When 100 g of boiling water at  ${\text{100}}^{0}C$  is added into a calorimeter containing 300 g of cold water at ${10}^{0}C$ , temperature of the mixture becomes  ${20}^{0}C$. Then a metallic blocker mass 1 kg at ${10}^{0}C$  is dipped into the mixture in the calorimeter. After reaching thermal equilibrium, the final temperature becomes  ${19}^{0}C$. What is the specific heat of the metal in C.G.S. unit?

 $m_{water\left(boiling\right)}=100g,c_{water}=1cal{g^{-1}}^0{C}^{-1},T_1={100}^{0}C$
 $m_{water\left(cold\right)}=300g,,T_2={10}^{0}C,T_{mix}={20}^{0}C$
Let mass of calorimeter is m and specific heat of material of calorimeter is c.
Heat lost by hot water = Heat gained by cold water + Heat gained by calorimeter
 $100\times 1\times (100-20)=300\times 1\times (20-10)+mc(20-10)$
 $8000=3000+10mc\Rightarrow mc=\mathrm{500.........}\left(i\right)$
Now mass of metal = 1 kg = 1000 g
Let $c_m$   be the specific heat of meta l.
The total mass of water in calorimeter is 400 g.
Heat lost by water + calorimeter = heat gained by metal 
 $400\times 1\times (20-19)+mc(20-19)=1000\times c_m\times (19-10)$
 $400+500=1000\times c_m\times 19\left[from\left(i\right)\right]$
 $c_m=0.1ca\mathrm{l}{g^{-1}}^0{C}^{-1}$