 Wet-ice bias in fusion calorimetry

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 isIf an analysis incorrectly treats all $m_a$ as solid and sets $Q_{\mathrm{env}}=0$, it obtainsThus $r=0.10$ 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.

For this model with $r=0$ and $Q_{\mathrm{env}}=0$, the balance predicts $T_f-T_m\simeq 8.829\mathrm{K}$, 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 $T_f-T_m\simeq 10.369\mathrm{K}$ and $L_{\mathrm{app}}=300.6\mathrm{kJ}\mathrm{k}{\mathrm{g}}^{-1}$. A net incoming $1\mathrm{kJ}$ shifts $L_{\mathrm{app}}$ downward by a further $20\mathrm{kJ}\mathrm{k}{\mathrm{g}}^{-1}$ 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 $L$.