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`).

Options

A. 0.495 kg
B. 0.595 kg
C. 0.695 kg
D. 0.795 kg

Answer

To solve the problem step by step, we will use the principle of calorimetry, which states that the heat lost by the hot body (the container) is equal to the heat gained by the cold body (the ice cube).

### Step 1: Identify the parameters
- Mass of the ice cube, \( m_{\text{ice}} = 0.1 \, \text{kg} \)
- Initial temperature of the ice, \( T_{\text{ice}} = 0^\circ C = 273 \, \text{K} \)
- Initial temperature of the container, \( T_{\text{container}} = 227^\circ C = 500 \, \text{K} \)
- Final temperature of the system, \( T_f = 27^\circ C = 300 \, \text{K} \)
- Latent heat of fusion of ice, \( L_f = 8 \times 10^4 \, \text{cal/kg} \)
- Specific heat of water, \( c_{\text{water}} = 10^3 \, \text{cal/kg.K} \)
- Specific heat of the container varies with temperature as \( s = A + BT \) where \( A = 100 \, \text{cal/kg.K} \) and \( B = 2 \times 10^{-2} \, \text{cal/kg.K}^2 \).

### Step 2: Calculate heat lost by the container
The heat lost by the container can be expressed as:
\[ Q_{\text{hot}} = m_{\text{container}} \int_{T_f}^{T_{\text{initial}}} s(T) \, dT \]
Substituting \( s(T) = A + BT \):
\[ Q_{\text{hot}} = m_{\text{container}} \int_{300}^{500} (A + BT) \, dT \]
Calculating the integral:
\[ Q_{\text{hot}} = m_{\text{container}} \left[ AT + \frac{B}{2} T^2 \right]_{300}^{500} \]
Substituting the limits:
\[ = m_{\text{container}} \left[ A(500) + \frac{B}{2}(500^2) - \left( A(300) + \frac{B}{2}(300^2) \right) \right] \]
Calculating \( A(500) \) and \( A(300) \):
\[ = 100 \times 500 - 100 \times 300 = 20000 \, \text{cal} \]
Calculating \( \frac{B}{2}(500^2) \) and \( \frac{B}{2}(300^2) \):
\[ \frac{2 \times 10^{-2}}{2} \left( 250000 - 90000 \right) = 0.01 \times 160000 = 1600 \, \text{cal} \]
Thus,
\[ Q_{\text{hot}} = m_{\text{container}} \left( 20000 + 1600 \right) = 21600 \, m_{\text{container}} \, \text{cal} \]

### Step 3: Calculate 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.
\[ Q_{\text{cold}} = m_{\text{ice}} L_f + m_{\text{ice}} c_{\text{water}} (T_f - T_{\text{ice}}) \]
Calculating each part:
1. Heat to melt the ice:
\[ Q_{\text{melt}} = 0.1 \times 8 \times 10^4 = 8000 \, \text{cal} \]
2. Heat to raise the temperature of the water:
\[ Q_{\text{heat}} = 0.1 \times 10^3 \times (27 - 0) = 0.1 \times 10^3 \times 27 = 2700 \, \text{cal} \]
Thus,
\[ Q_{\text{cold}} = 8000 + 2700 = 10700 \, \text{cal} \]

### Step 4: Equate heat lost and gained
Using the principle of calorimetry:
\[ Q_{\text{hot}} = Q_{\text{cold}} \]
\[ 21600 \, m_{\text{container}} = 10700 \]
Solving for \( m_{\text{container}} \):
\[ m_{\text{container}} = \frac{10700}{21600} \approx 0.495 \, \text{kg} \]

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

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