Q.A $100\ \text{g}$ piece of an unknown metal, initially at $150\,^\circ\text{C}$, is dropped into $200\ \text{g}$ of water at $25\,^\circ\text{C}$ contained in a calorimeter of negligible heat capacity. The mixture reaches a common final temperature of $30\,^\circ\text{C}$. Taking the specific heat capacity of water as $4200\ \text{J kg}^{-1}\text{K}^{-1}$, calculate the specific heat capacity of the metal.

Calorimetry is the measurement of the amount of heat released or absorbed by a thermodynamic system during a heating process. When a hot body is brought into contact with a cold body and no heat is allowed to escape to the surroundings, the heat lost by the hot body exactly equals the heat gained by the cold body: $Q_{gain} = -Q_{lost}$, or equivalently $Q_{gain}+Q_{lost}=0$ (heat lost taken as negative, heat gained as positive). This is usually measured using a calorimeter, an insulated vessel of water fitted with a thermometer and a stirrer. If a sample of mass $m_1$ and specific heat $s_1$ at temperature $T_1$ is dropped into water of mass $m_2$, specific heat $s_2$, at temperature $T_2$, both reach a common final (equilibrium) temperature $T_f$. Using $Q_{gain}=m_2s_2(T_f-T_2)$ and $Q_{lost}=m_1s_1(T_f-T_1)$ and equating them gives the final temperature $T_f = \dfrac{m_1s_1T_1+m_2s_2T_2}{m_1s_1+m_2s_2}.$ If equal masses of the same substance are mixed, this reduces to the simple average of the two temperatures; if both start at the same temperature, no heat exchange occurs and $T_f$ equals that common temperature.