2g ice at `0^(@)C` is mixed with 1 g steam at `100^(@)C`. Find the final temperature of the mixture.

Options

A. `0^(@)C`
B. `50^(@)`
C. `100^(@)C`
D. `gt100^(@)C`

Answer

To solve the problem of mixing 2 g of ice at 0°C with 1 g of steam at 100°C, we need to consider the heat exchange between the ice and steam, and the phase changes that occur.

### Step 1: Understand the heat exchange process
When the ice and steam are mixed, the ice will absorb heat and melt into water, while the steam will release heat and condense into water. We need to calculate the heat gained by the ice and the heat lost by the steam.

### Step 2: Calculate the heat required to melt the ice
The heat required to melt ice can be calculated using the formula:
\[ Q_{\text{ice}} = m_{\text{ice}} \cdot L_f \]
where:
- \( m_{\text{ice}} = 2 \, \text{g} \)
- \( L_f = 80 \, \text{cal/g} \) (latent heat of fusion for ice)

Calculating \( Q_{\text{ice}} \):
\[ Q_{\text{ice}} = 2 \, \text{g} \cdot 80 \, \text{cal/g} = 160 \, \text{cal} \]

### Step 3: Calculate the heat released by the steam
The heat released by the steam when it condenses can be calculated using the formula:
\[ Q_{\text{steam}} = m_{\text{steam}} \cdot L_v \]
where:
- \( m_{\text{steam}} = 1 \, \text{g} \)
- \( L_v = 540 \, \text{cal/g} \) (latent heat of vaporization for steam)

Calculating \( Q_{\text{steam}} \):
\[ Q_{\text{steam}} = 1 \, \text{g} \cdot 540 \, \text{cal/g} = 540 \, \text{cal} \]

### Step 4: Analyze the heat exchange
- The ice requires 160 cal to melt completely.
- The steam releases 540 cal when it condenses.

Since the heat released by the steam (540 cal) is greater than the heat required to melt the ice (160 cal), all the ice will melt, and there will still be excess heat from the steam.

### Step 5: Calculate the excess heat after melting the ice
After melting the ice, the remaining heat from the steam can be calculated as:
\[ Q_{\text{excess}} = Q_{\text{steam}} - Q_{\text{ice}} \]
\[ Q_{\text{excess}} = 540 \, \text{cal} - 160 \, \text{cal} = 380 \, \text{cal} \]

### Step 6: Determine the final temperature of the mixture
Now, we have 2 g of water (from melted ice) at 0°C and 1 g of water (from condensed steam) at 100°C. The total mass of water is:
\[ m_{\text{total}} = 2 \, \text{g} + 1 \, \text{g} = 3 \, \text{g} \]

The heat gained by the melted ice (2 g) to reach the final temperature \( T_f \) can be calculated using:
\[ Q_{\text{gained}} = m_{\text{ice}} \cdot c \cdot (T_f - 0) \]
where \( c = 1 \, \text{cal/g°C} \) (specific heat of water).

The heat lost by the condensed steam (1 g) as it cools down to \( T_f \) is:
\[ Q_{\text{lost}} = m_{\text{steam}} \cdot c \cdot (100 - T_f) \]

Setting the heat gained equal to the heat lost:
\[ 2 \cdot 1 \cdot (T_f - 0) = 1 \cdot 1 \cdot (100 - T_f) \]
\[ 2T_f = 100 - T_f \]
\[ 3T_f = 100 \]
\[ T_f = \frac{100}{3} \approx 33.33°C \]

### Final Answer
The final temperature of the mixture is approximately **33.33°C**.

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