T(t) = 75e^{-0.081t} + 20 gives the temperature of the coffee at time t - Leaving Cert Mathematics - Question b - 2021

To determine the time it takes for the coffee to reach 82°C, we start by setting up the equation:

$82 = 75e^{-0.081t} + 20$

Next, we isolate the exponential term:

$82 - 20 = 75e^{-0.081t}$ $62 = 75e^{-0.081t}$

Now, we divide both sides by 75:

rac{62}{75} = e^{-0.081t}

To solve for t, we take the natural logarithm of both sides:

$ext{ln} \left( \frac{62}{75} \right) = -0.081t$

Now, we can isolate t:

$t = \frac{\text{ln} \left( \frac{62}{75} \right)}{-0.081}$

Calculating the values:

1. Calculate the left-hand side:

   • $\frac{62}{75} \approx 0.8267$
   • $\text{ln}(0.8267) \approx -0.192$

2. Substitute back into the equation:

   • $t \approx \frac{-0.192}{-0.081} \approx 2.37037 ext{ minutes}$

3. Convert to seconds:

   • $2.37037 \times 60 \approx 142.22$ seconds

Thus, rounding to the nearest second:

Final Answer: The coffee takes approximately 142 seconds, or 2 minutes and 22 seconds, to reach the ideal drinking temperature.