= Mixing calorimetry
{title2=$m_xc_x(T_x-T_f)=(m_wc_w+C_{\rm cal})(T_f-T_w)+Q_{\rm loss,out}$}
Mixing calorimetry determines <heat capacities> by bringing bodies at different <temperatures> into thermal contact. The striking feature is that the final <temperature> is a <weighted arithmetic mean> with <heat capacities> as weights, rather than weights given by <mass>. The vessel participates in the exchange even though it is not the sample being studied.
For a hot nonreacting sample of <mass> $m_x$, initial <temperature> $T_x$ and unknown <specific heat capacity at constant pressure> $c_x$, let receiving <water> have <mass> $m_w$, initial <temperature> $T_w$ and known <specific heat capacity at constant pressure> $c_w$. Let the vessel and immersed parts start at $T_w$ and have combined <heat capacity> $C_{\rm cal}$, excluding that <water>. Assume a common final <temperature> $T_f$, approximately constant <pressure> and <heat capacities>, no <phase transition>, no <chemical reaction> and negligible external nonexpansion <work>. For signed outward <heat> loss $Q_{\rm loss,out}$, the combined <enthalpy> balance is
$$
m_xc_x(T_f-T_x)+(m_wc_w+C_{\rm cal})(T_f-T_w)=-Q_{\rm loss,out}.
$$
Hence
$$
c_x=\frac{(m_wc_w+C_{\rm cal})(T_f-T_w)+Q_{\rm loss,out}}{m_x(T_x-T_f)}.
$$
The loss is positive if the room receives net <heat>, negative if the room supplies it. If the vessel starts at a different <temperature> $T_{{\rm cal},i}$, replace its warming term by $C_{\rm cal}(T_f-T_{{\rm cal},i})$; do not assume the vessel shared the <water>'s initial <temperature>. For variable material properties, replace each $mc\Delta T$ by the corresponding <integral> $m\int c_p(T)\,dT$ and each vessel term by $\int C_{\rm cal}(T)\,dT$. The inferred constant $c_x$ then represents an average over the sample's cooling interval.<Chemistry 2e by Paul Flowers and others/5.2 Calorimetry>[Figure 5.14 and Example 5.4: exchange between metal and water and inference of unknown specific heat]{ref}
For a synthetic no-loss example, mix $0.100\,\mathrm{kg}$ of hot <water> at $80\,{}^\circ\mathrm C$ with $0.200\,\mathrm{kg}$ of cold <water> at $20\,{}^\circ\mathrm C$ in a vessel initially at $20\,{}^\circ\mathrm C$. Stipulate $c_w=4180\,\mathrm{J\,kg^{-1}\,K^{-1}}$ and $C_{\rm cal}=100\,\mathrm{J\,K^{-1}}$. The <enthalpy> balance gives
$$
T_f=\frac{(418)(80)+(836+100)(20)}{418+836+100}\,{}^\circ\mathrm C
\approx38.523\,{}^\circ\mathrm C.
$$
The hot <water> releases $17.337\,\mathrm{kJ}$: the cold <water> receives $15.485\,\mathrm{kJ}$ and the vessel $1.852\,\mathrm{kJ}$. Omitting the vessel predicts $40\,{}^\circ\mathrm C$, despite perfect insulation. These are computed numbers from stipulated properties, not laboratory observations.
\Image[/calorimetry-mixing-temperature-energy-balance.png]
{title=The vessel changes the mixing <temperature> by sharing the <heat> transfer}
{id=calorimetry-mixing-temperature-energy-balance}
{description=Original computed no-loss <water> example. Left: transfers sum to the hot portion's released <heat>. Right: increasing the initially cold vessel's <heat capacity> lowers the common final <temperature>; the marked point uses $C_{\rm cal}=100\,\mathrm{J\,K^{-1}}$. Each vessel setting is a separate modeled experiment.}
For an unknown-sample example using the same cold <water> and vessel, stipulate $m_x=0.100\,\mathrm{kg}$, $T_x=80\,{}^\circ\mathrm C$ and $c_x=450\,\mathrm{J\,kg^{-1}\,K^{-1}}$. With no loss, $T_f\approx22.752\,{}^\circ\mathrm C$. Inverting the full balance returns $450\,\mathrm{J\,kg^{-1}\,K^{-1}}$; ignoring vessel storage returns about $402\,\mathrm{J\,kg^{-1}\,K^{-1}}$. With a positive outward loss, ignoring that loss also biases the inferred $c_x$ downward under these hot-sample conditions. A small <water> <temperature> rise makes sensor offsets and loss estimates relatively influential; <propagation of uncertainty> must include the measured <temperatures>, <masses> and calibrated vessel term.
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