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
Thursday, February 14, 2013
An unknown solid with a mass of 125 grams is place inside a calorimeter. The initial temperature of the solid is 95 deg C. The calorimeter is made of copper with a mass of 75 grams and contains 130 grams of water. The water and calorimeter are at equilibrium initially at a temperature of 10 deg C. The final temperature after the calorimeter with the solid has reached equilibrium is 35 deg C. What is the heat capacity of the unknown solid.
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