Given:Mass of metal piece $(m_1) =420\ \mathrm{g}$Spécific heat capacity of metal piece $(C_1) =$ ?Initial temperature of metal piece $(T_1) =80^{\circ}\ \mathrm{C}$Mass of water $(m_2) =80\ \mathrm{g}$Specific heat capacity of water $(C_2)$$=4.2\,\mathrm{Jg}^{-1}\,{}^{\circ}\mathrm{C}^{-1}$Initial temperature of water $(T_2) =20^{\circ}\ \mathrm{C}$Mass of calorimeter $(m_3) =84\ \mathrm{g}$Specific heat capacity of calorimeter$\;=200\ \mathrm{Jkg}^{-1}\,{}^{\circ}\mathrm{C}^{-1}$$\;=\; \frac{200}{1000}\ \mathrm{Jg}^{-1}\,{}^\circ\mathrm{C}^{-1}$$\;=0.2\ \mathrm{Jg}^{-1}\,{}^{\circ}\mathrm{C}^{-1}$Final temperature of mixture $=30^{\circ}\ \mathrm{C}$According to the principal of calorimetry, Heat lost by metal piece (hot body) = Heat gained by water + Heat gained by calorimeter (cold body) $\;m_1 C_1 \Delta T_1= m_2 C_2 \Delta T_2$ $+m_3 C_3 \Delta T_3$ $\;420 \times C_1 \times (80-30)= 80 \times 4.2 \times (30-20)$ $+84 \times 0.2 \times (30-20)$ $\;21000 \times C_1= 3360+168$ $\;21000 C_1= 3528$ $\;C_1= \frac{3528}{21000}$ $\;\; =0.168\ \mathrm{Jg}^{-1}\,{}^{\circ}\mathrm{C}^{-1}$