Given:
Mass of metal piece $\left({\mathit{m}}_1\right)=420\mathrm{g}$
Spécific heat capacity of metal piece $\left({\mathrm{C}}_1\right)=$ ?
Initial temperature of metal piece $\left({\mathrm{T}}_1\right)={80}^{\circ }\mathrm{C}$
Mass of water $\left({\mathit{m}}_2\right)=80\mathrm{g}$
Specific heat capacity of water $\left({\mathrm{C}}_2\right)$
$=4.2{\mathrm{Jg}}^{-1}{}^{\circ }{\mathrm{C}}^{-1}$
Initial temperature of water $\left({\mathrm{T}}_2\right)={20}^{\circ }\mathrm{C}$
Mass of calorimeter $\left({\mathit{m}}_3\right)=84\mathrm{g}$
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)
${\mathit{m}}_1{\mathrm{C}}_1∆{\mathrm{T}}_1={\mathit{m}}_2{\mathrm{C}}_2∆{\mathrm{T}}_2$ $+{\mathit{m}}_3{\mathrm{C}}_3∆{\mathrm{T}}_3$
$420\times {\mathrm{C}}_1\times \left(80-30\right)=80\times 4.2\times \left(30-20\right)$
$+84\times 0.2\times \left(30-20\right)$
$21000\times {\mathrm{C}}_1=3360+168$
$21000{\mathrm{C}}_1=3528$
${\mathit{C}}_1=\frac{3528}{21000}$
$=0.168{\mathrm{Jg}}^{-1}{}^{\circ }{\mathrm{C}}^{-1}$