Specific Heat

A, 10.0 g piece of metal at 100 degree Celsius is transferred to a calorimeter containing 50.0 mL of water initially at 23.0 degree Celsius. Calculate the specific heat capacity of the calorimeter C, is 25.0 J/K. The final temperature, T, is 25.6 degrees Celsius.
the metal cooled from 100C down to 25.6C while the water & the calorimeter warmed from 23 C up to 25.6C heat lost by the metal = heat gained by water & by the calorimeter – m C dT metal = m C dT water & C dT…

alorimeter – [10(C)(25.6C -100)] = (50g)(4.184J/g-C)(25.6 – 23) & 25.0J/K(25.6 – 23C) 744(C) = 543.92 & 65 744 (C) = 608.92 C = 0.818 J/g-C

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