 Mixing calorimetry

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 $C_{\mathrm{cal}}$, 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 $Q_{\mathrm{loss},\mathrm{out}}$, the combined enthalpy balance isHenceThe loss is positive if the room receives net heat, negative if the room supplies it. If the vessel starts at a different temperature $T_{\mathrm{cal},i}$, replace its warming term by $C_{\mathrm{cal}}(T_f-T_{\mathrm{cal},i})$; 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 $\int C_{\mathrm{cal}}(T)dT$. The inferred constant $c_x$ then represents an average over the sample's cooling interval.^[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 $c_w=4180\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$ and $C_{\mathrm{cal}}=100\mathrm{J}{\mathrm{K}}^{-1}$. The enthalpy balance givesThe 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.

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 $c_x=450\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$. With no loss, $T_f\approx 22.752{}^{\circ }\mathrm{C}$. Inverting the full balance returns $450\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$; ignoring vessel storage returns about $402\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$. 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.