Thermodynamics - Carnot Cycle-esque question?

[ally2106]

Homework Statement

A reversible heat engine produces work from the temperature difference that exists between a mass of m = 9 kg of an ideal gas (c_v = 716 J/kgK, R = 287 J/kgK) in a rigid container and a heat reservoir at T_HR = 285 K. The only heat transfer interaction experienced by the container is with the heat engine. The gas in the container is initially at temperature T₁ = 772 K and pressure P₁ = 106 Pa. The reversible heat engine operates until the ideal gas and the heat reservoir are in thermal equilibirum (state 2). If the efficiency of the heat engine for process 1-2 can be defined as W/(Q_H)HE, calculate its value.Give your answer as a percent (without the % sign).

Homework Equations

• efficiency = (Q_H - Q_L)/Q_H = W/Q_H
• Q_L = MC_vT_HRln(T₂/T₁)
• E₂ - E₁ = Q_1-2 - W_1-2
• S₂ - S₁ = Q_L/ T_HR - change in entropy from or to a heat reservoir
• Q_H = MC_v(T2 - T1)

The Attempt at a Solution

If efficiency = (Q_H - Q_L)/Q_H[/B]
= 1 - (Q_L/Q_H)
= 1 - ((T_HRln(T₂/T₁))/(T₂-T₁))

I get an answer of 41.683538% but unsure as to what value to use for T₂? Would I just use T_HR (Temp of heat reservoir) seeing as it works at thermal equilibrium? I used the temperature of the reservoir as the T₂ value in my solution but not sure if this is correct?

 

[rude man]

Somewhat amazingly I got the same answer! (I mean the final expression, not the % number. I don't do numbers!). Vote of confidence, anyway!
How can T₂ be anything but the reservoir temperature T_HR? The reservoir has infinite heat capacity so at the end (i.e. "state 2") everything has to reach T = T_HR.
So far as I can see you got everything right. Congrats!