An unknown metal of mass $192\ \text{g}$ heated to a temperature of $100^\circ\text{C}$ was immersed into a brass calorimeter of mass $128\ \text{g}$ containing $240\ \text{g}$ of water at a temperature of $8.4^\circ\text{C}$. Calculate the specific heat of the unknown metal if water temperature stabilizes at $21.5^\circ\text{C}$. (Specific heat of brass is $394\ \text{J kg}^{-1}\text{K}^{-1}$)
Answer: (A) $916\ \text{J kg}^{-1}\text{K}^{-1}$
Heat gained by water and calorimeter (rise $13.1\ \text{K}$), with $c_w = 4200\ \text{J kg}^{-1}\text{K}^{-1}$:
$$(0.240\times4200 + 0.128\times394)\times13.1 = (1008 + 50.4)\times13.1 \approx 13865\ \text{J}$$
Heat lost by the metal (fall $78.5\ \text{K}$): $0.192\times c\times78.5 = 15.07\,c$.
$$c = \frac{13865}{15.07} \approx 920\ \text{J kg}^{-1}\text{K}^{-1}$$
This matches $916\ \text{J kg}^{-1}\text{K}^{-1}$ (the small difference comes from the value used for water's specific heat).
Solution by Sreeraj P, M.Sc Physics