# Steady State For An Air Conditioner

Here's the problem I'm trying to solve (from Blundell's Concepts In Thermal Physics):

I just don't understand what "steady state" exactly means here. Does it mean that the temperature of the house remains steady? If so, I've solved the problem. But I'm not sure about it. Perhaps it has to do something with the air conditioner's power? Please help me.

Well I was sure that I solved it, but now I doubt it.

Here's what I've done:

$Q_{1},Q_{2},Q,E>0 \\$

For the air conditionar:

$Q_{2}-Q{1}+E=0\Rightarrow \left\{\begin{matrix} E=Q_{1}-Q_{2}~(i)\\ Q_{1}=E+Q_{2}~(ii) \end{matrix}\right. \\$

For the house at steady state:

$Q-Q_{2}=0\Rightarrow Q=Q_{2}~(iii) \\$

From $(i)$:$\\$

$\eta =\left | \frac{-Q_{1}}{-E} \right |=\frac{Q_{1}}{Q_{1}-Q_{2}}=\frac{T_{1}}{T_{1}-T_{2}}~(I)\\$

From $(ii)~$and$~(iii)$:$\\$

$\eta =\left | \frac{-Q_{1}}{-E} \right |=\frac{Q_{2}+E}{E}=\frac{Q+E}{E}=\frac{A[T_{1}-T_{2}]+E}{E}~(II)\\$

From $(I)~$and$~(II)$:$\\$

$T_{2}=\frac{2AT_{1}+E+\sqrt{(2AT_{1}+E)^{2}-A^{2}T_{1}^{2}}}{2A}\\$

The parts that I doubt are the underlined equation and the final answer. Is -E really the work done by the system? I mean should I assume all the electrical energy consumed by the gas as work done on it? (I am using this form of the first law: Change in the internal energy = Q - W) And there are two answers (roots) for T2, which one should I choose?

This question has two parts. I don't understand the second part at all. Could somebody explain to me please? What is the roll of temperature of the thermostat here? Is the temperature of the house the same as the temperature of the thermostat? Really confused.