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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=QW##
Heat exchange: ##Q=C\cdot \Delta T##
This is how I solved part a) :
##Q_1=C\cdot (T_1T_i)## This quantity is negative because object 1 loses heat. (positive for the machine)
##Q_2=C\cdot (T_2T_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_1Q_2##
From the first law ##\Delta U=0 ⇒ Q=W ⇒ Q_2=Q_1W=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!
##Q_1=C\cdot (T_1T_i)## This quantity is negative because object 1 loses heat. (positive for the machine)
##Q_2=C\cdot (T_2T_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_1Q_2##
From the first law ##\Delta U=0 ⇒ Q=W ⇒ Q_2=Q_1W=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!