Problem — Ice lollies from a freezer

Problem

Estimate how many ice lollies (moulds of about $50$ g of water at $20\,^{\circ}$C brought to $-15\,^{\circ}$C) a $200$ W freezer can produce in $4$ hours, assuming an $80\%$ efficiency and neglecting the appliance’s own thermal losses. Data: $c_{\text{acqua}} = 4186$ J/(kg$\cdot$K), $c_{\text{ghiaccio}} = 2090$ J/(kg$\cdot$K), $L_f = 334$ kJ/kg.

Solution

Useful energy available. In $4$ hours the freezer delivers, at $80\%$ efficiency: $E = 0{,}80 \cdot 200 \cdot (4 \cdot 3600) = 0{,}80 \cdot 200 \cdot 14400 \approx 2{,}30 \cdot 10^{6} \ \text{J}$

Heat to remove from a single ice lolly ($m = 0{,}05$ kg), in three stages: $\begin{aligned} Q_1 &= m \, c_{\text{acqua}} \, \Delta T = 0{,}05 \cdot 4186 \cdot 20 = 4186 \ \text{J} \quad (20 \to 0\,^{\circ}\text{C}) \\ Q_2 &= m \, L_f = 0{,}05 \cdot 334000 = 16700 \ \text{J} \quad (\text{solidification}) \\ Q_3 &= m \, c_{\text{ghiaccio}} \, \Delta T = 0{,}05 \cdot 2090 \cdot 15 \approx 1568 \ \text{J} \quad (0 \to -15\,^{\circ}\text{C}) \end{aligned}$

Total per ice lolly: $Q_{\text{tot}} = Q_1 + Q_2 + Q_3 \approx 4186 + 16700 + 1568 \approx 22450 \ \text{J}$

Number of ice lollies: $N = \frac{E}{Q_{\text{tot}}} = \frac{2{,}30 \cdot 10^{6}}{22450} \approx 102$

$\ev{N \approx 100 \ \text{ice lollies}}$

Collegamenti

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