In order to solve this problem, it is necessary to apply the concepts related to density, as a function of volume and mass, as well as the principle of calorimetry.
By definition we know that Density is given as a function of mass in a given volume, mathematically it can be expressed as
[tex]\rho = \frac{m}{V}[/tex]
Where,
m = mass
V = Volume
Re-arrange to find the respective mass at each state we have
[tex]m = \rho V[/tex]
For state 1 we have that the mass is
[tex]m_1 = (1000kg/m^3)(0.391m^3) = 391kg[/tex]
For state 2
[tex]m_2 = (1000kg/m^3)(0.117m^3) = 117kg[/tex]
Now from calorimetry we know that heat change is given under
[tex]Q = mc_p\Delta T[/tex]
For energy conservation then,
[tex]m_1c_p\Delta T = m_2c_p\Delta T[/tex]
Since the specific heat is the same for the fluid then,
[tex]m_1\Delta T = m_2\Delta T[/tex]
[tex](391) (T-25\°C) = (117) (95\°C-T)[/tex]
[tex]T = 41.12\°C[/tex]
Therefore the final temperature of the mixture will be 41.12°C
