To calculate the mass of ice required to cool a fruit drink from 20°C to 5°C, we need to follow these steps:
- Calculate the energy lost by the drink when it cools down from 20°C to 5°C.
- Calculate the energy gained by the ice as it melts and then warms up to 5°C.
- Set the energy lost by the drink equal to the energy gained by the ice to find the mass of the ice.
Constants Needed
- Specific heat capacity of water (\(c_w\)): \(4.18 , \text{J/g°C}\)
- Latent heat of fusion for ice (\(L_f\)): \(334 , \text{J/g}\)
Step 1: Calculate the energy lost by the drink
- Volume of the drink: \(V = 150 , \text{cm}^3 = 150 , \text{g}\) (assuming the density of the drink is close to that of water, \(1 \text{ g/cm}^3\)).
- Temperature change: \[ \Delta T = T_{\text{initial}} - T_{\text{final}} = 20°C - 5°C = 15°C \]
- Energy lost by the drink (Q_d): \[ Q_d = m_d \cdot c_w \cdot \Delta T \] where \(m_d = 150 , \text{g}\). \[ Q_d = 150 , \text{g} \cdot 4.18 , \text{J/g°C} \cdot 15°C \] \[ Q_d = 150 \cdot 4.18 \cdot 15 = 9420 , \text{J} \]
Step 2: Calculate the energy gained by the ice
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Let \(m_i\) be the mass of the ice.
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The energy gained by the ice consists of two parts:
- The ice melts and becomes water at 0°C: \[ Q_{\text{melt}} = m_i \cdot L_f \]
- That water then warms up from 0°C to 5°C: \[ Q_{\text{heat}} = m_i \cdot c_w \cdot \Delta T = m_i \cdot 4.18 \cdot 5 \]
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Therefore, the total energy gained by the ice (Q_i): \[ Q_i = m_i \cdot L_f + m_i \cdot c_w \cdot 5 = m_i (334 + 4.18 \cdot 5) \] \[ Q_i = m_i (334 + 20.9) = m_i \cdot 354.9 \]
Step 3: Set \(Q_d = Q_i\) and solve for \(m_i\)
Setting the energy lost by the drink equal to the energy gained by the ice: \[ 9420 , \text{J} = m_i \cdot 354.9 , \text{J/g} \] \[ m_i = \frac{9420}{354.9} \] \[ m_i \approx 26.54 , \text{g} \]
Conclusion
The mass of ice required to cool 150 cm³ of the fruit drink from 20°C to 5°C is approximately 26.5 g.