Newton's Law of Cooling

If you notice that the water lost 15C of its initial 75C (above the room temp) in 3 mins then it lost 20% in those 3 mins and will therefore loose 20% of its excess temperature in the next 3 mins i.e. 12C bringing it down to 73C . Alas, in reality the cooling rate of hot water at temperatures near boiling is greatly increased by evaporation, convection and and even a some radiation all of which don't obey a simple linear conduction heat loss model.

Setting up and solving the separable equation gives $-\ln(T-T_0)=kt+c.$ Here, ignoring units, we have $T_0 = 25.$ At time $t=0,$ we have $T=100,$ which gives

$-\ln(100-25) = k\cdot 0 + c,$

so $c=-\ln(75).$ Then, at time $t=3,$ we have $T=85,$ so

$-\ln(85-25) = k\cdot 3 -\ln(75),$

so $k = \dfrac{\ln(75)-\ln(60)}{3} = \dfrac{\ln\left(\frac{5}{4}\right)}{3}.$ Finally, we can substitute $t=6$ to find

$-\ln(T-25) = \frac{\ln\left(\frac{5}{4}\right)}{3} \cdot 6 - \ln(75) = -\ln\left(75 \cdot \frac{4^2}{5^2}\right) = -\ln(48).$

Thus, $T-25 = 48,$ so the temperature after $3$ more minutes is $73^\circ\text{C}.\ _\square$