Problem — Ice cubes for lemonade

Problem

Estimate, to an order of magnitude, how many ice cubes are needed to cool a litre of lemonade from $25 \ ^{\circ}\text{C}$ to $5 \ ^{\circ}\text{C}$, assuming the cubes start at $0 \ ^{\circ}\text{C}$ and melt completely. (Assume a standard cube of about $20 \ \text{g}$, $c_{\text{acqua}} = 4186 \ \text{J/(kg·K)}$ and $L_f = 334 \ \text{kJ/kg}$.)

Solution

The lemonade, treated as roughly $1 \ \text{kg}$ of water, must release the heat $Q = 1 \cdot 4186 \cdot 20 \approx 83720 \ \text{J}$ Each $20 \ \text{g}$ cube absorbs two contributions: first it melts at $0 \ ^{\circ}\text{C}$, then the melt water warms up to $5 \ ^{\circ}\text{C}$: $Q_{\text{cubetto}} = m L_f + m c_w \Delta T = 0{,}02 \cdot 334000 + 0{,}02 \cdot 4186 \cdot 5 = 6680 + 419 \approx 7100 \ \text{J}$ The number of cubes needed is therefore $N = \frac{Q}{Q_{\text{cubetto}}} = \frac{83720}{7100} \approx 11{,}8$ $\ev{N \approx 12 \ \text{cubes}}$

Collegamenti

Argomenti: Termologia Concetti: Calore latente e cambiamenti di stato Competenze: Stime di Fermi