Question
How do I calculate the mass of ice required to cool 150 cm³ of a fruit drink from 20°C to 5°C?
Answers
GPT-4o mini
To calculate the mass of ice required to cool a fruit drink from 20°C to 5°C, we need to follow these steps:
1. **Calculate the energy lost by the drink** when it cools down from 20°C to 5°C.
2. **Calculate the energy gained by the ice** as it melts and then warms up to 5°C.
3. 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
1. **Volume of the drink**: \(V = 150 \, \text{cm}^3 = 150 \, \text{g}\) (assuming the density of the drink is close to that of water, \(1 \text{ g/cm}^3\)).
2. **Temperature change**:
\[
\Delta T = T_{\text{initial}} - T_{\text{final}} = 20°C - 5°C = 15°C
\]
3. **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
1. Let \(m_i\) be the mass of the ice.
2. 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:
\[
Q_{\text{heat}} = m_i \cdot c_w \cdot \Delta T = m_i \cdot 4.18 \cdot 5
\]
3. 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**.
1. **Calculate the energy lost by the drink** when it cools down from 20°C to 5°C.
2. **Calculate the energy gained by the ice** as it melts and then warms up to 5°C.
3. 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
1. **Volume of the drink**: \(V = 150 \, \text{cm}^3 = 150 \, \text{g}\) (assuming the density of the drink is close to that of water, \(1 \text{ g/cm}^3\)).
2. **Temperature change**:
\[
\Delta T = T_{\text{initial}} - T_{\text{final}} = 20°C - 5°C = 15°C
\]
3. **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
1. Let \(m_i\) be the mass of the ice.
2. 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:
\[
Q_{\text{heat}} = m_i \cdot c_w \cdot \Delta T = m_i \cdot 4.18 \cdot 5
\]
3. 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**.