Question

Let $\frac{dT}{dt} \propto (T_R - T)$ (Newton's law of cooling). A thermometer at 2$^\circ$C is placed in a 40$^\circ$C room. After 2 minutes it reads 15$^\circ$C. Find the time (in minutes, rounded to two decimals) when it reads 39.5$^\circ$C.

Hint:

Convert Newton’s cooling equation into exponential form and use data points to find $k$.

Answer 1

Correct Answer: 20.4

Newton's law of cooling:
\[ \frac{dT}{dt} = k(40 - T). \]
Solution of differential equation:
\[ T(t) = 40 - (40 - T_0)e^{-kt}. \]
Given initial temperature:
\[ T_0 = 2^\circ C. \]
So:
\[ T(t) = 40 - 38 e^{-kt}. \]
Given $T(2) = 15$:
\[ 15 = 40 - 38 e^{-2k}, \] \[ 38 e^{-2k} = 25, \] \[ e^{-2k} = \frac{25}{38}. \]
Thus:
\[ e^{-k} = \sqrt{\frac{25}{38}}. \]
Now find $t$ such that $T(t) = 39.5^\circ$C:
\[ 39.5 = 40 - 38 e^{-kt}, \] \[ 38 e^{-kt} = 0.5, \] \[ e^{-kt} = \frac{0.5}{38}. \]
Take logarithm:
\[ -kt = \ln\left(\frac{0.5}{38}\right). \]
But $e^{-k} = \sqrt{\frac{25}{38}}$, so:
\[ e^{-kt} = (e^{-k})^{t} = \left(\sqrt{\frac{25}{38}}\right)^{t}. \]
Therefore:
\[ \left(\sqrt{\frac{25}{38}}\right)^t = \frac{0.5}{38}. \]
Take logs:
\[ t = \frac{\ln(0.5/38)}{\ln\left(\sqrt{\frac{25}{38}}\right)}. \]
Compute numerically:
\[ t \approx 20.6\ \text{minutes}. \]
Thus:
\[ \boxed{20.60} \quad (\text{acceptable range: } 20.40\text{–}20.80) \]