Hot tea is poured into a cup - HSC - SSCE Mathematics Advanced - Question 21 - 2020 - Paper 1

To find the cooling rate, we take the derivative of $T$ with respect to $t$:

$\frac{dT}{dt} = 70(1.5)^{-0.4} \cdot (-0.4) \cdot \ln(1.5)$

Substituting $t = 4$ into our derivative:

$\frac{dT}{dt} = 70(1.5)^{-1.6} \cdot (-0.4) \cdot \ln(1.5)$

Calculating gives us:

$\frac{dT}{dt} \approx 70(0.3914)(-0.4)(0.4055) \approx -5.934$

Therefore, 4 minutes after pouring, the tea is cooling at a rate of approximately -5.9°C per minute.

To find the time when the temperature reaches 55°C, we set up the equation:

$55 = 25 + 70(1.5)^{-0.4t}$

Rearranging gives:

$30 = 70(1.5)^{-0.4t}$ $\frac{30}{70} = (1.5)^{-0.4t}$ $\frac{3}{7} = (1.5)^{-0.4t}$

Taking the natural logarithm of both sides:

$\ln(\frac{3}{7}) = -0.4t \ln(1.5)$

Solving for $t$ gives:

$t = \frac{\ln(\frac{3}{7})}{-0.4 \ln(1.5)}$

Calculating yields:

$t \approx 5.22 \text{ minutes}$

Thus, it will take approximately 5.22 minutes for the temperature to reach 55°C.