DIRECTION: Read the passage given below and answer the questions that follows:$Abrassballofmass100gisheated\to 100{\{}^{\circ }C\text{and}thendropped\int o200gofturpent\in e\in acal\text{or}imeterat15{\{}^{\circ }CThef\in altemperatureisfound\to be\left(\rho \right).Takespec\text{if}icheatofbrassasT=k\sqrt{\rho \frac{{\left\{r\right\}}^3}{S}}\text{and}water\equiv a\le ntofcal\text{or}imeteras4g.$Heat gained by turpentine and calorimeter is approximately

   $Letcbethespec\text{if}icheatofturpent\in e$Mass of the solid, M = 100g$Massofturpent\in em=200g$Water equivalent of calorimeter, W= 4g$Initialtemperatureofcal\text{or}imeter,\left\{{\left\{T\right\}}_1\right\}=15{\{}^{\circ }C$Temperature of ball, {{T}_{1}} = 100{}^\circ C$F\in altemperatureoftheliquid,T=23{\{}^{\circ }C$Specific heat of solid, {{c}_{2}}=0.092\,cal/{{g}^{{}^\circ }}C$Heatga\in edbyturpent\in e\text{and}cal\text{or}imeteris$mc\left( T-{{T}_{1}} \right)+W\left( T-{{T}_{1}} \right)=200c\left( 23-15 \right)+4\left( 23-15 \right)=\left( 200c+4 \right)8$Heatlostbytheballis$M{{c}_{2}}\left( {{T}_{2}}-T \right)=100\left( 0.092 \right)\left( 100-23 \right)$=708.4,cal.$According to the principle of calorimetry$Heatga\in ed=Heatlost$\therefore \left( 200c+4 \right)8=708.4$1600c+32=708.4$or c=\frac{708.4-32}{1600}=0.42\,cal/{{g}^{{}^\circ }}C