it so hard it is easier than you think

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1When 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?

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1work with the differences in the temperature, that is what decays to zero initial difference is \(205=15\) degrees, after 25 minutes the difference is \(1020=10\) degrees so the difference has decreased by \(\frac{10}{15}=\frac{2}{3}\)

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1i.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 \[15\left(\frac{2}{3}\right)^{\frac{t}{25}}\]

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1after another 25 minutes, at 50 minutes, it will decrease by another \(\frac{2}{3}\) from \(10\) to \(10\times \frac{2}{3}=\frac{20}{3}=6\tfrac{2}{3}\)

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1that of course means it is \(6\tfrac{2}{3}\) colder than the 20 degree room, so its temperature is \(206\tfrac{2}{3}=13\tfrac{1}{3}\)

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1b) 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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