Question

Two objects are placed in thermal contact and are allowed to come to equilibrium in isolation. the heat capacity of object a is three times the heat capacity of object b and the initial temperature of object a (ta) is twice the initial temperature of object b (tb). 1) what will the final temperature of the two-object system be?

Answer 1

Given:
Ca = 3Cb (1)
where
Ca = heat capacity of object A
Cb = heat capacity f object B

Also,
Ta = 2Tb (2)
where
Ta = initial temperature of object A
Tb = initial temperature of object B.

Let
Tf = final equilibrium temperature of both objects,
Ma = mass of object A,
Mb = mass of object B.

Assuming that all heat exchange occurs exclusively between the two objects, then energy balance requires that
Ma*Ca*(Ta - Tf) = Mb*Cb*(Tf - Tb) (3)

Substitute (1) and (2) into (3).
Ma*(3Cb)*(2Tb - Tf) = Mb*Cb*(Tf - Tb)
3(Ma/Mb)*(2Tb - Tf) = Tf - Tb

Define k = Ma/Mb, the ratio f the masses.
Then
3k(2Tb - Tf) = Tf - Tb
Tf(1+3k) = Tb(1+6k)
Tf = [(1+6k)/(1+3k)]*Tb

Answer:

`T_(f) =( (1+6k)/(1+3k) )T_(b)= (1)/(2)( (1+6k)/(1+3k))T_(a)`
where

`k= (M_(a))/(M_(b))`