An 18-g ice cube (at `0^(@)C`) is dropped into a glass containing 200 g of water at `25^(@)C` . If there is negligible heat exchange with the glass, what is the temperature after the ice melts?

Answer

To solve the problem, we will use the principle of conservation of energy, which states that 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, \( m_{ice} = 18 \, g \)
- Initial temperature of ice, \( T_{ice} = 0 \, °C \)
- Mass of water, \( m_{water} = 200 \, g \)
- Initial temperature of water, \( T_{water} = 25 \, °C \)
- Latent heat of fusion of ice, \( L = 80 \, cal/g \)
- Specific heat of water, \( c = 1 \, cal/g°C \)

2. **Calculate the Heat Gained by the Ice:**
- The ice first melts and then its temperature increases.
- Heat gained by the ice when it melts:
\[ Q_{gain} = m_{ice} \cdot L = 18 \, g \cdot 80 \, cal/g = 1440 \, cal \]
- After melting, the temperature of the water formed from the ice will increase from \( 0 \, °C \) to \( T \):
\[ Q_{gain, temp} = m_{ice} \cdot c \cdot (T - T_{ice}) = 18 \, g \cdot 1 \, cal/g°C \cdot (T - 0) = 18T \, cal \]
- Total heat gained by the ice:
\[ Q_{gain, total} = 1440 + 18T \]

3. **Calculate the Heat Lost by the Water:**
- The water cools down from \( 25 \, °C \) to \( T \):
\[ Q_{loss} = m_{water} \cdot c \cdot (T_{water} - T) = 200 \, g \cdot 1 \, cal/g°C \cdot (25 - T) = 200(25 - T) \, cal \]

4. **Set Heat Gained Equal to Heat Lost:**
\[ Q_{gain, total} = Q_{loss} \]
\[ 1440 + 18T = 200(25 - T) \]

5. **Solve for \( T \):**
- Expand the right side:
\[ 1440 + 18T = 5000 - 200T \]
- Rearranging gives:
\[ 18T + 200T = 5000 - 1440 \]
\[ 218T = 3560 \]
\[ T = \frac{3560}{218} \approx 16.33 \, °C \]

### Final Answer:
The temperature after the ice melts is approximately \( 16.33 \, °C \).

---