To solve the problem of finding the minimum amount of steam at 100°C required to melt 12 grams of ice completely, we can use the principle of conservation of energy. The heat lost by the steam will be equal to the heat gained by the ice.
### Step-by-Step Solution:
1. **Identify the Heat Gained by Ice**:
The heat gained by the ice to melt completely can be calculated using the formula:
\[
Q_{\text{ice}} = m_{\text{ice}} \times L_f
\]
where:
- \( m_{\text{ice}} = 12 \, \text{g} \) (mass of ice)
- \( L_f = 80 \, \text{cal/g} \) (latent heat of fusion of ice)
Therefore,
\[
Q_{\text{ice}} = 12 \, \text{g} \times 80 \, \text{cal/g} = 960 \, \text{cal}
\]
2. **Identify the Heat Lost by Steam**:
The steam will lose heat in two processes:
- First, it will condense into water at 100°C.
- Then, the water will cool down to 0°C.
The total heat lost by the steam can be calculated as:
\[
Q_{\text{steam}} = m_{\text{steam}} \times L_v + m_{\text{steam}} \times c \times (T_{\text{initial}} - T_{\text{final}})
\]
where:
- \( m_{\text{steam}} \) is the mass of steam (which we need to find),
- \( L_v = 540 \, \text{cal/g} \) (latent heat of vaporization of water),
- \( c = 1 \, \text{cal/g°C} \) (specific heat of water),
- \( T_{\text{initial}} = 100°C \),
- \( T_{\text{final}} = 0°C \).
Therefore, the heat lost by the steam is:
\[
Q_{\text{steam}} = m_{\text{steam}} \times 540 \, \text{cal/g} + m_{\text{steam}} \times 1 \, \text{cal/g°C} \times (100 - 0)
\]
\[
Q_{\text{steam}} = m_{\text{steam}} \times (540 + 100) = m_{\text{steam}} \times 640 \, \text{cal/g}
\]
3. **Set Heat Gained Equal to Heat Lost**:
According to the law of conservation of energy:
\[
Q_{\text{ice}} = Q_{\text{steam}}
\]
Thus,
\[
960 \, \text{cal} = m_{\text{steam}} \times 640 \, \text{cal/g}
\]
4. **Solve for \( m_{\text{steam}} \)**:
Rearranging gives:
\[
m_{\text{steam}} = \frac{960 \, \text{cal}}{640 \, \text{cal/g}} = 1.5 \, \text{g}
\]
### Final Answer:
The minimum amount of steam required to melt 12 grams of ice completely is \( 1.5 \, \text{g} \).
