Which thermodynamics variable is affected by an ice bath?

The problem:

A container of gas with a movable piston, initially at room temperature, is placed into an ice bath.  

Which variable in the first law of thermodynamics must be affected?

(A) Delta U

(B)  Q

(C)  W

(D) Delta U, Q, and W

(E)  None of them

A sample response from a reader:

If the volume of the container is kept constant then W = 0, so Q and Delta U must be affected.  If the temperature goes down then so must Delta U, but since W=0, then heat must be lost. I recognize the Delta U does not directly determine Q, that is hot does not equal heat added, but in this case I cant help but feel both are affected.

The correct answer, though, is (B):  Only Q, the heat added to the gas, must be affected by placing the gas into an ice bath.  So where is the mistake?

Look at the very first sentence, which contains a humongous "if":  IF the volume of the container is kept constant, then sure, I agree with the reasoning above.  But in the immortal words of Tia Carrere,* IF a frog had wings it wouldn't bump its arse when it hopped.

* What?  You don't recognize the reference?  Philistine, I sentence thee to watch Wayne's World, the 1992 film, 50 times.

Putting a gas in contact with an ice bath, hot plate, candle, or other means of heat transfer does just that -- transfers heat.  And heat transfer says nothing definite about temperature.

But, how can I remove heat from a gas without lowering the gas's temperature?  By doing work on the gas, of course... say the piston is compressed while the gas is in contact with the ice bath.  Say the ice bath removes 100 J of heat from the gas, but in compressing the piston I do 150 J of work on the gas.  Then, by the first law of thermodynamics, the internal energy of the gas INCREASES by 50 J.  Since internal energy alone relates to a gas's temperature, the gas temperature increases.

In this problem, the question is very clear that the container's volume can change.  The assumption of W=0 is unwarranted; W relates to the change in the container's volume, so is not necessarily zero here.  I love the way this problem stabs at the heart of the most common misconception in thermodynamics: the conflation of heat with temperature.