= Wet-ice bias in fusion calorimetry
{title2=$L_{\rm app}=(1-r)L-Q_{\rm env}/m_a$}
Surface <water> counted as <ice> needs warming but has already undergone <melting>. Its presence therefore biases a dry-ice estimate of <latent heat of fusion> downward. This is why drying the added <ice> is a quantitative control in <ice-melting calorimetry>, rather than cosmetic preparation. Let $m_a$ be the total added <mass> at $T_m$, of which a fraction $r$ is already <liquid> and $(1-r)$ remains <solid>. The correct constant-property balance is
$$
(m_wc_w+C_{\rm cal})(T_i-T_f)+Q_{\rm env}=(1-r)m_aL+m_ac_w(T_f-T_m).
$$
If an analysis incorrectly treats all $m_a$ as <solid> and sets $Q_{\rm env}=0$, it obtains
$$
L_{\rm app}=\frac{(m_wc_w+C_{\rm cal})(T_i-T_f)}{m_a}-c_w(T_f-T_m)=(1-r)L-\frac{Q_{\rm env}}{m_a}.
$$
Thus $r=0.10$ gives a $10\%$ low estimate with no heat leak. Unaccounted incoming <heat> also lowers the estimate; outgoing <heat> raises it. These signs follow directly from the displayed <energy> balance. Conversely, dry <ice> initially below $T_m$ absorbs additional <sensible heat>; neglecting that term overestimates $L$. These are different biases and may partially cancel, so agreement with a reference value alone does not validate the controls.
\Image[/fusion-latent-heat-calorimetry-bias.png]
{title=Count only the <solid> addition as fusion <mass>, and account for environmental <heat>}
{description=Original mixing schematic and analytic bias curves, not measured data. Stipulated model: $m_w=0.200\,\mathrm{kg}$, $m_a=0.050\,\mathrm{kg}$, $T_i-T_m=30\,\mathrm K$, $C_{\rm cal}=40\,\mathrm{J\,K^{-1}}$, $c_w=4180\,\mathrm{J\,kg^{-1}\,K^{-1}}$ and $L=334000\,\mathrm{J\,kg^{-1}}$. The leak curves use fixed total $Q_{\rm env}=+1000,0,-1000\,\mathrm J$ as separate hypothetical cases, not a fitted heat-transfer law.}
For this model with $r=0$ and $Q_{\rm env}=0$, the balance predicts $T_f-T_m\simeq8.829\,\mathrm K$, with $16.700\,\mathrm{kJ}$ consumed by fusion and $1.845\,\mathrm{kJ}$ by warming the melted <water>. At $r=0.10$ with the same zero leak, it predicts $T_f-T_m\simeq10.369\,\mathrm K$ and $L_{\rm app}=300.6\,\mathrm{kJ\,kg^{-1}}$. A net incoming $1\,\mathrm{kJ}$ shifts $L_{\rm app}$ downward by a further $20\,\mathrm{kJ\,kg^{-1}}$ because $m_a=0.050\,\mathrm{kg}$. The warmer final <water> in the wet case is evidence of a smaller fusion budget, not a larger true $L$.
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