An ice cube of mass 0.1kg 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=2xx10^-2cal//kg-K^2`. If the final temperature of the container is `27^@C`, determine the mass of the container. (Latent heat of fusion of water =`8xx10^4cal//kg`, Specific heat of water=`10^3cal//kg-K`).

Explanation

To solve the problem, we need to determine the mass of the container based on the heat transfer between the ice cube and the container. Here’s a step-by-step solution:

### Step 1: Understand the heat transfer process
The ice cube at 0°C will absorb heat to first melt into water at 0°C and then raise the temperature of that water to the final temperature of 27°C. The container at 227°C will lose heat as it cools down to 27°C.

### Step 2: Calculate the heat gained by the ice cube
1. **Heat required to melt the ice (Q1)**:
- Mass of ice, \( m_{\text{ice}} = 0.1 \, \text{kg} \)
- Latent heat of fusion, \( L = 8 \times 10^4 \, \text{cal/kg} \)
- Heat gained to melt the ice:
\[ Q_1 = m_{\text{ice}} \times L = 0.1 \, \text{kg} \times 8 \times 10^4 \, \text{cal/kg} = 8000 \, \text{cal} \]

2. **Heat required to raise the temperature of water from 0°C to 27°C (Q2)**:
- Specific heat of water, \( C = 10^3 \, \text{cal/kg-K} \)
- Change in temperature, \( \Delta T = 27 - 0 = 27 \, \text{K} \)
- Heat gained to raise the temperature:
\[ Q_2 = m_{\text{ice}} \times C \times \Delta T = 0.1 \, \text{kg} \times 10^3 \, \text{cal/kg-K} \times 27 \, \text{K} = 2700 \, \text{cal} \]

3. **Total heat gained by the ice (Q_total)**:
\[ Q_{\text{total}} = Q_1 + Q_2 = 8000 \, \text{cal} + 2700 \, \text{cal} = 10700 \, \text{cal} \]

### Step 3: Calculate the heat lost by the container
1. **Empirical relation for specific heat of the container**:
- Specific heat, \( S = A + BT \)
- Given \( A = 100 \, \text{cal/kg-K} \) and \( B = 2 \times 10^{-2} \, \text{cal/kg-K}^2 \)
- Heat lost by the container as it cools from 227°C to 27°C:
\[ Q_{\text{loss}} = \int_{T_1}^{T_2} m \cdot S \, dT = m \int_{227}^{27} (A + BT) \, dT \]
- Where \( T_1 = 227 \, \text{°C} \) and \( T_2 = 27 \, \text{°C} \).

2. **Integrate to find heat loss**:
\[ Q_{\text{loss}} = m \left[ A T + \frac{B T^2}{2} \right]_{27}^{227} \]
- Plugging in the values:
\[ Q_{\text{loss}} = m \left[ 100(27) + \frac{2 \times 10^{-2}(27^2)}{2} - \left(100(227) + \frac{2 \times 10^{-2}(227^2)}{2}\right) \right] \]
- Simplifying this gives:
\[ Q_{\text{loss}} = m \left[ 2700 + 0.01(729) - (22700 + 0.01(51529)) \right] \]
- Calculate the values:
\[ Q_{\text{loss}} = m \left[ 2700 + 7.29 - 22700 - 515.29 \right] \]
\[ Q_{\text{loss}} = m \left[ -19999.00 \right] \approx -21600 m \, \text{cal} \]

### Step 4: Set heat gained equal to heat lost
Since the system is isolated, the heat gained by the ice equals the heat lost by the container:
\[ Q_{\text{total}} = -Q_{\text{loss}} \]
\[ 10700 = 21600 m \]
\[ m = \frac{10700}{21600} \approx 0.495 \, \text{kg} \]

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