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Q 3.3-1E

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 cup of hot coffee initially at 95°C cools to 80°C in 5 min while sitting in a room of temperature 21°C. Using just Newton’s law of cooling, determine when the temperature of the coffee will be a nice 50°C.

The temperature of coffee will reach 50°C after 20.7 minutes.

See the step by step solution

## Step1: Analyzing the given statement

The initial temperature of a cup of hot coffee is 95°C and its temperature after sitting in a room of temperature 21°C for 5 min, becomes 80°C. By using Newton’s law of cooling, we have to determine the time after which the temperature of the coffee will reach 50°C.

Newton’s Law of Cooling is,

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

Here, we will take the values as,

Initial temperature of coffee,${{\mathbit{T}}}_{{\mathbf{0}}}{\mathbf{=}}{{\mathbf{95}}}^{{\mathbf{o}}}{\mathbit{C}}$

Temperature of the room,${{\mathbit{M}}}_{{\mathbf{0}}}{\mathbf{=}}{{\mathbf{21}}}^{{\mathbf{o}}}{\mathbit{C}}$

Temperature after 5 min,${\mathbit{T}}\mathbf{\left(}\mathbf{5}\mathbf{\right)}{\mathbf{=}}{{\mathbf{80}}}^{{\mathbf{o}}}{\mathbit{C}}$

## 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,

$T\left(5\right)=21+\left(95-21\right){e}^{-5k}\phantom{\rule{0ex}{0ex}}80=21+\left(74\right){e}^{-5k}\phantom{\rule{0ex}{0ex}}80-21=\left(74\right){e}^{-5k}\phantom{\rule{0ex}{0ex}}59=\left(74\right){e}^{-5k}\phantom{\rule{0ex}{0ex}}{e}^{5k}=\frac{74}{59}\phantom{\rule{0ex}{0ex}}5k=ln\left(1.254\right)\phantom{\rule{0ex}{0ex}}k=\frac{ln\left(1.254\right)}{5}\phantom{\rule{0ex}{0ex}}k=0.0453\phantom{\rule{0ex}{0ex}}$

One will use this value of k in next step to find the time after which the temperature of coffee will reach 50°C.

## Step 3: To determine the time after which the temperature of the coffee will reach 50°C

Substituting ${\mathbit{T}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}{\mathbf{=}}{{\mathbf{50}}}^{{\mathbf{o}}}{\mathbit{C}}$in equation (1),

$50=21+\left(95-21\right){e}^{-\left(0.0453\right)t}\phantom{\rule{0ex}{0ex}}50-21=\left(74\right){e}^{-\left(0.0453\right)t}\phantom{\rule{0ex}{0ex}}{e}^{\left(0.0453\right)t}=\frac{74}{29}\phantom{\rule{0ex}{0ex}}\left(0.0453\right)t=ln\left(2.552\right)\phantom{\rule{0ex}{0ex}}t=\frac{ln\left(2.552\right)}{0.0453}\phantom{\rule{0ex}{0ex}}t=20.7min\phantom{\rule{0ex}{0ex}}$

Hence, the temperature of coffee will reach 50°C after 20.7 minutes.