Question 1067180
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Calculate the specific heat of a metal (in calories/gram-degree C) from the following data. 
A container made of the metal has a mass of 3.80 kg and contains 16.3 kg of water. A 1.40 kg piece of the same metal, 
initially at a temperature of 150 degrees C, is placed in the water. 
The container and water initially have a temperature of 15 degrees C, and the final temperature of the entire system is 18 degrees C.
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A balance equation (the balance of thermal energy) is


{{{C[m]*3.8*(18-15) + C[w]*16.3*(18-15)}}} = {{{C[m]*1.4*(150-18)}}}.      (1)


    It says "amount of heat absorbed by the system to raise its temperature from 15°C to 18°C is the same as the amount of heat 
    lost by the metal piece when it cooled from 150°C to 15°C).


Here {{{C[m]}}} is the specific heat of the metal  ( in calories/(gram-degree C) ), which is under the question),

     {{{C[w]}}} is the specific heat of water  ( = 1 calorie/(gram-degree C) ).


From (1), express  {{{C[m]}}} = {{{(16.3*(18-15))/(1.4*(150-18) - 3.8*(18-15))}}} = 0.282 calorie/(gram-degree C)
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