Please anyone help me!!

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Please anyone help me!!

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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?
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@Cha1234 That question is asking for different values.

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but this site has really confused answer
can anyone explain this to me here?
I learned this in basic physics, I assume solving the same way. Have you been taught Newton's Law of Cooling?\[\frac{ dT }{ dt }=k(T-20)\]
can anyone explain this to me that how I m getting this answer ln[(20 - T)/(20 - 5)] = - 0.01622
when I m calculating it I m getting 1.10623
don't make it so hard it is easier than you think
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?
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 \(\frac{10}{15}=\frac{2}{3}\)
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 \[15\left(\frac{2}{3}\right)^{\frac{t}{25}}\]
after another 25 minutes, at 50 minutes, it will decrease by another \(\frac{2}{3}\) from \(10\) to \(10\times \frac{2}{3}=\frac{20}{3}=6\tfrac{2}{3}\)
that of course means it is \(6\tfrac{2}{3}\) colder than the 20 degree room, so its temperature is \(20-6\tfrac{2}{3}=13\tfrac{1}{3}\)
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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