A 10 g ice cube is dropped into 45 g of water kept in a glass. If the water was initially at a temperature of `28^(@)C` and the temperature of ice`-15^(@)C`, find the final temperature (in `^(@)C`) of water.
(Specific heat of ice `=0.5" "cal//g-""^(@)C" and "L=80" "cal//g)`

Options

A. 14
B. 7
C. 28
D. None

Answer

To solve the problem, we need to find the final temperature of the water after a 10 g ice cube at -15°C is added to 45 g of water at 28°C. We will consider the heat gained by the ice and the heat lost by the water to find the equilibrium temperature.

### Step-by-Step Solution:

1. **Identify the heat gained by the ice:**
- The ice first warms up from -15°C to 0°C.
- The heat gained by the ice (Q1) can be calculated using the formula:
\[ Q_1 = m_{\text{ice}} \cdot c_{\text{ice}} \cdot \Delta T \]
- Where:
- \( m_{\text{ice}} = 10 \, \text{g} \)
- \( c_{\text{ice}} = 0.5 \, \text{cal/g°C} \)
- \( \Delta T = 0 - (-15) = 15 \, \text{°C} \)
- Substituting the values:
\[ Q_1 = 10 \cdot 0.5 \cdot 15 = 75 \, \text{cal} \]

2. **Identify the heat required to melt the ice:**
- The heat required to melt the ice (Q2) is given by:
\[ Q_2 = m_{\text{ice}} \cdot L \]
- Where:
- \( L = 80 \, \text{cal/g} \)
- Substituting the values:
\[ Q_2 = 10 \cdot 80 = 800 \, \text{cal} \]

3. **Identify the heat gained by the melted ice (now water at 0°C):**
- After melting, the water from the ice will warm up to the final temperature \( \theta \).
- The heat gained by the melted ice (Q3) can be calculated as:
\[ Q_3 = m_{\text{ice}} \cdot c_{\text{water}} \cdot ( \theta - 0 ) \]
- Where:
- \( c_{\text{water}} = 1 \, \text{cal/g°C} \)
- Substituting the values:
\[ Q_3 = 10 \cdot 1 \cdot \theta = 10\theta \, \text{cal} \]

4. **Identify the heat lost by the warm water:**
- The warm water will cool down from 28°C to the final temperature \( \theta \).
- The heat lost by the warm water (Q4) can be calculated as:
\[ Q_4 = m_{\text{water}} \cdot c_{\text{water}} \cdot (28 - \theta) \]
- Where:
- \( m_{\text{water}} = 45 \, \text{g} \)
- Substituting the values:
\[ Q_4 = 45 \cdot 1 \cdot (28 - \theta) = 45(28 - \theta) \, \text{cal} \]

5. **Set up the heat balance equation:**
- At thermal equilibrium, the heat gained by the ice and the melted ice equals the heat lost by the warm water:
\[ Q_1 + Q_2 + Q_3 = Q_4 \]
- Substituting the expressions we found:
\[ 75 + 800 + 10\theta = 45(28 - \theta) \]

6. **Simplify and solve for \( \theta \):**
- Combine terms:
\[ 875 + 10\theta = 1260 - 45\theta \]
- Rearranging gives:
\[ 10\theta + 45\theta = 1260 - 875 \]
\[ 55\theta = 385 \]
\[ \theta = \frac{385}{55} = 7 \, \text{°C} \]

### Final Answer:
The final temperature of the water mixture is **7°C**.