A `1.0g ` sample of substance `A` at `100^(@)C` is added to `100 mL` of `H_(2)O` at `25^(@)C`. Using separate `100 mL` portions of `H_(2)O`, the procedure is repeated with substance `B` and then with substance `C`. How will the final temperatures of the water compare ? `{:("Substance","Specific heat"),(A,0.60 J g^(-1) "^(@)C^(-1)),(B,0.40 J g^(-1) "^(@)C^(-1)),(C,0.20 J g^(-1) "^(@)C^(-1)):}`A. `T_(C ) gt T_(B) gt T_(A)`B. `T_(B ) gt T_(A) gt T_(C )`C. `T_(A ) gt T_(B) gt T_(C )`D. `T_(A ) = T_(B) = T_(C )`

A `1.0g ` sample of substance `A` at `100^(@)C` is added to `100 mL` o

1.	A `1.0g ` sample of substance `A` at `100^(@)C` is added to `100 mL` of `H_(2)O` at `25^(@)C`. Using separate `100 mL` portions of `H_(2)O`, the procedure is repeated with substance `B` and then with substance `C`. How will the final temperatures of the water compare ? `{:("Substance","Specific heat"),(A,0.60 J g^(-1) "^(@)C^(-1)),(B,0.40 J g^(-1) "^(@)C^(-1)),(C,0.20 J g^(-1) "^(@)C^(-1)):}`A. `T_(C ) gt T_(B) gt T_(A)`B. `T_(B ) gt T_(A) gt T_(C )`C. `T_(A ) gt T_(B) gt T_(C )`D. `T_(A ) = T_(B) = T_(C )`
Answer» Correct Answer - C Heat evolved or absorbed, `Q=m s Deltat` where `m=` mass of the substance, `s=` specific heat, `Deltat=` temperative difference. Therefore more the specific heat more is the heat content and hence more the rise in temperature. Thus correct order of final temperature `T_(A) gt T_(B) gt T_(C )`.

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