Two objects exchange heat through a cyclical thermal machine

[ValeForce46]

Homework Statement

Two object (thermal capacity $C=3\cdot10^3 \frac{J}{K}$) are initially at the same temperature $T_i=450 K$, and they're linked through a cyclical thermal machine.

a) One of the two object is cooled down at the temperature $T_1=300 K$ and the work done by the machine is $W=6\cdot10^4 J$. Calculate the temperature $T_2$ of the second object when the first reachs $T_1$.

b)Assume, now, that the thermal machine is reversible and the first object reachs $T_1=250 K$. How much work did the machine do?

Relevant Equations

First law of thermodynamics: $\Delta U=Q-W$
Heat exchange: $Q=C\cdot \Delta T$

This is how I solved part a) :
$Q_1=C\cdot (T_1-T_i)$ This quantity is negative because object 1 loses heat. (positive for the machine)
$Q_2=C\cdot (T_2-T_i)$ This one is positive because the object 2 absorbs heat.(negative for the machine)
Then the exchanged heat FOR THE MACHINE is $Q=-Q_1-Q_2$
From the first law $\Delta U=0 ⇒ Q=W ⇒ Q_2=-Q_1-W=4.44\cdot 10^5 J$
$T_2=\frac{Q_2}{C}+T_i=548 K$. Am I right?
For part b)... Do I have to use the relation $\frac{Q_1}{Q_2}=\frac{T_1}{T_2}$?
I don't really know... Help me!

 

[Chestermiller]

Do I have to use the relation $\frac{Q_1}{Q_2}=\frac{T_1}{T_2}$?

Yes.