Liquid A, B and C are at temperature `15^(@)C, 25^(@)C and 35^(@)C` respectively. When equal masses of A and B are mixed, the temperature becomes `21^(@)C`. Temperature of the mixture of equal masses of B and C becomes `32^(@)C`. (i) Show that the ratio of the specific heats of A and C is `2:7`. (ii) What will be the final temperature of the mixture of equal masses of A and C?

Explanation

Let the specific heats of A, B and C be `s_(A), s_(B)` and `s_(C)` respectively.
When liquid A of mass m and liquid B of mass m are mixed, heat lost by B = heat gained by A
`"i.e., "mxxs_(B)xx(25-21)=mxxs_(A)xx(21-15)`
`"or, "s_(A)6=s_(B).4`
`"or, "(s_(A))/(s_(B))" ...(1)"`
When liquid B of mass m and liquid C of mass m are mixed, heat lost by C = heat gained by B.
`"i.e., "mxxs_(C)xx(35-32)=mxxs_(B)xx(32-25)`
`"or, "s_(C).3=s_(B).7`
When liquid B of mass m and liquid C of mass m are mixed, heat lost by C = heat gained by B.
`"i.e., "mxxs_(C)xx(35-32)=mxxs_(B)xx(32-25)`
`"or, "s_(C).3=s_(B).7`
`"or, "(s_(B))/(s_(C))=(3)/(7)" ...(2)"`
From (1) and (2), we get
`(s_(A))/(s_(B))xx(s_(B))/(s_(C))=(2)/(3)xx(3)/(7) or, (s_(A))/(s_(C))=(2)/(7)`
`therefore" "s_(A):s_(C)=2:7`
(ii) Let the final temperaure of the mixture of liquid A of mass m and liquid C of mass m be `t^(@)C`.
So, heat gained by A = heat lost by C
`therefore" "mxxs_(A)xx(t-15)=mxxs_(C)xx(35-t)" or, "t=30.56^(@)C`.
`therefore` Temperature of the mixture is `30.56^(@)C`.