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Q 3.3-4E

Expert-verified
Found in: Page 107

### Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069

# A red wine is brought up from the wine cellar, which is a cool 10°C, and left to breathe in a room of temperature 23°C. If it takes 10 min for the wine to reach 15°C, when will the temperature of the wine reach 18°C?

The temperature of the wine will reach 18°C after 19.7 minutes.

See the step by step solution

## Step 1: Analyzing the given statement

The initial temperature of red wine is 10°C and left to breathe in a room of temperature 23°C. It takes 10 min for the wine to reach 15°C. By using Newton’s law of cooling, we have to determine the time after which the temperature of the wine will reach 18°C.

Newton’s Law of Cooling is,

$T\left(t\right)={M}_{0}+\left({T}_{0}-{M}_{0}\right){e}^{-kt}······\left(1\right)$

Here, we will take the values as,

Initial temperature,${T}_{0}={10}^{o}C$,

The temperature of the room, ${M}_{0}={23}^{o}C$

Temperature after 10 min,$T\left(10\right)={15}^{o}C$

## Step 2: To find the value of k in the formula of Newton’s Law of cooling

Using the given values in equation (1), to find the value of k,

$T\left(10\right)=23+\left(10-23\right){e}^{-10k}\phantom{\rule{0ex}{0ex}}15=23+\left(-13\right){e}^{-10k}\phantom{\rule{0ex}{0ex}}15-23=\left(-13\right){e}^{-10k}\phantom{\rule{0ex}{0ex}}-8=\left(-13\right){e}^{-10k}\phantom{\rule{0ex}{0ex}}{e}^{10k}=\frac{13}{8}\phantom{\rule{0ex}{0ex}}10k=ln\left(1.625\right)\phantom{\rule{0ex}{0ex}}k=\frac{ln\left(1.625\right)}{10}\phantom{\rule{0ex}{0ex}}k=0.0485\phantom{\rule{0ex}{0ex}}$

One will use this value of k in next step to find the time after which the temperature of the wine will reach 18°C.

## Step 3: To determine the time after which the temperature of the wine will reach 18°C

Substituting $T\left(t\right)={18}^{o}C$ in equation (1),

$18=23+\left(10-23\right){e}^{-\left(0.0485\right)t}\phantom{\rule{0ex}{0ex}}18-23=\left(-13\right){e}^{-\left(0.0485\right)t}\phantom{\rule{0ex}{0ex}}{e}^{\left(0.0485\right)t}=\frac{13}{5}\phantom{\rule{0ex}{0ex}}\left(0.0485\right)t=ln\left(2.6\right)\phantom{\rule{0ex}{0ex}}t=\frac{ln\left(2.6\right)}{0.0485}\phantom{\rule{0ex}{0ex}}t=19.7min\phantom{\rule{0ex}{0ex}}$

Hence, the temperature of the wine will reach 18°C after 19.7 minutes.