= Calorimeter calibration
{title2=$C_{\rm eff}=m_wc_w+C_{\rm cal}$}
Calorimeter calibration establishes the response of the actual measurement assembly to a known <energy> input. It determines a participating <heat capacity>, not just the empty vessel's <mass>. For an electrical calibration with an immersed <resistor>, measure <voltage> $V(t)$ across that <resistor> and <electric current> $I(t)$ through it. If that electrical <work> is deposited inside the chosen boundary, its input is
$$
E_{\rm el}=\int_{t_0}^{t_1}V(t)I(t)\,dt
=C_{\rm eff}(T_1-T_0)+Q_{\rm loss,out},\qquad
C_{\rm eff}=m_wc_w+C_{\rm cal}.
$$
For a covered receiver at constant <pressure>, $C_{\rm eff}$ describes <enthalpy> storage; for a rigid fixed-<volume (thermodynamics)> assembly, use the matching <heat capacity at constant volume>. Negligible endpoint mechanical storage and no escaping material are assumed. Thus $E_{\rm el}/(T_1-T_0)$ overestimates $C_{\rm eff}$ if outward loss is ignored. Account for lead losses outside the boundary and maintain the same <water> charge and immersed hardware in calibration and sample runs.
An alternative calibration mixes known hot and cold <water> portions. With the vessel initially at the cold portion's <temperature> $T_c$, the <mixing calorimetry> balance gives
$$
C_{\rm cal}=\frac{m_hc_w(T_h-T_f)-Q_{\rm loss,out}}{T_f-T_c}-m_cc_w.
$$
The previous synthetic <water> example returns $100\,\mathrm{J\,K^{-1}}$. The sign matters: attributing room losses to the vessel makes this estimate too large. With both $C_{\rm cal}$ and the loss unknown, one final <temperature> supplies only one equation; separate cooling information or controlled independent runs are needed.
A <calorimeter heat-loss correction> can make the electrical balance quantitative. For a second synthetic example, stipulate $0.100\,\mathrm{kg}$ of <water> with $c_w=4180\,\mathrm{J\,kg^{-1}\,K^{-1}}$ and immersed hardware with $C_{\rm cal}=82\,\mathrm{J\,K^{-1}}$, so $C_{\rm eff}=500\,\mathrm{J\,K^{-1}}$. Supply $10\,\mathrm W$ for $300\,\mathrm s$, starting at a room <temperature> of $20\,{}^\circ\mathrm C$, with loss coefficient $G=1\,\mathrm{W\,K^{-1}}$. The model below gives $T_1\approx24.512\,{}^\circ\mathrm C$, storage $2255.942\,\mathrm J$ and outward loss $744.058\,\mathrm J$. The uncorrected ratio gives about $665\,\mathrm{J\,K^{-1}}$; subtracting the loss recovers $500\,\mathrm{J\,K^{-1}}$.
\Image[/calorimeter-electrical-calibration-cooling-balance.png]
{title=Electrical input splits into thermal storage and room losses during calibration}
{id=calorimeter-electrical-calibration-cooling-balance}
{description=Original synthetic computed traces for the stated constant-<heat capacity>, constant-loss model. The heater switches off at $300\,\mathrm s$; stored <energy> then declines while cumulative outward <heat> increases. The no-loss comparison reaches $26\,{}^\circ\mathrm C$. No points represent experimental measurements.}
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