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Sintering joins particles into a coherent solid by heating, sometimes with applied pressure, without completely melting the material. It is used to consolidate powders of ceramics and metals.

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 $L$.