Answer
See explanation.
Work Step by Step
Heat entered the balloon, and work was done by the air inside the balloon. The first effect raises the balloon’s internal energy and the second lowers it. During the expansion, the internal energy increased, because the air ended up hotter, so the heat added is greater than the work done.
$\Delta U=Q-W$, and with $\Delta U \gt 0$, $Q\gt W$.
After the balloon has returned to room temperature, the internal energy change is zero, because the system is at its initial state. The net heat equals the net work.
$\Delta U=Q-W$, and with $\Delta U=0$, $Q=W$.
