question_answer

A 50g lead bullet, sp. Heat 0.02 is initially at \[{{30}^{o}}C\]. It is fired vertically upwards with a speed of 840 m/s and on returning to the starting level strikes a cake of ice at \[{{0}^{o}}C\]. How much ice is melted.

Answer:

                Here,     \[m=50g=50\times {{10}^{-3}}kg\].                         \[c=0.02\], \[{{\theta }_{1}}={{30}^{o}}C\]                         \[\upsilon =840\,m/s\], \[{{\theta }_{2}}={{0}^{o}}C\], \[M=?\] \[L=80\,cal./g\] In going up, KE is converted into P.E. and in coming down, P.E. is reconverted into same amount of K.E. \[\therefore \]   Energy converted into heat = \[\frac{1}{2}\,m{{\upsilon }^{2}}\,joule\,=\frac{1}{2}\frac{m{{\upsilon }^{2}}}{J}cal\]. Heat lost by bullet to come to \[{{0}^{O}}C=c\,m\,\Delta \theta =0.02\times 50\times \left( 30-0 \right)=30cals\]. Heat spent in melting M gram of ice\[=M\times L\] Acc. to principle of energy conservation, \[\frac{m{{\upsilon }^{2}}}{2J}+30=M\times L\] \[\frac{50\times {{10}^{-3}}{{\left( 840 \right)}^{2}}}{2\times 4.2}+30=M\times 80\] \[4200\div 30=80M\] \[\therefore \]   \[M=\frac{4230}{80}=52.88\,gram\]