Question
One liter of water at 58◦C is used to make iced tea. How many grams of ice at 0 ◦C must be added to lower the temperature of the tea to 13 ◦C? The specific heat of water is 1 cal/g ·◦ C and latent heat of ice is 79.7 cal/g. Answer in units of g.
Answers
GPT-4o mini
To find the mass of ice required to lower the temperature of the tea from 58°C to 13°C, we need to apply the principle of conservation of energy, specifically that the heat lost by the water will be equal to the heat gained by the ice.
1. **Calculate the heat lost by the water**:
The heat lost by the water can be calculated using the formula:
\[
Q_{\text{lost}} = m_{\text{water}} \cdot c \cdot \Delta T_{\text{water}}
\]
Where:
- \( m_{\text{water}} = 1000 \, g \) (since 1 liter of water = 1000 grams)
- \( c = 1 \, \text{cal/g°C} \) (specific heat of water)
- \( \Delta T_{\text{water}} = T_{\text{initial}} - T_{\text{final}} = 58°C - 13°C = 45°C \)
Now, substituting the values:
\[
Q_{\text{lost}} = 1000 \, g \cdot 1 \, \text{cal/g°C} \cdot 45°C = 45000 \, \text{cal}
\]
2. **Calculate the heat gained by the ice**:
The ice will first need to melt and then warm up to 13°C. The total heat gained by the ice can be split into two parts:
- Heat gained when the ice melts to become water (\(Q_{\text{melt}}\)):
\[
Q_{\text{melt}} = m_{\text{ice}} \cdot L_f
\]
- Heat gained by the melted ice (now water) to warm up to 13°C (\(Q_{\text{warm}}\)):
\[
Q_{\text{warm}} = m_{\text{ice}} \cdot c \cdot \Delta T_{\text{ice}}
\]
Where:
- \( m_{\text{ice}} \) is the mass of the ice
- \( L_f = 79.7 \, \text{cal/g} \) (latent heat of fusion for ice)
- \( \Delta T_{\text{ice}} = 13°C - 0°C = 13°C \)
Combining these two heat quantities gives us:
\[
Q_{\text{gained}} = m_{\text{ice}} \cdot L_f + m_{\text{ice}} \cdot c \cdot \Delta T_{\text{ice}}
\]
\[
Q_{\text{gained}} = m_{\text{ice}} \cdot 79.7 \, \text{cal/g} + m_{\text{ice}} \cdot 1 \, \text{cal/g°C} \cdot 13°C
\]
\[
Q_{\text{gained}} = m_{\text{ice}} \cdot (79.7 + 13) \, \text{cal/g}
\]
\[
Q_{\text{gained}} = m_{\text{ice}} \cdot 92.7 \, \text{cal/g}
\]
3. **Set the heat lost equal to the heat gained**:
\[
45000 \, \text{cal} = m_{\text{ice}} \cdot 92.7 \, \text{cal/g}
\]
4. **Solve for \( m_{\text{ice}} \)**:
\[
m_{\text{ice}} = \frac{45000 \, \text{cal}}{92.7 \, \text{cal/g}} \approx 485.54 \, g
\]
Thus, the mass of ice needed is approximately **485.54 grams**. Rounding to a reasonable precision, we can express the final answer as:
\[
\boxed{486 \, g}
\]
1. **Calculate the heat lost by the water**:
The heat lost by the water can be calculated using the formula:
\[
Q_{\text{lost}} = m_{\text{water}} \cdot c \cdot \Delta T_{\text{water}}
\]
Where:
- \( m_{\text{water}} = 1000 \, g \) (since 1 liter of water = 1000 grams)
- \( c = 1 \, \text{cal/g°C} \) (specific heat of water)
- \( \Delta T_{\text{water}} = T_{\text{initial}} - T_{\text{final}} = 58°C - 13°C = 45°C \)
Now, substituting the values:
\[
Q_{\text{lost}} = 1000 \, g \cdot 1 \, \text{cal/g°C} \cdot 45°C = 45000 \, \text{cal}
\]
2. **Calculate the heat gained by the ice**:
The ice will first need to melt and then warm up to 13°C. The total heat gained by the ice can be split into two parts:
- Heat gained when the ice melts to become water (\(Q_{\text{melt}}\)):
\[
Q_{\text{melt}} = m_{\text{ice}} \cdot L_f
\]
- Heat gained by the melted ice (now water) to warm up to 13°C (\(Q_{\text{warm}}\)):
\[
Q_{\text{warm}} = m_{\text{ice}} \cdot c \cdot \Delta T_{\text{ice}}
\]
Where:
- \( m_{\text{ice}} \) is the mass of the ice
- \( L_f = 79.7 \, \text{cal/g} \) (latent heat of fusion for ice)
- \( \Delta T_{\text{ice}} = 13°C - 0°C = 13°C \)
Combining these two heat quantities gives us:
\[
Q_{\text{gained}} = m_{\text{ice}} \cdot L_f + m_{\text{ice}} \cdot c \cdot \Delta T_{\text{ice}}
\]
\[
Q_{\text{gained}} = m_{\text{ice}} \cdot 79.7 \, \text{cal/g} + m_{\text{ice}} \cdot 1 \, \text{cal/g°C} \cdot 13°C
\]
\[
Q_{\text{gained}} = m_{\text{ice}} \cdot (79.7 + 13) \, \text{cal/g}
\]
\[
Q_{\text{gained}} = m_{\text{ice}} \cdot 92.7 \, \text{cal/g}
\]
3. **Set the heat lost equal to the heat gained**:
\[
45000 \, \text{cal} = m_{\text{ice}} \cdot 92.7 \, \text{cal/g}
\]
4. **Solve for \( m_{\text{ice}} \)**:
\[
m_{\text{ice}} = \frac{45000 \, \text{cal}}{92.7 \, \text{cal/g}} \approx 485.54 \, g
\]
Thus, the mass of ice needed is approximately **485.54 grams**. Rounding to a reasonable precision, we can express the final answer as:
\[
\boxed{486 \, g}
\]