Question
One liter of water at 58◦C is used to make iced tea. How much 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 solve for the mass of ice required to lower the temperature of the water, we can use the principle of conservation of energy, where the heat lost by the water will be equal to the heat gained by the ice as it melts and warms up to the final temperature.
1. **Calculate Heat Lost by Water:**
The heat lost by the water (Q_water) as it cools from 58°C to 13°C can be calculated using the formula:
\[
Q_{\text{water}} = m_{\text{water}} \cdot c_{\text{water}} \cdot \Delta T
\]
where:
- \(m_{\text{water}} = 1000 \, \text{g}\) (since 1 liter of water = 1000 g)
- \(c_{\text{water}} = 1 \, \text{cal/g·°C}\)
- \(\Delta T = T_{\text{initial}} - T_{\text{final}} = 58 \, \text{°C} - 13 \, \text{°C} = 45 \, \text{°C}\)
Substituting the values in:
\[
Q_{\text{water}} = 1000 \, \text{g} \cdot 1 \, \text{cal/g·°C} \cdot 45 \, \text{°C} = 45000 \, \text{cal}
\]
2. **Calculate Heat Gained by Ice:**
The heat gained by the ice (Q_ice) when it melts and warms to the final temperature (13°C) is given by two components:
- The heat required to melt the ice (Q_melt)
- The heat gained by the water from melted ice warming up to 13°C (Q_warm)
The total heat gained by the ice can be expressed as:
\[
Q_{\text{ice}} = m_{\text{ice}} \cdot L + m_{\text{ice}} \cdot c_{\text{water}} \cdot (T_{\text{final}} - T_{\text{initial, ice}})
\]
where:
- \(L = 79.7 \, \text{cal/g}\) is the latent heat of fusion for ice
- \(T_{\text{final}} = 13 \, \text{°C}\)
- \(T_{\text{initial, ice}} = 0 \, \text{°C}\)
Combining, the heat gained by the ice is:
\[
Q_{\text{ice}} = m_{\text{ice}} \cdot 79.7 \, \text{cal/g} + m_{\text{ice}} \cdot 1 \, \text{cal/g·°C} \cdot (13 \, \text{°C} - 0 \, \text{°C})
\]
Simplifying yields:
\[
Q_{\text{ice}} = m_{\text{ice}} \cdot 79.7 + m_{\text{ice}} \cdot 13 = m_{\text{ice}} \cdot (79.7 + 13) = m_{\text{ice}} \cdot 92.7
\]
3. **Setting Up the Equation:**
Setting the heat lost by the water equal to the heat gained by the ice:
\[
45000 \, \text{cal} = m_{\text{ice}} \cdot 92.7
\]
4. **Solving for Mass of Ice:**
We can solve for \(m_{\text{ice}}\):
\[
m_{\text{ice}} = \frac{45000 \, \text{cal}}{92.7 \, \text{cal/g}} = 485.53 \, \text{g}
\]
Thus, the mass of ice needed is approximately:
\[
\boxed{486 \, \text{g}}
\]
1. **Calculate Heat Lost by Water:**
The heat lost by the water (Q_water) as it cools from 58°C to 13°C can be calculated using the formula:
\[
Q_{\text{water}} = m_{\text{water}} \cdot c_{\text{water}} \cdot \Delta T
\]
where:
- \(m_{\text{water}} = 1000 \, \text{g}\) (since 1 liter of water = 1000 g)
- \(c_{\text{water}} = 1 \, \text{cal/g·°C}\)
- \(\Delta T = T_{\text{initial}} - T_{\text{final}} = 58 \, \text{°C} - 13 \, \text{°C} = 45 \, \text{°C}\)
Substituting the values in:
\[
Q_{\text{water}} = 1000 \, \text{g} \cdot 1 \, \text{cal/g·°C} \cdot 45 \, \text{°C} = 45000 \, \text{cal}
\]
2. **Calculate Heat Gained by Ice:**
The heat gained by the ice (Q_ice) when it melts and warms to the final temperature (13°C) is given by two components:
- The heat required to melt the ice (Q_melt)
- The heat gained by the water from melted ice warming up to 13°C (Q_warm)
The total heat gained by the ice can be expressed as:
\[
Q_{\text{ice}} = m_{\text{ice}} \cdot L + m_{\text{ice}} \cdot c_{\text{water}} \cdot (T_{\text{final}} - T_{\text{initial, ice}})
\]
where:
- \(L = 79.7 \, \text{cal/g}\) is the latent heat of fusion for ice
- \(T_{\text{final}} = 13 \, \text{°C}\)
- \(T_{\text{initial, ice}} = 0 \, \text{°C}\)
Combining, the heat gained by the ice is:
\[
Q_{\text{ice}} = m_{\text{ice}} \cdot 79.7 \, \text{cal/g} + m_{\text{ice}} \cdot 1 \, \text{cal/g·°C} \cdot (13 \, \text{°C} - 0 \, \text{°C})
\]
Simplifying yields:
\[
Q_{\text{ice}} = m_{\text{ice}} \cdot 79.7 + m_{\text{ice}} \cdot 13 = m_{\text{ice}} \cdot (79.7 + 13) = m_{\text{ice}} \cdot 92.7
\]
3. **Setting Up the Equation:**
Setting the heat lost by the water equal to the heat gained by the ice:
\[
45000 \, \text{cal} = m_{\text{ice}} \cdot 92.7
\]
4. **Solving for Mass of Ice:**
We can solve for \(m_{\text{ice}}\):
\[
m_{\text{ice}} = \frac{45000 \, \text{cal}}{92.7 \, \text{cal/g}} = 485.53 \, \text{g}
\]
Thus, the mass of ice needed is approximately:
\[
\boxed{486 \, \text{g}}
\]