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)).
http://hyperphysics.phy-astr.gsu.edu/hbase/Tables/sphtt.html
http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/spht.html
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