In this situation, we ignore the heat loss between the calorimeter and the environment. Then the heat that an unknown solid lost is the same as the heat that copper and water gain.
Denote the heat capacity of the unknown solid as `C,` then the solid lost `Delta T_s * C` Joules of heat, where `Delta T_s = 95 - 35 = 60` (degrees).
Denote the specific heat of copper as `c_c = 0.386 J / (g * ^@C)` and the specific heat of water as `c_w = 4.186 J / (g * ^@C).` The temperature change for them is `35 - 10 = 25 (^@C),` thus the heat gain is
`25*(75*c_c + 130*c_w) =14294.5 (J).`
The final equation is `60 C =14294.5` and thus
`C =14294.5 / 60 approx238.242 (J)/ (^@C).` This is the answer.
The mass `m_s` of the solid would be used to find the specific heat of the solid. It is `c_s = C/m_s approx1.91 (J/(g*^@C)).`