A heated metal ball is dropped into a liquid - Edexcel - A-Level Maths Pure - Question 5 - 2006 - Paper 4

Setting $T = 300$, we have:

$300 = 400e^{-0.05t} + 25.$
Subtract $25$ from both sides:

$275 = 400e^{-0.05t}.$
Dividing by $400$ gives:

$e^{-0.05t} = \frac{275}{400} = 0.6875.$
Taking the natural logarithm of both sides:

$-0.05t = \ln(0.6875).$
Now solve for $t$:

$t = \frac{-\ln(0.6875)}{0.05} \approx 7.49.$
Thus, the value of $t$ is approximately $7.49$.

To find the rate of change of temperature, we differentiate $T$ with respect to $t$:

$\frac{dT}{dt} = -20e^{-0.05t}.$
Now, substituting $t = 50$ into this derivative:

$\frac{dT}{dt}\bigg|_{t=50} = -20e^{-0.05 \times 50} = -20e^{-2.5} \approx -1.64 \degree C/min.$
Thus, the rate at which the temperature of the ball is decreasing at $t = 50$ is approximately $-1.64 \degree C/min$.

From the temperature formula,

$T = 400e^{-0.05t} + 25,$
we see that as $t$ approaches infinity, $e^{-0.05t}$ approaches $0$. Thus,

$T \rightarrow 25 \degree C$
as $t \rightarrow \infty$.
This means that the temperature of the ball will always be greater than $25 \degree C$ and will never fall to $20 \degree C$.