Answer
$57^\circ C$
see graph.
Work Step by Step
Use the Newton's Law of Cooling $T(t)=T_s+D_0e^{-kt}$ with the given conditions
$T_s=20, D_0=100-20=80,T(15)=75$, we have $75=20+80e^{-15k}$ which gives
$k=-ln(\frac{75-20}{80})/15\approx0.0313$ thus the equation becomes
$T(t)=20+80e^{-0.0313t}$ which can be graphed and shown in the figure.
After another 10 minutes $t=25$ and $T(25)=20+80e^{-0.0313\times25}\approx57^\circ C$
