Answer:
Approximately [tex]50.5\; {\rm ^{\circ} C}[/tex].
Explanation:
Assuming that there is no energy exchange with the outside. The energy that the hot water released should be equal to the energy that the cold water gained.
Let [tex]m_{1} = 50\; {\rm g}[/tex] denote the mass of the hot water initially at [tex]T_{1} = 99\; {\rm ^{\circ} C }[/tex], and let [tex]m_{2} = 60\; {\rm g}[/tex] denote the mass of the cold water initially at [tex]T_{2} = 10\; {\rm ^{\circ} C}[/tex].
Let [tex]T[/tex] denote the temperature of the mixture. Note that [tex]T[/tex] needs to be between the initial temperature of the hot water and that of the cold water.
Temperature change of the hot water: [tex](T_{1} - T)[/tex].
Energy that the hot water released: [tex](m_{1})\, (T_{1} - T)\, (\text{specific heat})[/tex].
Temperature change of the cold water: [tex]T - T_{2}[/tex].
Energy that the cold water gained: [tex](m_{2})\, (T - T_{2})\, (\text{specific heat})[/tex].
Assuming that there is no energy exchange with the outside:
[tex](m_{1})\, (T_{1} - T)\, (\text{specific heat}) = (m_{2})\, (T - T_{2})\, (\text{specific heat})[/tex].
Rearrange and simplify to find [tex]T[/tex]:
[tex](m_{1})\, (T_{1} - T) = (m_{2})\, (T - T_{2})[/tex].
Note that since this equation involves only temperature differences, [tex]T[/tex], [tex]T_{1}[/tex], and [tex]T_{2}[/tex] can all be measured in degree celsius instead of degrees Kelvin.
[tex]\begin{aligned}T &= \frac{m_{1}\, T_{1} + m_{2}\, T_{2}}{m_{1} + m_{2}} \\ &= \frac{(50)\, (99) + (60)\, (10)}{50 + 60}\; {\rm ^{\circ} C} \\ &\approx 50.5\; {\rm ^{\circ} C}\end{aligned}[/tex].
In other words, the temperature of the mixture would be approximately [tex]50.5\; {\rm ^{\circ} C}[/tex].
