How many grams of ice at `0^(@)C` should be mixed with 240 g of water at `40^(@)C` so that the ice completely melts and the final temperature is of `0^(@)C`?

Options

A. 120 g
B. 240 g
C. 360 g
D. 480 g

Answer

To solve the problem of how many grams of ice at \(0^\circ C\) should be mixed with 240 g of water at \(40^\circ C\) so that the ice completely melts and the final temperature is \(0^\circ C\), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Heat Transfer**:
- The heat lost by the warm water will be equal to the heat gained by the ice.
- The warm water cools down from \(40^\circ C\) to \(0^\circ C\), and the ice at \(0^\circ C\) absorbs heat to melt into water at \(0^\circ C\).

2. **Write the Heat Loss Equation**:
- The heat lost by the water can be calculated using the formula:
\[ Q_{\text{water}} = m_{\text{water}} \cdot c_{\text{water}} \cdot \Delta T \]
- Where:
- \(m_{\text{water}} = 240 \, \text{g}\) (mass of water)
- \(c_{\text{water}} = 1 \, \text{cal/g}^\circ C\) (specific heat of water)
- \(\Delta T = 40^\circ C - 0^\circ C = 40^\circ C\) (change in temperature)

3. **Calculate the Heat Lost by Water**:
\[ Q_{\text{water}} = 240 \, \text{g} \cdot 1 \, \text{cal/g}^\circ C \cdot 40^\circ C = 9600 \, \text{cal} \]

4. **Write the Heat Gain Equation**:
- The heat gained by the ice to melt can be calculated using the formula:
\[ Q_{\text{ice}} = m_{\text{ice}} \cdot L_f \]
- Where:
- \(m_{\text{ice}}\) is the mass of the ice (which we need to find)
- \(L_f = 80 \, \text{cal/g}\) (latent heat of fusion of ice)

5. **Set the Heat Lost Equal to Heat Gained**:
\[ Q_{\text{water}} = Q_{\text{ice}} \]
\[ 9600 \, \text{cal} = m_{\text{ice}} \cdot 80 \, \text{cal/g} \]

6. **Solve for the Mass of Ice**:
\[ m_{\text{ice}} = \frac{9600 \, \text{cal}}{80 \, \text{cal/g}} = 120 \, \text{g} \]

### Final Answer:
The mass of ice required is **120 grams**.

---