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)):}`

Options

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

C

Explanation

`q=msDeltaT`
` s prop(1)/(DeltaT)`
Higher is the temperature of given solution, lesser is the temperature difference, so higher is the specific heat. Order of specific heat is `A gt B gt C`. Hence order of temperature will be:
`T_(A) gt T_(B) gt T_(C)`.