QPaperGen

The differential equation is:
\[
\frac{d}{dt}T(t) = -k(T(t) - 25)
\]

This is a first-order linear differential equation. Rearranging:
\[
\frac{dT}{dt} + kT = 25k
\]
The integrating factor is \( e^{kt} \).\[\]

Multiplying both sides by integrating factor: \(\quad e^{kt}\frac{dT}{dt} + ke^{kt}T = 25k e^{kt}\)

\[
\Rightarrow \frac{d}{dt}(T e^{kt}) = 25k e^{kt}
\]
Integrate both sides:
\[
T e^{kt} = 25 e^{kt} + C
\]
\[
T(t) = 25 + Ce^{-kt}
\]
Initial condition: \( T(0) = 85 \)
\[
85 = 25 + C \cdot 1 \implies C = 60
\]
Therefore, the solution is:
\[
\boxed{T(t) = 25 + 60 e^{-kt}}
\]
Where \( k \) is a constant.
\[\]

(ii)We use the solution from above:
\[
T(t) = 25 + 60 e^{-0.03 t}
\]
Set \( T(t) = 40 \):
\[
40 = 25 + 60 e^{-0.03 t}
\]
\[
15 = 60 e^{-0.03 t}
\]
\[
\frac{15}{60} = e^{-0.03 t}
\]
\[
\frac{1}{4} = e^{-0.03 t}
\]
Take natural log on both sides:
\[
\ln \left(\frac{1}{4}\right) = -0.03t
\]
\[
-\ln 4 = -0.03t
\]
\[
\ln 4 = 0.03t
\]
Substitute \( \ln 4 = 1.3863 \):
\[
t = \frac{1.3863}{0.03} = 46.21
\]
\(\textbf{Final answer:}\)
\[
\boxed{t \approx 46.2 \text{ minutes}}
\]