5 g of ice at `0^(@)C` is dropped in a beaker containing 20 g of water at `40^(@)C` . The final temperature will be

Options

A. `32^(@)C`
B. `16^(@)C`
C. `8^(@)C`
D. `24^(@)C`

Answer

To solve the problem of finding the final temperature when 5 g of ice at 0°C is dropped into 20 g of water at 40°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 Values:**
- Mass of ice, \( m_{\text{ice}} = 5 \, \text{g} \)
- Initial temperature of ice, \( T_{\text{ice}} = 0 \, ^\circ C \)
- Mass of water, \( m_{\text{water}} = 20 \, \text{g} \)
- Initial temperature of water, \( T_{\text{water}} = 40 \, ^\circ C \)
- Latent heat of fusion of ice, \( L_f = 336 \, \text{J/g} \)
- Specific heat of water, \( c_{\text{water}} = 4.186 \, \text{J/g} \cdot ^\circ C \)

2. **Calculate the Heat Gained by Ice:**
The ice will first absorb heat to melt into water at 0°C, and then it will absorb more heat to increase its temperature to the final temperature \( T_f \).
- Heat gained by ice to melt:
\[ Q_{\text{melt}} = m_{\text{ice}} \times L_f = 5 \, \text{g} \times 336 \, \text{J/g} = 1680 \, \text{J} \]
- Heat gained by melted ice (now water) to reach \( T_f \):
\[ Q_{\text{heat}} = m_{\text{ice}} \times c_{\text{water}} \times (T_f - T_{\text{ice}}) = 5 \, \text{g} \times 4.186 \, \text{J/g} \cdot ^\circ C \times (T_f - 0) = 20.93 \, \text{J/}^\circ C \times T_f \]

3. **Calculate the Heat Lost by Water:**
The water will lose heat as it cools down to the final temperature \( T_f \):
\[ Q_{\text{lost}} = m_{\text{water}} \times c_{\text{water}} \times (T_{\text{water}} - T_f) = 20 \, \text{g} \times 4.186 \, \text{J/g} \cdot ^\circ C \times (40 - T_f) = 83.72 \, \text{J/}^\circ C \times (40 - T_f) \]

4. **Set Up the Energy Balance Equation:**
According to the principle of conservation of energy:
\[ Q_{\text{gained by ice}} = Q_{\text{lost by water}} \]
Therefore, we can write:
\[ 1680 + 20.93 T_f = 83.72(40 - T_f) \]

5. **Expand and Rearrange the Equation:**
\[ 1680 + 20.93 T_f = 3348.8 - 83.72 T_f \]
Rearranging gives:
\[ 20.93 T_f + 83.72 T_f = 3348.8 - 1680 \]
\[ 104.65 T_f = 1668.8 \]

6. **Solve for \( T_f \):**
\[ T_f = \frac{1668.8}{104.65} \approx 15.95 \, ^\circ C \]

### Final Answer:
The final temperature \( T_f \) is approximately \( 16 \, ^\circ C \).