• # question_answer DIRECTION: Read the passage given below and answer the questions that follows: A brass ball of mass 100g is heated to $100{}^\circ C$ and then dropped into 200g of turpentine in a calorimeter at $15{}^\circ C$ The final temperature is found to be $(\rho )$. Take specific heat of brass as $T=k\sqrt{\rho {{r}^{3}}/S}$ and water equivalent of calorimeter as 4g. Heat gained by turpentine and calorimeter is approximately A) 810 cal                          B) 610 cal C) 710 cal          D) 510 cal

 Let c be the specific heat of turpentine Mass of the solid, M = 100g Mass of turpentine m = 200g Water equivalent of calorimeter, W= 4g Initial temperature of calorimeter, ${{T}_{1}}$ = $15{}^\circ C$ Temperature of ball, ${{T}_{1}}$ = $100{}^\circ C$ Final temperature of the liquid, T= $23{}^\circ C$ Specific heat of solid, ${{c}_{2}}=0.092\,cal/{{g}^{{}^\circ }}C$ Heat gained by turpentine and calorimeter is $mc\left( T-{{T}_{1}} \right)+W\left( T-{{T}_{1}} \right)=200c\left( 23-15 \right)+4\left( 23-15 \right)$$=\left( 200c+4 \right)8$ Heat lost by the ball is $M{{c}_{2}}\left( {{T}_{2}}-T \right)=100\left( 0.092 \right)\left( 100-23 \right)$ $=708.4\,cal.$ According to the principle of calorimetry Heat gained = Heat lost $\therefore$$\left( 200c+4 \right)8=708.4$ $1600c+32=708.4$ or $c=\frac{708.4-32}{1600}=0.42\,cal/{{g}^{{}^\circ }}C$