Three metallic spheres A,B and C have their masses in the ratio 1:2:3, specific heat capacities in the ratio 6:3:4. When the initial temperature of the spheres are measrued in Celsius scale, the ratio of their temperature is found to be 1:2:3. Initially the two spheres A and B are brought into contact. when equilibrium temperature is attained, sphere B is brought into contact with 'C'. Determine the ratio of the final temperatures of A,B and C as measured in Celsius scale.

Explanation

Let the masses of A, B and C be m, 2m and 3m the specific heat capacities be 6s, 3s and 4s, and the initial temperatures be 1, 2 and 3t.
Since the temperature of A is less than the temperature of B, heat flows from B to A when these two are brought into contact.
If `theta_(1)` is the equilibrium temperature, `tlttheta_(1)lt2t`
Heat lost by hot body = heat gained by cold body
`:.(2m)(3s)(2t-theta_(1))=(m)(6s)(theta_(1)-t)`
`6(2t-theta_(1))=6(theta_(1)-t)`
`2theta_(1)=3t`
`theta_(1)=(3t)/(2)`
(You may observe here that since `m_(1)s_(1)=m_(2)s_(2),Deltat_(1)=Deltat_(2)` and `theta_(1)` is the average of t and 2t)
Thus, `t_(A)=t_(B)=(3t)/(2)`
When B and C are brought into contact the equilibrium temperature `theta_(2)` will be greater than that of `B(=(3t)/(2))` and less than that of C (= 3t)
Once again heat lost by hot body (C) = heat gained by cold body (B)
I,e,. `(3m)(4s)(3t-theta_(2))=(2m)(3s)(theta_(2)-(3t)/(2))`
`2(3t-theta_(2))=(theta_(2)-(3t)/(2))`
`3theta_(2)=6t+(3t)/(2)`
`theta_(2)=(5t)/(2)`
i.e., `t_(B)=t_(C)=(5t)/(2)`
`:.t_(A):t_(B):t_(C)=(3t)/(2):(5t)/(2):(5t)/(2)=3:5:5`