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//kgK and B = 2xx (10^-2) cal//kg`. 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 step by step, we need to calculate the heat lost by the container and the heat gained by the ice cube, and then set them equal to each other according to the principle of calorimetry.

### Step 1: Calculate the heat lost by the container

The heat lost by the container can be expressed as:
\[ Q_{\text{lost}} = -\int_{T_i}^{T_f} m_C \cdot s(T) \, dT \]
where:
- \( m_C \) is the mass of the container,
- \( s(T) = A + BT \) is the specific heat of the container,
- \( T_i = 227^\circ C = 500 \, K \) (initial temperature in Kelvin),
- \( T_f = 27^\circ C = 300 \, K \) (final temperature in Kelvin),
- \( A = 100 \, \text{cal/kgK} \),
- \( B = 2 \times 10^{-2} \, \text{cal/kgK} \).

Substituting the values, we have:
\[ Q_{\text{lost}} = -m_C \int_{500}^{300} (100 + 0.02T) \, dT \]

### Step 2: Evaluate the integral

Calculating the integral:
\[ \int (100 + 0.02T) \, dT = 100T + 0.01T^2 \]
Now, we evaluate it from 500 to 300:
\[ \left[ 100T + 0.01T^2 \right]_{500}^{300} = \left(100 \times 300 + 0.01 \times 300^2\right) - \left(100 \times 500 + 0.01 \times 500^2\right) \]

Calculating the values:
\[ = (30000 + 900) - (50000 + 2500) = 30900 - 52500 = -21600 \, \text{cal} \]
Thus, the heat lost by the container is:
\[ Q_{\text{lost}} = -m_C \cdot (-21600) = 21600 m_C \, \text{cal} \]

### Step 3: Calculate the heat gained by the ice

The heat gained by the ice consists of two parts: the heat required to melt the ice and the heat required to raise the temperature of the resulting water.

1. **Heat to melt the ice**:
\[ Q_{\text{melt}} = m_{\text{ice}} \cdot L = 0.1 \, \text{kg} \cdot 8 \times 10^4 \, \text{cal/kg} = 8000 \, \text{cal} \]

2. **Heat to raise the temperature of water**:
The water will be heated from \(0^\circ C\) to \(27^\circ C\):
\[ Q_{\text{heat}} = m_{\text{ice}} \cdot c_{\text{water}} \cdot \Delta T = 0.1 \, \text{kg} \cdot 10^3 \, \text{cal/kgK} \cdot (27 - 0) = 0.1 \cdot 1000 \cdot 27 = 2700 \, \text{cal} \]

Adding both parts together gives:
\[ Q_{\text{gained}} = Q_{\text{melt}} + Q_{\text{heat}} = 8000 + 2700 = 10700 \, \text{cal} \]

### Step 4: Set heat lost equal to heat gained

According to the principle of calorimetry:
\[ Q_{\text{lost}} = Q_{\text{gained}} \]
Substituting the values:
\[ 21600 m_C = 10700 \]

### Step 5: Solve for the mass of the container

Now, we can solve for \( m_C \):
\[ m_C = \frac{10700}{21600} \approx 0.495 \, \text{kg} \]

### Conclusion

The mass of the container is approximately \( 0.495 \, \text{kg} \).

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