1.0g sample of subtance 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) gtT_(B)gtT_(C)`
D. `T_(A) =T_(B)=T_(C)`
`{:(“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) gtT_(B)gtT_(C)`
D. `T_(A) =T_(B)=T_(C)`
Correct Answer – C
Specific heat is the amount of heat required to raise the temperature of 1 g of substance through `1^(@)C` or it is the heat released when temperature of 1 g of the substance falls through `1^(@)C`. Greater the specific heat of the substance, greater is th heat released which is absorbed by water and hence greater is the rise in temperature of water or greater is the temperature ofwater. As specific heats ofA, B and C are in the order `A gt B gt C`, therefore, `T_(A) gt T_(B) gt T_(C)`