- #1

ValeForce46

- 40

- 3

- 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!

##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!