= Solution
Replace the feedback resistor by a capacitor $C_f$, retaining the input resistor $R_i$. The <virtual ground> makes the input charge flow onto the feedback capacitor rather than into the ideal amplifier. Integrating its current gives
$$
V_o(t)-V_o(0)=-\frac1{C_f}\int_0^t I_i(t')\,dt',\qquad
\boxed{\frac{\Delta V_o}{Q_{\rm in}}=-\frac1{C_f}.}
$$
If the calorimeter delivers $Q_{\rm in}=\kappa\mathcal E$ for deposited energy $\mathcal E$, then $\Delta V_o=-\kappa\mathcal E/C_f$. A reset establishes the initial output; a practical large resistor across $C_f$ limits long-time drift, but is omitted from the ideal integration interval.
\Image[/past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2013-charge-integrator.png]
{title=Ideal inverting charge integrator for the calorimeter signal}
{description=The original feedback resistor is replaced by C_f; the input resistor remains for the subsequent voltage-driven integrator calculation.}
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