Per calcular la quantitat de gel necessària, utilitzem les equacions següents:
Energia a dissipar per la beguda:
\[ Q_{\text{beguda}} = m_{\text{beguda}} \cdot c_{\text{beguda}} \cdot \Delta T \] \[ Q_{\text{beguda}} = 0.33 \cdot 4186 \cdot (25 - 5) = 27,628.8 \, \text{J} \]Energia absorbida pel gel:
\[ Q_{\text{gel}} = m_{\text{gel}} \cdot L_f + m_{\text{gel}} \cdot c_{\text{aigua}} \cdot 5 \] Substituint: \[ Q_{\text{gel}} = m_{\text{gel}} \cdot (334,000 + 4186 \cdot 5) \]Equilibri tèrmic:
\[ Q_{\text{beguda}} = Q_{\text{gel}} \] Resolent per \(m_{\text{gel}}\): \[ m_{\text{gel}} = \frac{27,628.8}{334,000 + 20,930} = 0.074 \, \text{kg} = 74 \, \text{g} \]La capacitat calorífica de l'alcohol és menor que la de l'aigua. Suposem una capacitat calorífica mitjana ponderada:
\[ c_{\text{mixt}} = x_{\text{aigua}} c_{\text{aigua}} + x_{\text{alcohol}} c_{\text{alcohol}} \] On: - \(x_{\text{aigua}} = 0.55\), \(x_{\text{alcohol}} = 0.45\) - \(c_{\text{aigua}} = 4186\,\text{J/kg·°C}\), \(c_{\text{alcohol}} = 2400\,\text{J/kg·°C}\) Calculant: \[ c_{\text{mixt}} = (0.55)(4186) + (0.45)(2400) = 3402\,\text{J/kg·°C} \] Energia a dissipar: \[ Q_{\text{beguda}} = m_{\text{beguda}} \cdot c_{\text{mixt}} \cdot (25 - 5) \] Substituint: \[ Q_{\text{beguda}} = 0.33 \cdot 3402 \cdot 20 = 22,452\,\text{J} \] Energia absorbida pel gel: Igual que abans: \[ Q_{\text{gel}} = m_{\text{gel}} (334,000 + 20,930) \] Equilibri tèrmic: Resolent per \(m_{\text{gel}}\): \[ m_{\text{gel}} = \frac{22,452}{334,000 + 20,930} = 0.061\,\text{kg} = 61\,\text{g} \]