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 of finding the final temperature when a 10 g ice cube at -15°C is dropped into 45 g of water at 28°C, we will use the principle of conservation of energy. The heat lost by the water will be equal to the heat gained by the ice.

### Step-by-Step Solution:

1. **Identify the given data:**
- Mass of ice, \( m_i = 10 \, \text{g} \)
- Initial temperature of ice, \( T_i = -15 \, ^\circ C \)
- Mass of water, \( m_w = 45 \, \text{g} \)
- Initial temperature of water, \( T_w = 28 \, ^\circ C \)
- Specific heat of ice, \( c_i = 0.5 \, \text{cal/g} \cdot ^\circ C \)
- Latent heat of fusion of ice, \( L = 80 \, \text{cal/g} \)
- Specific heat of water, \( c_w = 1 \, \text{cal/g} \cdot ^\circ C \)

2. **Calculate the heat required to raise the temperature of ice to 0°C:**
\[ Q_1 = m_i \cdot c_i \cdot (0 - T_i) = 10 \cdot 0.5 \cdot (0 - (-15)) = 10 \cdot 0.5 \cdot 15 = 75 \, \text{cal} \]

3. **Calculate the heat required to melt the ice at 0°C:**
\[ Q_2 = m_i \cdot L = 10 \cdot 80 = 800 \, \text{cal} \]

4. **Total heat absorbed by the ice:**
\[ Q_{\text{ice}} = Q_1 + Q_2 = 75 + 800 = 875 \, \text{cal} \]

5. **Calculate the heat lost by the water when it cools down to the final temperature \( T_f \):**
\[ Q_{\text{water}} = m_w \cdot c_w \cdot (T_w - T_f) = 45 \cdot 1 \cdot (28 - T_f) = 45(28 - T_f) \]

6. **Set the heat gained by the ice equal to the heat lost by the water:**
\[ Q_{\text{ice}} = Q_{\text{water}} \]
\[ 875 = 45(28 - T_f) \]

7. **Solve for \( T_f \):**
\[ 875 = 1260 - 45T_f \]
\[ 45T_f = 1260 - 875 \]
\[ 45T_f = 385 \]
\[ T_f = \frac{385}{45} \approx 8.56 \, ^\circ C \]

### Final Answer:
The final temperature \( T_f \) of the water is approximately \( 8.56 \, ^\circ C \).