Fermi estimate: ice in a drink

Problem

How many grams of ice at $0\ ^\circ\text{C}$ must be added to a 200 ml drink at $20\ ^\circ\text{C}$ to bring it to $5\ ^\circ\text{C}$? Consider both the melting and the subsequent warming of the meltwater.

Solution

Heat released by the drink. Treated as water, $m = 0{,}2\ \text{kg}$, $c_\text{acqua} = 4186\ \text{J/(kg·K)}$, cooling from 20 to 5 °C:

$Q_\text{ced}=m\,c_\text{acqua}\,\Delta T=0{,}2\cdot 4186\cdot(20-5)=12558\ \text{J}$

Heat absorbed by the ice. Each kilogram of ice first melts ($L_f = 334000\ \text{J/kg}$), then the meltwater warms from 0 to 5 °C:

$Q_\text{ass}=m_\text{gh}\left(L_f+c_\text{acqua}\cdot 5\right)=m_\text{gh}\left(334000+20930\right)=m_\text{gh}\cdot 354930$

Balance. Equating heat released and absorbed:

$m_\text{gh}=\frac{12558}{354930}\approx 0{,}0354\ \text{kg}$

$\ev{m_\text{gh}\approx 35\ \text{g (about two ice cubes)}}$

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Topics: Thermodynamics Concepts: Latent heat and phase changes Skills: Fermi estimates · Calorimetric equation