An ice cube of mass 0.1 kg at `0^@C` is placed in an isolated container which is at `227^@C`. The specific heat s of the container varies with temperature T according to the empirical relation `s=A+BT`, where `A= 100 cal//kg-K and B = 2xx (10^-2) cal//kg-K^2`. If the final temperature of the container is `27^@C`, determine the mass of the container.
(Latent heat of fusion for water = `8xx (10^4) cal//kg`, specific heat of water `=10^3 cal//kg-K`).

Explanation

To solve the problem, we need to find the mass of the container using the principle of conservation of energy. The heat lost by the container will be equal to the heat gained by the ice cube.

### Step-by-Step Solution:

1. **Identify the Heat Gained by the Ice Cube:**
The ice cube at `0°C` will first absorb heat to melt into water at `0°C`, and then it will further absorb heat to reach the final temperature of `27°C`. The total heat gained by the ice can be calculated as:
\[ Q_{\text{ice}} = m_{\text{ice}} \cdot L_f + m_{\text{ice}} \cdot c_w \cdot \Delta T \]
where:
- \( m_{\text{ice}} = 0.1 \, \text{kg} \)
- \( L_f = 8 \times 10^4 \, \text{cal/kg} \) (latent heat of fusion)
- \( c_w = 10^3 \, \text{cal/kg-K} \) (specific heat of water)
- \( \Delta T = 27 - 0 = 27 \, \text{K} \)

Substituting the values:
\[ Q_{\text{ice}} = 0.1 \cdot (8 \times 10^4) + 0.1 \cdot (10^3) \cdot (27) \]
\[ Q_{\text{ice}} = 8000 + 2700 = 10700 \, \text{cal} \]

2. **Identify the Heat Lost by the Container:**
The heat lost by the container can be expressed as:
\[ Q_{\text{container}} = m_{\text{container}} \cdot \int_{T_i}^{T_f} s(T) \, dT \]
where:
- \( T_i = 227°C = 500K \)
- \( T_f = 27°C = 300K \)
- \( s(T) = A + BT \) where \( A = 100 \, \text{cal/kg-K} \) and \( B = 2 \times 10^{-2} \, \text{cal/kg-K}^2 \)

The integral becomes:
\[ \int_{500}^{300} (100 + 2 \times 10^{-2} T) \, dT \]

Calculating the integral:
\[ = \left[ 100T + \frac{2 \times 10^{-2}}{2} T^2 \right]_{500}^{300} \]
\[ = \left[ 100(300) + 0.01(300^2) \right] - \left[ 100(500) + 0.01(500^2) \right] \]
\[ = (30000 + 0.01 \cdot 90000) - (50000 + 0.01 \cdot 250000) \]
\[ = (30000 + 900) - (50000 + 2500) \]
\[ = 30900 - 52500 = -21600 \, \text{cal} \]

Since we are interested in the magnitude of heat lost:
\[ Q_{\text{container}} = 21600 \cdot m_{\text{container}} \]

3. **Setting Heat Gained Equal to Heat Lost:**
According to the conservation of energy:
\[ Q_{\text{ice}} = Q_{\text{container}} \]
\[ 10700 = 21600 \cdot m_{\text{container}} \]

4. **Solving for the Mass of the Container:**
\[ m_{\text{container}} = \frac{10700}{21600} \approx 0.495 \, \text{kg} \]

### Final Answer:
The mass of the container is approximately **0.495 kg**.