The rate of change of the temperature, $T^{ ext{°C}}$ of a mug of coffee is given by dT dt = \frac{1}{25}T - k, \quad 0 \leq t \leq 50 - $t$ is the elapsed time, in minutes, after the coffee is poured into the mug - $k$ is a constant - initially, the temperature of the coffee is 100°C - 10 minutes later the temperature has fallen to 82°C - Scottish Highers Maths - Question 15 - 2015

Integrating the left side gives: $T = \frac{1}{25}Tt - kt + C$

Rearranging results in: $T - \frac{1}{25}Tt = -kt + C$

To find the constant $C$, we know that initially, at $t = 0$, $T = 100°C$: $100 = 0 - k(0) + C \implies C = 100$

After 10 minutes, the temperature falls to 82°C, so we set up the equation: $82 = \frac{1}{25}(82) \cdot 10 - 10k + 100$

This simplifies to: $-10k = 82 - 32.8 - 100 \implies k = \frac{51.8}{10} = 5.18$

Substituting $k$ back into the expression for $T$, we get: $T = \frac{1}{25}(Tt) - (5.18)t + 100$

This can also be simplified to: $T(t) = \frac{100 - 5.18t}{25}t$