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Jaweria Group TitleBest ResponseYou've already chosen the best response.0
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?
 one year ago

Cha1234 Group TitleBest ResponseYou've already chosen the best response.0
http://answers.yahoo.com/question/index?qid=20090326150117AA7jVrX :)
 one year ago

chrismoon Group TitleBest ResponseYou've already chosen the best response.0
@Cha1234 That question is asking for different values.
 one year ago

Jaweria Group TitleBest ResponseYou've already chosen the best response.0
but this site has really confused answer
 one year ago

Jaweria Group TitleBest ResponseYou've already chosen the best response.0
can anyone explain this to me here?
 one year ago

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

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

Jaweria Group TitleBest ResponseYou've already chosen the best response.0
when I m calculating it I m getting 1.10623
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
don't make it so hard it is easier than you think
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
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?
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
work 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 \(\frac{10}{15}=\frac{2}{3}\)
 one year ago

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

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

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

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
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 \(t\)
 one year ago
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