A heated metal ball is dropped into a liquid - Edexcel - A-Level Maths Pure - Question 6 - 2006 - Paper 4

To find the value of $t$ when $T = 300$, we set up the equation:

\implies 300 - 25 = 400e^{-0.05t} \ \implies 275 = 400e^{-0.05t} \ \implies e^{-0.05t} = \frac{275}{400} \ \ \implies e^{-0.05t} = 0.6875.$$ Taking the natural logarithm of both sides gives: $$-0.05t = \ln(0.6875)\ \implies t = \frac{\ln(0.6875)}{-0.05} \ \ \approx 7.49.$$ Thus, $t \approx 7.49$ minutes.

To find the rate at which the temperature is decreasing, we need to differentiate the temperature function:

$\frac{dT}{dt} = -20e^{-0.05t}.$

Now substituting $t = 50$:

= -20e^{-2.5} \ \ \approx -20 \times 0.0821 \ \ \approx -1.64 \text{ }^\circ C/min.$$ Thus, the rate at which the temperature is decreasing at $t = 50$ is approximately **1.64 °C/min**.

From the equation for temperature:

$T = 400e^{-0.05t} + 25,$

as $t \to \infty$, $e^{-0.05t} \to 0$. Therefore, $T$ approaches:

$T \to 25 \text{ }^\circ C.$

Thus, as time progresses, the temperature of the ball will approach 25 °C, but it will never fall below this value. Hence, the temperature can never fall to 20 °C.