1gm of ice at `0^(@)C` is mixed with 1 gm of water at `100^(@)C` the resulting temperature will be

Options

A. `5^(@)C`
B. `0^(@)C`
C. `10^(@)C`
D. `oo`

Answer

To solve the problem of finding the resulting temperature when 1 gm of ice at 0°C is mixed with 1 gm of water at 100°C, we can follow these steps:

### Step 1: Understand the heat exchange
When the ice melts and the warm water cools down, the heat lost by the warm water will equal the heat gained by the melting ice. This is based on the principle of conservation of energy.

### Step 2: Identify the heat gained by the ice
The heat gained by the ice consists of two parts:
1. The heat required to melt the ice (latent heat of fusion).
2. The heat required to raise the temperature of the melted ice (now water) from 0°C to the final temperature \( t \).

The formula for the heat gained by the ice is:
\[ Q_{\text{gained}} = m_{\text{ice}} \cdot L_f + m_{\text{ice}} \cdot c \cdot (t - 0) \]
Where:
- \( m_{\text{ice}} = 1 \, \text{gm} \)
- \( L_f = 80 \, \text{cal/g} \) (latent heat of fusion)
- \( c = 1 \, \text{cal/g°C} \) (specific heat of water)
- \( t \) is the final temperature.

Substituting the values:
\[ Q_{\text{gained}} = 1 \cdot 80 + 1 \cdot 1 \cdot t = 80 + t \]

### Step 3: Identify the heat lost by the water
The heat lost by the water as it cools down from 100°C to the final temperature \( t \) is given by:
\[ Q_{\text{lost}} = m_{\text{water}} \cdot c \cdot (100 - t) \]
Where:
- \( m_{\text{water}} = 1 \, \text{gm} \)

Substituting the values:
\[ Q_{\text{lost}} = 1 \cdot 1 \cdot (100 - t) = 100 - t \]

### Step 4: Set up the equation for heat exchange
According to the principle of calorimetry:
\[ Q_{\text{gained}} = Q_{\text{lost}} \]
Thus, we have:
\[ 80 + t = 100 - t \]

### Step 5: Solve for \( t \)
Now, we can solve the equation:
1. Rearranging gives:
\[ 80 + t + t = 100 \]
\[ 2t = 100 - 80 \]
\[ 2t = 20 \]
\[ t = 10 \, °C \]

### Conclusion
The resulting temperature when 1 gm of ice at 0°C is mixed with 1 gm of water at 100°C is **10°C**.

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