Martha Richmond

2023-04-01

When a cold drink is taken from a refrigerator, its temperature is 5 degree C. After 25 minutes in a 20 degree C room its temperature has increased to 10 degree C. What is the temperature of the drink after 50 minutes?

Lena Navarro

To solve this problem, we can use the concept of exponential decay, which relates temperature change over time to the difference between the initial temperature and the ambient temperature.
Let's denote the initial temperature of the cold drink as ${T}_{0}$, the ambient temperature as ${T}_{a}$, and the time as $t$.
Given information:
${T}_{0}=5$ degrees Celsius (initial temperature of the cold drink)
${T}_{a}=20$ degrees Celsius (ambient temperature after 25 minutes)
$t=50$ minutes (total time)
We can use the exponential decay formula to find the temperature of the drink after 50 minutes. The formula is given by:
$T\left(t\right)={T}_{a}+\left({T}_{0}-{T}_{a}\right)·{e}^{-kt}$,
where $k$ is a constant that depends on the specific situation.
To find the constant $k$, we can use the given information after 25 minutes:
$T\left(25\right)=10$ degrees Celsius.
Plugging these values into the formula, we have:
$10=20+\left(5-20\right)·{e}^{-25k}$.
Let's solve this equation for $k$:
$10-20=-15·{e}^{-25k}$.
Dividing both sides by $-15$:
$\frac{2}{3}={e}^{-25k}$.
Now, we can solve for $k$ by taking the natural logarithm (ln) of both sides:
$ln\left(\frac{2}{3}\right)=ln\left({e}^{-25k}\right)$.
Using the property that $ln\left({e}^{x}\right)=x$, we can simplify the right side:
$ln\left(\frac{2}{3}\right)=-25k$.
Dividing both sides by $-25$:
$k=\frac{1}{25}·ln\left(\frac{2}{3}\right)$.
Now that we have the value of $k$, we can find the temperature of the drink after 50 minutes:
$T\left(50\right)=20+\left(5-20\right)·{e}^{-k·50}$.
Plugging in the values, we get:
$T\left(50\right)=20+\left(5-20\right)·{e}^{-\left(\frac{1}{25}·ln\left(\frac{2}{3}\right)\right)·50}$.
Simplifying this expression, we have:
$T\left(50\right)=20+\left(-15\right)·{e}^{-2·ln\left(\frac{2}{3}\right)}$.
Using the property that ${e}^{ln\left(x\right)}=x$, we can further simplify:
$T\left(50\right)=20-15·{\left(\frac{2}{3}\right)}^{2}$.
Calculating the value inside the parentheses:
$T\left(50\right)=20-15·\frac{4}{9}$.
Performing the multiplication:
$T\left(50\right)=20-\frac{20}{3}$.
Combining the terms:
$T\left(50\right)=\frac{60}{3}-\frac{20}{3}=\frac{40}{3}\approx 13.3$.
Therefore, the temperature of the drink after 50 minutes is approximately 13.3 degrees Celsius.

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