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

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

$300 = 400 e^{-0.06t} + 25.$

This simplifies to:

$275 = 400 e^{-0.06t}.$

Dividing both sides by 400 gives:

e^{-0.06t} = rac{275}{400}.

Taking the natural logarithm of both sides:

-0.06t = ext{ln}rac{275}{400}.

Solving for $t$:

t = rac{- ext{ln}(0.6875)}{0.06} ext{ minutes}.

Calculating this value:

$t ext{ } ext{ } ext{ } ext{ } t ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } = 7.49 ext{ minutes (to 3 significant figures)}.$

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

rac{dT}{dt} = -24 e^{-0.06t}.

Substituting $t = 50$:

rac{dT}{dt} = -24 e^{-0.06(50)}.

Calculating $e^{-3}$:

$e^{-3} ext{ } ext{ } = 0.04979.$

Thus,

rac{dT}{dt} = -24 imes 0.04979 ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } = -1.19 ext{ °C/min}.

Rounding to 3 significant figures:

Rate of decrease = -1.19 °C/min.

From the equation:

$T = 400 e^{-0.06t} + 25,$

we observe that as $t o ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } + ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } + ext{ } ext{ } ext{ } ext{ } + ext{ } ext{ } ext{ } ext{ } ext{ } + ext{ } ext{ } ext{ } ext{ } o ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } { ightarrow ext{ } ext{ } ext{ } + ext{ } ext{ } ext{ } ext{ } ext{ }}{ ightarrow ext{ } 25}.$$

The term $400 e^{-0.06t}$ approaches 0 as $t$ increases, and thus $T$ will approach 25 °C but never actually reach it. Therefore, the temperature of the ball can never fall below 25 °C, making it impossible for it to reach 20 °C.